The power liens that run through your neighborhood carry alternating currents that reverse direction 120 times per second. As current changes so does magnetic field. If you put a loop of wire up near the power line to extract power by tapping the magnetic field, sketch how you would orient the coil of wire next to a power line to develop the max emf in the coilo If magnetic flux through a loop changes, induced emf is produced
o If the area of the loop is parallel to the field, the flux through the loop is minimum
• So no emf is produced
o To get the most of the flux through the loop you place it closer to the power lines and in orientations so the plane of the loop also contains power lines
• Flux would be max and result in greater induced emf as field goes from max to zero to max and then in other direction
Answers
Answer:
the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
Explanation:
Faraday's law is
ε = - dΦ / dt
where Ф magnetic flow
the flow is
Ф = B. dA = B dA cos θ
therefore, to obtain the maximum energy, the cosine function must be maximum, for this the direction of the fluctuating magnetic field and the normal direction to the area must be parallel.
The magnetic field in a cable through which current flows is circular, so the loop must be perpendicular to the wire, this is the normal to the area of the loop parallel to the wire to induce the maximum electromotive force
an outline is shown in the attachment
the correct answer is b
Related Questions
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibration amplitude of 0.05mm at 2 Hz (a typical resonant frequency of a building). a) If there is no damping, how much transmitted vibration (amplitude of displacement) does the microscope experience
Answers
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope [tex]\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}][/tex]
where;
[tex]\epsilon = 0[/tex]
[tex]W_n = \sqrt { \frac{k}{m}}[/tex]
= [tex]\sqrt { \frac{100*32.2}{200}}[/tex]
= 4.0124
replacing them into the above equation and making X the subject of the formula:
[tex]X =[/tex] [tex]Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]5.676*10^{-3} \ mm[/tex]
Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
You are given a long length of string and an oscillator that can shake one end of the string at any desired frequency. The oscillator has a display that indicates the frequency. You are asked to design an experiment to study how the velocity of waves on the string depends on the string's tension. You do not have any way to measure time with sufficient accuracy to help in your investigation.
A. Describe your experimental setup and procedure, including any additional pieces of equipment you would need and the kind of data you would record. Include enough detail that another student could follow and complete the experiment successfully.
B. Describe how you would analyze your data to obtain information about the wave velocity's dependence on tension.
Answers
Answer:
Explanation:
a.
AIM :
TO STUDY HOW VELOCITY OF WAVES ON THE STRING DEPENDS ON THE STRING'S TENSION.
APPARATUS:
Oscillator, long strings , some masses( to create tension in string) and the support ( rectangular wooden piece).
EXPERIMENTAL SETUP:
1. Measure the length of the string and mass of the weights used.
2. Connect one end of string to the oscillator.
3. Place the support below string on table such that the string is in same line without touching table.
4. After the support, the string should hang freely.
5. The other end of string is connected with some small measured masses which should be hanging.
PROCEDURE:
1. Note down the length of string and mass of weights.
2. Adjust the frequency in the oscillator which creates standing waves in the string.
3. Start from lower frequency and note down the lowest frequency at which mild sound is heard or when string forms one loop while oscillating.
4. Calculate the wavelength using of waves using length of string.
5. Calculate the velocity using frequency and wavelength.
6. Calculate linear mass density.
8. Repeat the procedure with different masses.
7. plot a graph with tension in y axis and linear mass density in x axis.
8. Find slope and compare with velocity.
Linear mass density
µ = m/l(kg-1)
tension
T = m x 9.8N
wave length
ƛ = 2L
b.
We can analyze the data by comparing slope of the graph, tension Vs linear mass density with velocity which is constant for constant length.
Write the slope value in terms of value of velocity and find the relationship between velocity and string's tension.
The expected result is
slope = v²
T ∝ V²
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is 1.1 m, while the mass of each is 0.11 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 8.3 mm. Determine the current in the rods.
Answers
Answer:
I = 215.76 A
Explanation:
The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.
Fm = u_o*I^2*L/2*π*d
where I is the current in each rod, d = 0.0082 m is the distance 27rId
between each, L = 0.85 m is the length of each rod.
Fm = 4π*10^-7*I^2*1.1/2*π*0.0083
The mass of each rod is m = 0.0276 kg
F_m = mg
4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8
I = 215.76 A
note:
mathematical calculation maybe wrong or having little bit error but the method is perfectly fine
The current in the rods is approximately 132.75 A.
To determine the current in the rods, we need to use the formula for the force between two parallel current-carrying conductors, which is given by:
[tex]\[ F = \frac{\mu_0 I^2 L}{2 \pi d} \][/tex]
where:
F is the force between the two rods,
[tex]\( \mu_0 \)[/tex] is the permeability of free space [tex](\( 4\pi \times 10^{-7} \) T\·m/A)[/tex],
I is the current in each rod,
L is the length of the rods (1.1 m),
d is the distance between the rods (8.3 mm or [tex]\( 8.3 \times 10^{-3} \) m[/tex]).
The rods repel each other, and the force of repulsion must equal the gravitational force on the floating rod for it to levitate. The gravitational force is given by:
[tex]\[ F_g = m g \][/tex]
where:
m is the mass of the rod (0.11 kg),
g is the acceleration due to gravity (approximately 9.81 m/s²).
