Answer:
v = 3.7 m/s
Explanation:
As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:
m. g. h = 1/2 m v²
The only unknown (let alone the speed) in the equation , is the height from which the swing is released.
At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.
As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:
h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m
Replacing in the energy conservation equation, and solving for v, we get:
v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s
An astronaut is a short distance away from her space station without a tether rope. She has a large wrench. What should she do with the wrench to move toward the space station?
a. Throw it directly toward the space station.
b. Throw it directly away from the space station.
c. Throw it opposite to the direction of the station's orbit.
d. Throw it parallel to the direction of the station's orbit.
e. Throw it toward the station without letting go of it.
Answer: b. Throw it directly away from the space station.
Explanation:
According to Newton's third law of motion, when two bodies interact between them, appear equal forces and opposite senses in each of them.
To understand it better:
Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude but in the opposite direction on the first.
So, if the astronaut throws the wrench away from the space station (in the opposite direction of the space station), according to Newton's third law, she will be automatically moving towards the station and be safe.
Final answer:
To move toward the space station, the astronaut should throw the wrench directly away from it. By doing so, the astronaut imparts a momentum in the opposite direction, causing herself to move towards the space station.
Explanation:
The astronaut should throw the wrench directly away from the space station. When the astronaut throws the wrench in the opposite direction of the space station, the principle of conservation of momentum comes into play. According to this principle, the total momentum of a closed system remains constant. By throwing the wrench away from the space station, the astronaut imparts a momentum in the opposite direction, causing herself to move towards the space station. This is similar to how an ice skater moves faster when she throws her arms outward while spinning.
A diffraction grating that has 4500 lines/cm is illuminated by light that has a single wavelength. If a second order maximum is observed at an angle of 42° with respect to the central maximum, what is the wavelength of this light?
Answer:
The wavelength is 742.7 nm.
Explanation:
Given that,
Grating = 4500 lines/cm
Angle = 42°
Order number =2
We need to calculate the distance
[tex]d=\dfrac{1\times10^{-2}}{4500}[/tex]
[tex]d=2.22\times10^{-6}\ m[/tex]
We need to calculate the wavelength
Using diffraction formula
[tex]d\sin\theta=m\times\lambda[/tex]
[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]
[tex]\lambda=\dfrac{2.22\times10^{-6}\times\sin42}{2}[/tex]
[tex]\lambda=7.427\times10^{-7}[/tex]
[tex]\lambda=742.7\ nm[/tex]
Hence, The wavelength is 742.7 nm.
The wavelength of this light was found to be 7.08 × 10⁻⁴ m.
To find the wavelength of light that results in a second-order maximum at a 42° angle when striking a diffraction grating with 4500 lines per centimeter, we use the formula for diffraction grating: d sin θ = mλ, where d is the distance between adjacent grating lines, θ is the angle of the maximum, m is the order of the maximum, and λ is the wavelength of the light. First, calculate the distance between grating lines (d) as the reciprocal of the grating's line density (4500 lines/cm or 4.5×105 lines/m), yielding d = 1/4.5×105 m. Then, substituting the given values into the formula with m = 2 and θ = 42°, solve for the wavelength λ.
λ = (1/4.5×105) * sin(42°) / 2
= 7.08 × 10⁻⁴ m
It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used toassess eligibility can be described by Hooke’s law and is depressed 0.75 cm byitsmaximum load of 120 kg. (a) What is the spring’s effective spring constant? (b) Aplayer stands on the scales and depresses it by 0.48 cm. Is he eligible to play on thisunder-85-kg team?
To solve this problem it is necessary to apply the equations given in Newton's second law as well as Hooke's Law.
Since Newton's second law we have that force is
F = mg
Where,
m = mass
g = gravity
From the hook law, let us know that
F = -kx
Where
k = Spring constant
x = Displacement
Re-arrange to find k,
[tex]k = -\frac{F}{x}[/tex]
PART A ) We can replace the Newton definitions here, then
[tex]k = -\frac{mg}{x}[/tex]
Replacing with our values we have that
[tex]k = \frac{120*9.8}{-0.75*10^{-2}}[/tex]
[tex]k = 1.57*10^5N/m[/tex]
Therefore the required value of the spring constant is [tex]1.57*10^5N/m[/tex]
PART B) We can also equating both equation to find the mass, then
[tex]mg = -kx[/tex]
[tex]m = \frac{-kx}{g}[/tex]
Replacing with tour values we have
[tex]m = \frac{1.568*10^5(-9.48*10^{-2})}{9.8}[/tex]
[tex]m = 76.8Kg[/tex]
Therefore the mass of the player can be of 76.8Kg, then the player is eligible to play because the mass is less than 85Kg
Final answer:
The spring's effective force constant is 156,800 N/m. The player is eligible to play on the under-85-kg team.
Explanation:
Using Hooke's law, the effective force constant of the spring can be calculated by dividing the maximum load by the displacement. In this case, the spring is depressed by 0.75 cm under a maximum load of 120 kg.
To find the effective force constant, we can use the formula:
F = k * x
Where F is the force, k is the force constant, and x is the displacement.
So, the effective force