Setting the magnetic force equal to the gravitational force, we get:
[tex]\[ \frac{\mu_0 I^2 L}{2 \pi d} = m g \][/tex]
Solving for I, we have:
[tex]\[ I^2 = \frac{m g 2 \pi d}{\mu_0 L} \][/tex]
[tex]\[ I = \sqrt{\frac{m g 2 \pi d}{\mu_0 L}} \][/tex]
[tex]\[ I = \sqrt{\frac{0.11 \text{ kg} \times 9.81 \text{ m/s}^2 \times 2 \pi \times 8.3 \times 10^{-3} \text{ m}}{4\pi \times 10^{-7} \text{ T·m/A} \times 1.1 \text{ m}}} \][/tex]
[tex]\[ I = \sqrt{\frac{0.11 \times 9.81 \times 2 \times 8.3 \times 10^{-3}}{4\pi \times 10^{-7} \times 1.1}} \][/tex]
[tex]\[ I = \sqrt{\frac{0.11 \times 9.81 \times 16.6 \times 10^{-3}}{4\pi \times 10^{-7} \times 1.1}} \][/tex]
[tex]\[ I = \sqrt{\frac{1.77558 \times 10^{-2}}{4\pi \times 10^{-7}}} \][/tex]
[tex]\[ I = \sqrt{\frac{1.77558 \times 10^{-2}}{1.256637 \times 10^{-6}}} \][/tex]
[tex]\[ I = \sqrt{1413.55} \][/tex]
[tex]\[ I \approx 132.75 \text{ A} \][/tex]
A heat conducting rod, 0.90 m long, is made of an aluminum section, 0.20 m long, and a copper section, 0.70 m long. Both sections have a cross-sectional area of 0.0004 m^2. The aluminum end and the copper end are maintained at temperatures of 30*C and 230*C respectively. The thermal conductivities of aluminum and copper are 205 and 385 W/m ? K, respectively. The temperature of the aluminum-copper junction in the rod, in *C, is closest to:
Answers
Answer:
is closest to 100*C temperature at the aluminum-copper junction
Explanation:
The expression for calculating the resistance of each rod is given by
[tex]R =\frac{ L}{ kA}[/tex]
Now; for Aluminium
[tex]R_{al} =\frac{ 0.20 }{ 205*0.0004}[/tex]
[tex]R_{al}[/tex] = 2.439
For Copper
[tex]R_{Cu}=\frac{0.70}{385*0.0004}[/tex]
[tex]R_{Cu} = 4.545[/tex]
Total Resistance [tex]R = R_{al} + R_{Cu}[/tex]
= 2.439 + 4.545
= 6.9845
Total temperature difference = 230*C + 30*C
= 200 *C
The Total rate of heat flow is then determined which is = [tex]\frac{ total \ temp \ difference}{total \ resistance }[/tex]
=[tex]\frac{200}{ 6.9845 }[/tex]
= 28.635 Watts
However. the temperature difference across the aluminium = Heat flow × Resistance of aluminium
= 28.635 × 2.349
= 69.84 *C
Finally. for as much as one end of the aluminium is = 30 *C , the other end is;
=30*C + 69.84*C
= 99.84 *C
which is closest to 100*C temperature at the aluminum-copper junction
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasing at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.900 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
Required:
Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.
Answers
Final answer:
To find the magnitude of the emf induced in the loop after 8.00 seconds has passed since the circumference started to decrease, we can use Faraday's law of electromagnetic induction. We calculate the rate of change of magnetic flux through the loop based on the changing area of the loop, and then determine the magnitude of the emf induced in the loop. The emf induced is -253.30 V, indicating that the induced current flows in a direction that opposes the change in magnetic flux.
Explanation:
To find the magnitude of the emf induced in the loop after 8.00 seconds has passed since the circumference started to decrease, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the emf induced in a loop is equal to the rate of change of magnetic flux through the loop. In this case, as the loop shrinks, its area decreases, resulting in a decrease in magnetic flux.
We know that the circumference of the loop is decreasing at a constant rate of 15.0 cm/s. Using the formula for the circumference of a circle, we can determine the radius of the circle at the given time: r = C / (2*pi), where C is the circumference and pi is a mathematical constant approximately equal to 3.14159. Substituting the given values, we get r = 168 cm / (2*3.14159) = 26.79 cm.
Next, we can calculate the area of the loop as a function of time using the equation A = pi*r^2. Substituting the value of the radius at 8.00 seconds, we get A = 3.14159 * (26.79 cm)^2 = 2252.68 cm^2.
Since the magnetic field is perpendicular to the loop and uniform in magnitude, we can calculate the rate of change of magnetic flux as: dPhi/dt = B*dA/dt, where B is the magnitude of the magnetic field and dA/dt is the rate of change of the area.
Finally, we can calculate the magnitude of the emf induced in the loop as: EMF = -dPhi/dt. The negative sign indicates that the induced current flows in a direction that opposes the change in magnetic flux. Substituting the given values, we get EMF = -0.900 T * (2252.68 cm^2) / 8.00 s = -253.30 V.
At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until a circuit breaker trips at time t= 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.a.) Through what total angle did the wheel turn between t= 0 and the time it stopped? answer in radb.) At what time does the wheel stop? (secs)c.)What was the wheel's angular acceleration as it slowed down?Express your answer in radians per second per second.
Answers
Answer:
a) [tex]\Delta \theta = 617.604\,rad[/tex], b) [tex]\Delta t = 10.392\,s[/tex], c) [tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]
Explanation:
a) The final angular speed at the end of the acceleration stage is:
[tex]\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)[/tex]
[tex]\omega = 111.5\,\frac{rad}{s}[/tex]
The angular deceleration is:
[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}[/tex]
[tex]\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}[/tex]
[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]
The change in angular position during the acceleration stage is:
[tex]\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}[/tex]
[tex]\theta = 177.604\,rad[/tex]
Finally, the total change in angular position is:
[tex]\Delta \theta = 440\,rad + 177.604\,rad[/tex]
[tex]\Delta \theta = 617.604\,rad[/tex]
b) The time interval of the deceleration interval is:
[tex]\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }[/tex]
[tex]\Delta t = 7.892\,s[/tex]
The time required for the grinding wheel to stop is:
[tex]\Delta t = 2.50\,s + 7.892\,s[/tex]
[tex]\Delta t = 10.392\,s[/tex]
c) The angular deceleration of the grinding wheel is:
[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]
Answer:
Ф_T =544
The wheel will stop after 10 s.
α_z = -11.15 rad/s^2
Explanation:
The angular acceleration is constant. Thus, we will apply the equations of rotation with constant angular acceleration model.
(a) In order to calculate the total angle, we will divide the entire interval from t = 0 to the time the wheel stops into two intervals.
From t = 0 to t = 2 s:
Ф-Ф_o =1/2(w_0z+w_z)t (1)
We will calculate w_z first:
w_z = w_0x +α_xt
w_z = 24 + (35)(2.5)
w_z = 111.5
Substitute w_x into Eq.1
Ф-Ф_o = 1/2(24+111.5)(2)
Ф-Ф_o = 136 rad
We can calculate it directly from the following equation:
Ф-Ф_o = w_0x*t+1/2a_xt^2
Ф-Ф_o = 24*2.5+1/2*35*2.5
Ф-Ф_o =103.75 rad
Thus, the total angle the wheel turned between t = 0 and the time it stopped:
Ф_T =103.75 +440
Ф_T =544
(b)
We will take the interval from when the circuit breaker trips until the wheel comes to a stop.
w_oz = 111.5 rad/s
wz = 0 the wheel will stop at the end. 1
Ф-Ф_o =1/2(w_0z+w_z)t
440 = 1/2(111.5+0)*t
t = 8s
Adding t of the first interval to t of the second interval :
t_T = 2 + 8 =10
The wheel will stop after 10 s.
(c)
We will take the interval from when the circuit breaker trips until the wheel comes to a stop.
w_oz = 111.5 rad/s
w_z = 0
w_x =w_oz+α_z*t
α_z = -11.15 rad/s^2
A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two pieces. Piece A is propelled in the forward direction with a speed of 0.43 c relative to the original nucleus. Piece B is sent backward at 0.34 c relative to the original nucleus. Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.
Answers
Answer with Explanation:
We are given that
Velocity of uranium=v=0.94 c
Speed of piece A relative to the original nucleus=[tex]u'_A=0.43c[/tex]
Speed of piece B relative to the original nucleus=[tex]u'_B=0.34c[/tex]
Velocity of piece A observed by observer
[tex]u_A=\frac{u'_A+v}{1+\frac{u'_A v}{c^2}}[/tex]
Substitute the values
[tex]u_A=\frac{0.43c+0.94c}{1+\frac{0.43c\times 0.94c}{c^2}}[/tex]
[tex]u_A=\frac{1.37c}{1+0.4042}=0.98c[/tex]
Velocity of piece B observed by observer
[tex]u_B=\frac{0.34c+0.94c}{1+\frac{0.34c\times 0.94c}{c^2}}[/tex]
[tex]u_B=\frac{1.28c}{1+0.3196}[/tex]
[tex]u_B=0.97 c[/tex]
The velocity of piece A and piece B as measured by an observer in the laboratory are not same.
A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base of the hill. (a) At what rate was it rotating at the base of the hill? (b) What was the rotational kinetic energy then? (Ignore rolling friction and assume total mechanical energy is conserved). [Hint: Soccer ball is a thin walled h
Answers
Answer with Explanation:
We are given that
Diameter=d=22.6 cm
Mass,m=426 g=[tex]426\times 10^{-3} kg[/tex]
1 kg=1000 g
Radius,r=[tex]\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m[/tex]
1m=100 cm
Height,h=5m
[tex]I=\frac{2}{2}mr^2[/tex]
a.By law of conservation of energy
[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh[/tex]
[tex]\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh[/tex]
[tex]v=\omega r[/tex]
[tex]gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2[/tex]
[tex]\omega^2=\frac{6}{5r^2}gh[/tex]
[tex]\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s[/tex]
Where [tex]g=9.8m/s^2[/tex]
b.Rotational kinetic energy=[tex]\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J[/tex]
Rotational kinetic energy=8.35 J
A ray of light passes from one material into a material with a higher index of refraction. Determine whether each of the following quantities increases, decreases, or remains unchanged. HINT (a) The ray's angle with the normal. increases decreases remains unchanged (b) The light's wavelength. increases decreases remains unchanged (c) The light's frequency. increases decreases remains unchanged (d) The light's speed. increases decreases remains unchanged (e) The photon energy. increases decreases remains unchanged
Answers
Answer:
a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged
Explanation:
When a ray of light passes from one transparent material to another, it must comply with the law of refraction
n1 sin θ₁ = n2 sin θ₂
Where index v1 is for the incident ray and index 2 for the refracted ray
With this expression let's examine the questions
a) They indicate that the refractive index increases,
sin θ₂ = n₁ / n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases
Answer is decrease
b) The wave velocity eta related to the wavelength and frequency
v = λ f
The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.
The speed decreases with the indicated refraction increases and therefore the wavelength decreases
λ = λ₀ / n
The answer is decrease
c) from the previous analysis the frequency remains unchanged
d) the refractive index is defined by
n = c / v
So if n increases, the speed must decrease
The answer is decrease
e) the energy of the photon is given by the Planck equation
E = h f
Since the frequency does not change, the energy does not change either
Answer remains unchanged
a) The Angle te also decreases, b) Decrease, c) Unchanged, d) The speed decrease, e) Unchanged
What is the Law of Refraction?
When a glimmer of light perishes from one translucent material to another, it must capitulate with the law of refraction
n1 sin θ₁ = n2 sin θ₂
Where index v1 is for the happening ray and index 2 is for the refracted ray With this expression let's discuss.
a) They indicate that the refractive index increases,
sin θ₂ = n₁ / n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
As m is more significant than n1 the quantity on the ownership is less than one, the whole quantity in parentheses diminishes so the Angle te also decreases
Solution is decrease
b) The wave velocity eta connected to the wavelength and frequency
v = λ f
The frequency does not modify since the passage from one medium to the other is a procedure of forced oscillation and resonance whereby the frequency in the two mediums is the same.
The speed decreases with the foreshadowed refraction increases and thus the wavelength decreases
So, λ = λ₀ / n
The answer is to decrease
c) from the earlier analysis the frequency remains unchanged
d) When the refractive index is defined by
n is = c / v
Then, if n increases, the speed must decrease
The solution is to decrease
e) When the energy of the photon is given by the Planck equation
E is = hf. Since When the frequency does not transform, the energy does not change either Solution is remains unchanged
Find more information about Law of Refraction here:
https://brainly.com/question/16050768
PROBLEM-10 GRAVITATIONAL POTINTIAL ENERGY In a typical professional downhill ski race, athletes start 820 m above where they cross the finish line. Assume that they start at rest, friction is negligible, and their mass is 80 kg. What is the magnitude of the change in gravitational potential energy of the Earth-athlete system over the course of the race? (in J )
Answers
Answer:
The magnitude of the change in gravitational potential energy of the earth-athlete system over the course of the race is 643,536Joules
Explanation
Potential energy is one of the form of mechanical energy and it is defined as the energy possessed by a body due to virtue of its position. When the body is under gravity, it possesses an energy called the gravitational potential energy.
Gravitational potential energy is expressed as shown:
GPE = mass × acceleration due to gravity × height
Given mass of athlete = 80kg
height covered = 820m
acceleration due to gravity = 9.81m/s
GPE = 80×9.81×820
GPE = 643,536Joules
You have a summer job working at a company developing systems to safely lower large loads down ramps. Your team is investigating a magnetic system by modeling it in the laboratory. The safety system is a conducting bar that slides on two parallel conducting rails that run down the ramp. The bar is perpendicular to the rails and is in contact with them. At the bottom of the ramp, the two rails are connected together. The bar slides down the rails through a vertical uniform magnetic field. The magnetic field is supposed to cause the bar to slide down the ramp at a constant velocity even when friction between the bar and the rails is negligible. Before setting up the laboratory model, your task is to calculate the constant velocity of the bar sliding down the ramp on rails in a vertical magnetic field as a function of the mass of the bar, the strength of the magnetic field, the angle of the ramp from the horizontal, the length of the bar which is the same as the distance between the tracks, and the resistance of the bar. Assume that all of the other conductors in the system have a much smaller resistance than the bar.
If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.
Answers
Answer:
Explanation:
find the solution below
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as . You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as What is the speed of the police car? The speed of sound in a
Answers
The complete question is:
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as 1340 Hz. You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as 1300 Hz. What is the speed of the police car? The speed of sound in air is 343m/s.
Answer:
V_s = 30 m/s
Explanation:
The change in frequency observation occur due to doppler effect is given by the equation;
f_o = [(V ± V_o)/(V ∓ V_s)]f_s
Where;
f_o is observed frequency
f_source is frequency of the source
V is speed of sound
V_o is velocity of the observer
V_s is velocity of the source
Now, When the police is coming to you , you hear a higher frequency and thus, we'll use the positive sign on the numerator and negative sign on denominator.
Thus,
f_o = [(V + V_o)/(V - V_s)]f_s
Plugging in relevant values, we have;
1340 = [(343 + 35)/(343 - V_s)]f_s
1340 = [(378)/(343 - V_s)]f_s - - (eq1)
when the police is passing you , you hear a lesser frequency, and thus, we'll use the negative sign on the numerator and positive sign on denominator. thus;
f_o = [(V - V_o)/(V + V_s)]f_s
Plugging in the relevant values to get;
1300 = [(343 - 35)/(343 + V_s)]f_s
1300 = [(308)/(343 + V_s)]f_s - - eq2
Divide eq2 by eq1 with f_s canceling out to give
1340/1300 = [(378)/(343 - V_s)]/[(308)/(343 + V_s)]
V_s = 30 m/s
A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is suspended from the bar 3/4ths of its length from the wall. The angle indicated is 60°. What is the tension in the cable? What is the horizontal force provided by the hinge?
Answers
Answer:
The tension is [tex]T= \frac{11}{2\sqrt{3} } Mg[/tex]
The horizontal force provided by hinge [tex]Fx= \frac{11}{4\sqrt{3} } Mg[/tex]
Explanation:
From the question we are told that
The mass of the beam is [tex]m_b =M[/tex]
The length of the beam is [tex]l = L[/tex]
The hanging mass is [tex]m_h = 3M[/tex]
The length of the hannging mass is [tex]l_h = \frac{3}{4} l[/tex]
The angle the cable makes with the wall is [tex]\theta = 60^o[/tex]
The free body diagram of this setup is shown on the first uploaded image
The force [tex]F_x \ \ and \ \ F_y[/tex] are the forces experienced by the beam due to the hinges
Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero
So
[tex]\sum F =0[/tex]
Now about the x-axis the moment is
[tex]F_x -T cos \theta = 0[/tex]
=> [tex]F_x = Tcos \theta[/tex]
Substituting values
[tex]F_x =T cos (60)[/tex]
[tex]F_x= \frac{T}{2} ---(1)[/tex]
Now about the y-axis the moment is
[tex]F_y + Tsin \theta = M *g + 3M *g ----(2)[/tex]
Now the torque on the system is zero because their is no rotation
So the torque above point 0 is
[tex]M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0[/tex]
[tex]\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0[/tex]
[tex]\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}[/tex]
[tex]T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }[/tex]
[tex]T= \frac{11}{2\sqrt{3} } Mg[/tex]
The horizontal force provided by the hinge is
[tex]F_x= \frac{T}{2} ---(1)[/tex]
Now substituting for T
[tex]F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}[/tex]
[tex]Fx= \frac{11}{4\sqrt{3} } Mg[/tex]
The answer is:[tex]\frac{4Mg}{\sqrt{3}}.[/tex]
To solve this problem, we will analyze the forces acting on the uniform beam and apply the conditions for rotational and translational equilibrium. Here's the step-by-step solution:
1. Identify the forces acting on the beam:
- The weight of the beam, [tex]\( W_b = Mg \)[/tex], acting at the center of mass of the beam, which is at [tex]\( \frac{L}{2} \)[/tex] from the wall.
- The weight of the mass suspended from the bar, [tex]\( W_s = 3Mg \), acting at \( \frac{3L}{4} \)[/tex] from the wall.
- The tension in the cable, [tex]\( T \)[/tex], acting at an angle[tex]\( \theta = 60^\circ \)[/tex] with the horizontal.
- The horizontal force provided by the hi-nge,[tex]\( F_h \)[/tex].
2. Break down the tension in the cable into its horizontal and vertical components:
- The horizontal component of the tension,[tex]\( T_h = T \cos(60^\circ) \).[/tex]
- The vertical component of the tension,[tex]\( T_v = T \sin(60^\circ) \)[/tex].
3. Apply the condition for translational equilibrium in the vertical direction:
- The sum of the vertical forces must be zero:[tex]\( T_v + F_h = W_b + W_s \).[/tex]
- Substituting the expressions for the weights, we get: [tex]\( T \sin(60^\circ) = Mg + 3Mg \).[/tex]
- Simplifying, [tex]\( T \sin(60^\circ) = 4Mg \)[/tex].
4. Solve for the tension [tex]\( T \)[/tex]:
[tex]- \( T = \frac{4Mg}{\sin(60^\circ)} \).- Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), we have \( T = \frac{4Mg}{\frac{\sqrt{3}}{2}} = \frac{8Mg}{\sqrt{3}} \).[/tex]
5. Apply the condition for rotational equilibrium about the hi-nge:
- The sum of the torques about the hi-nge must be zero.
- The torque due to the weight of the beam is [tex]\( \tau_b = W_b \left(\frac{L}{2}\right) = Mg \left(\frac{L}{2}\right) \)[/tex].
- The torque due to the weight of the suspended mass is[tex]\( \tau_s = W_s \left(\frac{3L}{4}\right) = 3Mg \left(\frac{3L}{4}\right) \).[/tex]
- The torque due to the tension in the cable is[tex]\( \tau_T = T_v \left(\frac{L}{2}\right) \)[/tex] (since the tension acts at the end of the beam).
- Setting the sum of the torques to zero: [tex]\( \tau_T = \tau_b + \tau_s \).[/tex]
- Substituting the expressions for the torques, we get: [tex]\( T \sin(60^\circ) \left(\frac{L}{2}\right) = Mg \left(\frac{L}{2}\right) + 3Mg \left(\frac{3L}{4}\right) \)[/tex]
- We already know that [tex]\( T \sin(60^\circ) = 4Mg \)[/tex], so this equation is satisfied, confirming that our value for [tex]\( T \)[/tex] is correct.
6. Solve for the horizontal force [tex]\( F_h \)[/tex]:
- From the vertical equilibrium equation, [tex]\( T \sin(60^\circ) = 4Mg \)[/tex] , we have already found [tex]\( T \)[/tex].
- The horizontal component of the tension is [tex]\( T_h = T \cos(60^\circ) \)[/tex].
- Since there are no other horizontal forces, [tex]\( F_h = T_h \)[/tex].
- Substituting the value of[tex]\( T \), we get \( F_h = \frac{8Mg}{\sqrt{3}} \cos(60^\circ) \).[/tex]
- Since [tex]\( \cos(60^\circ) = \frac{1}{2} \), we have \( F_h = \frac{8Mg}{\sqrt{3}} \cdot \frac{1}{2} = \frac{4Mg}{\sqrt{3}} \).[/tex]
Therefore, the tension in the cable is [tex]\( \boxed{\frac{8Mg}{\sqrt{3}}} \)[/tex] and the horizontal force provided by the hi-nge is[tex]\( \boxed{\frac{4Mg}{\sqrt{3}}} \).[/tex]
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
Answers
Answer:
a) The speed of the ball is 2.47 m/s (in -x direction)
b) The speed of the block, both just after the collision is 1.236 m/s (in +x direction)
Explanation:
Please look at the solution in the attached Word file.
A piece of glass has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and glass are 1.40 and 1.50, respectively, what is the thickness of the film if we see a bright reflection in the visible range
Answers
Answer:
200 nm
Explanation:
Refractive index of gasoline = 1.4
Wavelength = 560 nm
t = Thickness of film
m = Order = 1
Wavelength is given by
[tex]\lambda=\dfrac{560}{1.4}=400\ nm[/tex]
We have the relation
[tex]2t=m\lambda\\\Rightarrow t=\dfrac{m\lambda}{2}\\\Rightarrow t=\dfrac{1\times 400}{2}\\\Rightarrow t=200\ nm[/tex]
The thickness of the film is 200 nm
Determine the location of the center of mass of a "L" whose thin vertical and horizontal members have the same length L and the same mass M. Use the formal definition to find the x and y coordinates, and check your result by doing the calculation with respect to two different origins, one in the lower left corner at the intersection of the horizontal and vertical members and one at the top of the vertical member.
(a) Origin at the lower left
x = ?
y = ?
(b) Origin at the top of the vertical member
x = ?
y = ?
Answers
Answer:
a) x_{cm} = L / 2 , y_{cm}= L/2, b) x_{cm} = L / 2 , y_{cm}= L/2
Explanation:
The center of mass of a body is the point where all external forces are applied, it is defined by
[tex]x_{cm}[/tex] = 1 / M ∑ [tex]x_{i} m_{i}[/tex] = 1 /M ∫ x dm
[tex]y_{cm}[/tex] = 1 / M ∑ [tex]y_{i} m_{i}[/tex] = 1 / M ∫ y dm
where M is the total body mass
Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis
a) let's start with the reference zero at the left end of the horizontal rod
let's use the concept of linear density
λ = M / L = dm / dl
since the rod is on the x axis
dl = dx
dm = λ dx
let's calculate
x_{cm} = M ∫ x λ dx = λ / M ∫ x dx
x_{cm} = λ / M x² / 2
we evaluate between the lower integration limits x = 0 and upper x = L
x_{cm} = λ / M (L² / 2 - 0)
we introduce the value of the density that is cosntnate
x_{cm} = (M / L) L² / 2M
x_{cm} = L / 2
We repeat the calculation for verilla verilla
λ = M / L = dm / dy
y_{cm} = 1 / M ∫ y λ dy
y_{cm} = λ M y² / 2
[tex]y_{cm}[/tex] = M/L 1/M (L² - 0)
y_{cm}= L/2
b) we repeat the calculation for the origin the reference system is top of the vertical rod
horizontal rod
x_{cm} = 1 / M ∫λ x dx = λ/M x² / 2
we evaluate between the lower limits x = 0 and the upper limit x = -L
x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M
x_{cm} = L / 2
vertical rod
y_{cm} = 1 / M ∫y dm
y_{cm} = λ / M ∫y dy
y_{cm} = λ / M y2 / 2
we evaluate between the integration limits x = 0 and higher x = -L
y_{cm} = (M / L) 1 / M ((-L)²/2 -0)
y_{cm} = L / 2
To find the center of mass of the 'L' structure, use the formal definition and consider the midpoint of each member. The x-coordinate is halfway between the ends and the y-coordinate is halfway between the top and bottom ends.
Explanation:To determine the location of the center of mass of the 'L' structure, we can use the formal definition of center of mass. Since the thin vertical and horizontal members have the same length and mass, the center of mass of each member is located at its midpoint. The x-coordinate of the center of mass is halfway between the x-coordinate of the left end and the x-coordinate of the right end of the 'L'. The y-coordinate of the center of mass is halfway between the y-coordinate of the bottom end and the y-coordinate of the top end of the 'L'.
(a) Origin at the lower left:
x = -L/4, y = L/4
(b) Origin at the top of the vertical member:
x = L/4, y = -L/4
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Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region
Answers
Answer:
[tex]2.429783984\times 10^{-14}\ A[/tex]
Explanation:
Velocity of electron = 6020 m/s
Velocity of proton = 1681 m/s
Electron space = 0.0476 m
Proton space = 0.0662 m
e = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
Number of electrons passing per second
[tex]n_e=\dfrac{6020}{0.0476}\\\Rightarrow n_e=126470.588[/tex]
Number of protons passing per second
[tex]n_p=\dfrac{1681}{0.0662}\\\Rightarrow n_p=25392.749[/tex]
Current due to electrons
[tex]I_e=n_ee\\\Rightarrow I_e=126470.588\times 1.6\times 10^{-19}\\\Rightarrow I_e=2.0235\times 10^{-14}\ A[/tex]
Current due to protons
[tex]I_p=n_pe\\\Rightarrow I_p=25392.749\times 1.6\times 10^{-19}\\\Rightarrow I_p=4.06283984\times 10^{-15}\ A[/tex]
Total current
[tex]I=2.0235\times 10^{-14}+4.06283984\times 10^{-15}\\\Rightarrow I=2.429783984\times 10^{-14}\ A[/tex]
The average current is [tex]2.429783984\times 10^{-14}\ A[/tex]
Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would require quantum mechanics to describe, that is, would show both particle and wave properties? Which objects would act like everyday objects, and be adequately described by classical mechanics? object quantum or classical? A virus with a mass of 9.4 x 10-17 g, 280. nm wide, moving at 0.50 µm/s. classical quantum A buckyball with a mass of 1.2 x 10-21 g, 0.7 nm wide, moving at 37. m/s. classical quantum A mosquito with a mass of 1.0 mg, 6.3 mm long, moving at 1.1 m/s. classical quantum A turtle with a mass of 710. g, 22. cm long, moving at 2.8 cm/s. classical quantum
Answers
All objects are described using classical mechanics, with the exception of the buckyball, which requires a quantum mechanical description due to its small mass and high speed that result in a significant de Brogli wavelength.
Explanation:Whether an object shows particle-wave duality, requiring quantum mechanics to describe, or behaves like everyday objects, described by classical mechanics, is determined by the de Broglie wavelength of the object. Calculating the de Broglie wavelength (λ) can be accomplished using the equation λ = h/mv, where h is Planck's constant and m and v are the object's mass and velocity, respectively.
For the virus, buckyball, mosquito, and turtle, the calculated de Broglie wavelengths are so small compared to their physical dimensions that quantum mechanical effects would be negligible. Hence, all these objects can be described using classical mechanics. The buckyball is the only exception. Owing to its tiny mass and high speed, its de Broglie wavelength becomes significant in relation to its size, requiring quantum mechanical description.
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Objects that are atomic or subatomic in size require quantum mechanics for their description, while larger objects can be adequately explained by classical mechanics. In the provided cases, a virus and a buckyball would require quantum mechanics, while a mosquito and a turtle would be described by classical mechanics.
Explanation:To determine whether an object would be adequately described by classical mechanics or require quantum mechanics, we need to consider its size, mass, and speed. Under the domain of quantum mechanics are typically objects that are atomic or subatomic in size, meaning they exhibit wave-particle duality due to their small size and significant motion.
1. A virus: Being atomic in size, a virus would require quantum mechanics for its description. Specifically, it would demonstrate both particle and wave properties due to its small size.
2. A buckyball: Given its subatomic mass, size, and significant speed, a buckyball falls into the quantum domain, displaying both particle and wave properties.
3. A mosquito and 4. A turtle : These are larger, macroscopic objects. Their motion and behavior can be adequately described using classical mechanics as they primarily exhibit particle properties, their wave-like characteristics being effectively obscured due to their larger mass and size. Their behaviors conform to the laws defined under classical mechanics and do not exhibit the quantum nature that smaller objects do.
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Problem 2: A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.11 m. Attached to the disk are four low mass rods of radius b = 0.14 m, each with a small mass m = 0.4 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 21 N for a time of ∆t=0.2 s. (Mom. of inertia of a disk I_(disk,CM)=1/2 MR^2 and point mass is I_(point mass)=mb^2.) Part 1. What is the angular speed of the apparatus? Part 2: What is the angle through which the apparatus turns in radians or degrees?
Answers
Answer:
the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]
the angle through which the apparatus turns in radians or degrees is : [tex]\theta = 68.54^0[/tex]
Explanation:
Given that :
mass of uniform disk M = 1.2 kg
Radius R = 0.11 m
lower radius (b) = 0.14 m
small mass (m) = 0.4 kg
Force (F) = 21 N
time (∆t) =0.2 s
Moment of Inertia of [tex]I_{disk, CM}[/tex] = [tex]\frac{1}{2}MR^2[/tex]
= [tex]\frac{1}{2}*1.2*0.11^2[/tex]
= 0.00726 kgm²
Point mass [tex]I_{point \ mass}[/tex] = mb²
But since four low rods are attached ; we have :
[tex]I_{point \ mass}[/tex] = 4 × mb²
= 4 × 0.4 (0.14)²
= 0.03136 kgm²
Total moment of Inertia = [tex]I_{disk, CM}[/tex] + [tex]I_{point \ mass}[/tex]
= (0.00726 + 0.03136) kgm²
= 0.03862 kgm²
Assuming ∝ = angular acceleration = constant;
Then; we can use the following kinematic equations
T = FR
T = 2.1 × 0.11 N
T = 2.31 N
T = I × ∝
2.31 = 0.03862 × ∝
∝ = [tex]\frac{2.31}{0.03862}[/tex]
∝ = 59.81 rad/s²
Using the formula [tex]\omega_f = \omega_i + \alpha \delta T[/tex] to determine the angular speed of the apparatus; we have:
[tex]\omega_f =0 + 59.81*0.2[/tex] since [tex]( w_i \ is \ at \ rest ; the n\ w_i = 0 )[/tex]
[tex]\omega_f =11.962 \ rad/s[/tex]
∴ the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]
b) Using the formula :
[tex]\theta = \omega_i t + \frac{1}{2}*\alpha*(t)^2\\\\\theta = 0 *0.2 + \frac{1}{2}*59.81*(0.2)^2 \\ \\ \theta = 29.905 *(0.2)^2 \\ \\ \theta = 1.1962 rads \ \ \ ( to \ degree; \ we \ have) \\ \\ \theta = 68.54^0[/tex]
Thus, the angle through which the apparatus turns in radians or degrees is : [tex]\theta = 68.54^0[/tex]
To solve the problem, compute the angular acceleration using the torque and moment of inertia. Given the angular acceleration and the time, calculate the angular velocity. Finally, use the angular acceleration and time to compute the angle turned through.
Explanation:The first part of this problem involves calculating angular acceleration, which is the rate of change of angular velocity. Using the formula, angular acceleration (α) = Torque (τ) / Moment of Inertia (I), the torque can be calculated using the formula Torque = Force * Radius, and the moment of inertia is calculated using the given formulas for disk and point masses.
For the disk, I_(disk,CM) = 1/2 M R^2 and for the point masses, I_(point mass) = 4 * m * b^2. Summing these gives the total moment of inertia. The angular acceleration is then obtained by dividing the torque by the total moment of inertia.
Having the angular acceleration and the time, we can calculate the angular velocity (ω) using the formula ω = α * Δt.
For the second part, the angle through which the apparatus turns can be calculated using the formula θ = 0.5 * α * (Δt)^2, as the initial angular velocity was zero.
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A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. Answer in units of m/s.
Answers
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
To practice Tactics Box 14.1 Identifying and analyzing simple harmonic motion. A complete description of simple harmonic motion must take into account several physical quantities and various mathematical relations among them. This Tactics Box summarizes the essential information needed to solve oscillation problems of this type. The position of a 60 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. Determine the velocity at t=0.40s. Assume that the oscillating mass described in Part A is attached to a spring. What would the spring constant k of this spring be? What is the total energy E of the mass described in the previous parts?
Answers
To solve for the velocity at t=0.40s of a mass in SHM, differentiate the position function with respect to time, and use the resulting velocity function. The spring constant can be calculated using Hooke's law, and the total energy can be found through the energy relationship in SHM.
Explanation:The student is asking about the characteristics of simple harmonic motion (SHM) related to a mass attached to a spring. To find the velocity at a specific time t for the mass undergoing SHM, we need to differentiate the position function x(t) with respect to time. The provided position function is x(t)=(2.0 cm)cos(10t), implying that the velocity function will be v(t)=-ω A sin(ωt), where ω is the angular frequency, and A is the amplitude. To calculate the spring constant k and total energy E, we will employ both Hooke's law, F = -kx, and the relationship between the total energy in SHM and the displacement from the equilibrium position, given by E = ½ k A².
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The velocity at [tex]t=0.40s[/tex] is [tex]0.1514m/s[/tex].
The spring constant is [tex]6N/m[/tex].
The total energy of the mass is 0.0012J.
Determining Velocity, Spring Constant, and Total Energy in Simple Harmonic Motion
The position of a 60 g oscillating mass is given by the function: [tex]x(t) = (2.0 cm)cos(10t)[/tex], where time t is in seconds. Here's how we solve for the velocity at [tex]t = 0.40 s[/tex], the spring constant k, and the total energy E of the system.
Step-by-Step Solution
Velocity at [tex]t = 0.40 s[/tex]:Given:[tex]x(t) = (0.02 m)cos(10t)[/tex] (converted cm to meters)
Velocity v(t) is the derivative of x(t):
[tex]v(t) = dx/dt = - (0.02 m)(10)sin(10t)[/tex]
Substitute [tex]t = 0.40 s[/tex]:
[tex]v(0.40 s) = - (0.02 m)(10)sin(10 \times 0.40) = -0.2 sin(4) = -0.2(-0.7568) = 0.1514 m/s[/tex].Spring Constant (k):
The angular frequency, w, is given by [tex]w = 10 s^{-1}[/tex]. Using the formula [tex]w = \sqrt{(k/m)}[/tex], we can solve for k:
[tex]w^2 = k/m[/tex]
[tex](10)^2 = k/(0.06 kg)[/tex] (mass converted to kg)
[tex]100 = k/0.06[/tex]
[tex]k = 100 \times 0.06 = 6 N/m.[/tex]Total Energy (E):
The total energy in a simple harmonic oscillator is given by [tex]E = 0.5 \times k \times A^2,[/tex]
where A is the amplitude (0.02 m), k is the spring constant (6 N/m):
[tex]E = 0.5 \times 6 N/m \times (0.02 m)^2 = 0.5 \times 6 \times 0.0004 = 0.0012 J[/tex].
You have been sent to a new home. The homeowner reports that sometimes the electric furnace trips the 240-V, 60-A circuit breaker connected to it. Upon examination, you find that the furnace contains three 5000-W heating elements designed to turn on in stages. For example, when the thermostat calls for heat, the first 5000-W element turns on. After some period of time, the second element will turn on, and then, after another time delay, the third element will turn on. What do you think the problem is, and what would be your recommendation for correcting it? Explain your answer.
Answers
The problem is likely an excessive combined power draw from the three heating elements, exceeding the circuit breaker's capacity; the recommendation is to upgrade the circuit breaker or adjust the furnace's operation to stay within the breaker's limit.
We have,
The problem seems to be related to the load on the circuit exceeding the capacity of the circuit breaker.
When the furnace is turned on and all three 5000-W heating elements turn on in stages, the combined power consumption becomes 3 * 5000 W = 15000 W.
This is a substantial load that exceeds the circuit breaker's capacity, which is likely 240 V * 60 A = 14400 W (due to the product of voltage and current rating).
As a result, the circuit breaker is tripping to protect the circuit from overloading, as the total power drawn from the furnace exceeds its rated capacity.
To correct this issue and prevent the circuit breaker from tripping, you could consider the following recommendations:
- Check the Circuit Breaker Rating:
Confirm the rating of the circuit breaker connected to the furnace. If it's indeed 60 A, you might need to upgrade the circuit breaker to a higher rating that can handle the combined power consumption of all three heating elements.
- Reduce Load:
Alternatively, you could reconfigure the furnace to operate only one or two heating elements at a time to reduce the load and stay within the circuit breaker's capacity.
This may involve adjusting the furnace's internal settings or installing additional controls to manage the heating elements' activation.
- Consider Energy Management:
Implement an energy management system that staggers the activation of the heating elements over time.
This would ensure that the power demand doesn't exceed the circuit breaker's capacity during startup.
- Professional Electrician:
Thus,
The problem is likely an excessive combined power draw from the three heating elements, exceeding the circuit breaker's capacity; the recommendation is to upgrade the circuit breaker or adjust the furnace's operation to stay within the breaker's limit.
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The issue is likely that the electric furnace surpasses the 60-A limit of the circuit breaker when all heating elements are active, causing it to trip. A potential solution could be to upgrade the circuit breaker to a higher amperage rating or adjust the furnace so that it never exceeds 60 A.
Explanation:The problem the homeowner is encountering with their electric furnace is likely due to the furnace exceeding the capacity of the 240-V, 60-A circuit breaker when all three heating elements activate. Each 5000-W heating element at 240 V consumes about 20.8 A of current. If all three elements are powered simultaneously, the total current draw is 62.4 A, which surpasses the 60-A circuit breaker limit and causes it to trip.
One potential solution would be to upgrade the circuit breaker to one with a higher amperage rating. However, any modifications should comply with local electricity standards and should be carried out by a qualified electrician. Alternatively, the furnace could be rewired or adjusted to ensure that the cumulative draw never surpasses 60 A at any given time.
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2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust the length of the air column in the tube. A tuning fork of frequency 330 Hz is sounded and held at the upper end as the tube is lifted out of the water until the first resonant sound is heard. The length of tube above the water level when the fundamental note sounds is 26 cm. a. Calculate the speed of sound in the air column in the tube. b. Determine another position for the tube that will result in a resonance with this tuning fork. c. Determine the first position (as we again lower the tube into the water) that will resonate with a 256 Hz tuning fork.
Answers
Answer:
Explanation:
This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows
l = λ / 4 where l is length of tube and λ is wave length.
here l = .26 m
λ = .26 x 4 = 1.04 m
frequency of sound = 330 Hz
velocity of sound = frequency x wave length
= 330 x 1.04
= 343.2 m /s
b )
Next overtone will be produced at 3 times the length
so next length of air column = 3 x 26
= 78 cm
c )
If frequency of sound = 256 Hz
wavelength = velocity / frequency
= 343.2 / 256
= 1.34 m
= 134 cm
length of air column for resonance
= wavelength / 4
134/4
= 33.5 cm
In class we learned that you can detect The Big C (molecular weight of 303 grams/mol) by mass spectrometry. The mass spec has a magnetic field of 0.3 T and the singly charged ions in the sample are moving at 50,000 m/s. How far from the entrance in m do the Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory?
Answers
Answer:
1.04m
Explanation:
the distance is determined by the diameter of the trajectory.
you can find the radius of the trajectory by using the following formula:
[tex]r=\frac{m_cv}{qB}[/tex]
mc : mass of the Big C = 0.303kg/mol/(6.02*10^{23}/mol)=5.03*10^{-25}kg
v: velocity of the ion = 50000m/s
B: magnetic constant = 0.3T
q: 1.6*10^{-19}C
By replacing you obtain:
[tex]r=\frac{(5.03*10^{-25}kg)(50000m/s)}{(1.6*10^{-19}C)(0.3T)}=0.52m[/tex]
the diameter wiil be:
d=2r=1.04m
the ion strikes the detector from 1.04m to the entrance of the spectrometer
The Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory will be "1.04 m" far.
SpectrometerAccording to the question,
Big C's mass, [tex]m_c[/tex] = [tex]\frac{0.303}{6.02\times 10^{23}}[/tex]
= 5.03 × 10⁻²⁵ kg
Ion's velocity, v = 50000 ms
Magnetic constant, B = 0.3 T
Charge, q = 1.6 × 10⁻¹⁹ C
We know that,
→ r = [tex]\frac{m_c v}{qB}[/tex]
or,
The radius, r = [tex]\frac{5.03\times 10^{-25}\times 50000}{1.6\times 10^{-19}\times 0.3 }[/tex]
= 0.52 m or,
Diameter, d = 2 × r
= 2 × 0.52
= 1.04 m
Thus the above approach is correct.
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An electric generator contains a coil of 99 turns of wire, each forming a rectangular loop 73.9 cm by 34.9 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 2.96 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1200 rev/min about an axis perpendicular to the magnetic field?
Answers
Answer:
12078.46 V
Explanation:
Applying,
E₀ = BANω.................... Equation 1
Where: E₀ = maximum emf, B = magnetic Field, A = Area of the coil, N = Number of turns of the coil. ω = angular velocity
Given: N = 99 turns, B = 2.96 T, ω = 1200 rev/min = (1200+0.10472) = 125.664 rad/s
A = L×W, where L= Length = 93.9 cm = 0.939 m, W = width = 34.9 cm = 0.349 m
A = (0.939×0.349) = 0.328 m²
Substitute into equation 1
E₀ = 99(2.96)(0.328)(125.664)
E₀ = 12078.46 V
Hence the maximum value of the emf produced = 12078.46 V
Answer:
The maximum emf induced in the loop is 9498.268 V
Explanation:
Given;
number of turns of coil, N = 99 turns
area of the rectangular loop, A = 0.739 m x 0.349 m = 0.2579 m²
magnetic field strength, B = 2.96 T
angular speed of the loop, ω = 1200 rev/min
angular speed of the loop, ω (rad/s) = (2π x 1200) / 60 = 125.68 rad/s
The maximum value of the emf produced is calculated using the formula below;
ξ = NABω
Substitute the given values and calculate the maximum emf induced;
ξ = (99)(0.2579)(2.96)(125.68)
ξ = 9498.268 volts
Therefore, the maximum emf induced in the loop is 9498.268 V
As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over the cause of the sparking, with one stating that the microwaves "herd" electrons into "pointy" areas of the metal object, from which the electrons jump from one part of the object to another. The other friend says that the sparks are caused by the photoelectric effect. Prove or disprove the latter idea using basic physics. Suppose the typical work function of the metal is roughly 4.910×10−19 J.4.910×10−19 J. Calculate the maximum wavelength in angstroms of the radiation that will eject electrons from the metal.
Answers
Answer:
[tex]4.048\times 10^{-7}\ m[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
E = Energy = [tex]4.91\times 10^{-19}\ J[/tex]
Wavelength ejected is given by
[tex]\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m[/tex]
The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is [tex]4.048\times 10^{-7}\ m[/tex]
The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 9.00 s, at which time it is turning at 7.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 13.0 s. Through how many revolutions does the tub rotate while it is in motion
Answers
Answer:
Explanation:
7 rev /s = 7 x 2π rad /s
angular velocity = 14π rad /s
Angular acceleration α = increase in angular velocity / time
= 14π - 0 / 9
α = 4.8844 rad / s
θ = 1/2 α x t² θ is angle of rotation , t is time
= 1/2 x 4.8844 x 9²
= 197.8182 rad
2π n = 197.8182
n = 31.5 rotation
During acceleration , no of rotation made = 31.5
During deceleration : -----
deceleration =
Angular deceleration α = decrease in angular velocity / time
= 14π - 0 / 13
α = 3.3815 rad / s
θ = 1/2 α x t² θ is angle of rotation , t is time
= 1/2 x 3.3815 x 13²
= 285.736 rad
2π n = 285.736
n = 45.5 rotation
total rotation
= 45.5 + 31.5
= 77 .
g Show that the first derivative of the magnitude of the net magnetic field of the coils (dB/dx) vanishes at the mid- point P regardless of the value of s. Why would you expect this to be true from symmetry? (b) Show that the second derivative (d2 B/dx2 ) also vanishes at P, provided s R. This accounts for the uniformity of B near P for this particular coil separation.
Answers
Answer:
Hi Carter,
The complete answer along with the explanation is shown below.
I hope it will clear your query
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Explanation:
The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:
B = NμοiR² / 2(R²+Z²)³÷²
where
R is the radius of the loop
N is the number of turns
i is the current.
Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore
B = NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]
Its derivative with respect to x is
dB /dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]
When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB /dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷² ] =0
independent of the value of s.
A toy car has a 1.5 A current, and its resistance is 2 . How much voltage does the car require?
Answers
Answer:
3 Volts
Explanation:
Using Ohm's Law
Voltage=current x resistance
V = I x R
V = 1.5 x 2
V = 3 V
Find the charge q(t) on the capacitor and the current i(t) in the given LRC-series circuit. L = 1 h, R = 100 Ω, C = 0.0004 f, E(t) = 20 V, q(0) = 0 C, i(0) = 3 A q(t) = C i(t) = A
Find the maximum charge on the capacitor. (Round your answer to four decimal places.)
Answers
Answer:
The maximum charge around the capacitor is 0.03170189C
Explanation:
See attached file
In an RLC series circuit with given values for resistance, inductance, and capacitance, we can calculate the charge on the capacitor five cycles later and fifty cycles later using the equation q(t) = q(0) * e^(-(R/L)t)
Explanation:In an RLC series circuit with a resistance of 7.092 ohms, an inductance of 10 mH, and a capacitance of 3.0 µF, the charge on the capacitor can be calculated using the equation:
q(t) = q(0) * e^(-(R/L)t)
Given that the initial charge on the capacitor is 8.0 µC, we can calculate the charge five cycles later and fifty cycles later using this equation.
(a) To find the charge five cycles later, we need to find the time period of one cycle. The time period can be calculated using the formula:
T = 2π√(LC)
Once we have the time period, we can calculate the time it takes for five cycles to pass and substitute it in the equation to find the charge.
(b) To find the charge fifty cycles later, we can use the same approach as in part (a), but substituting the time it takes for fifty cycles to pass.
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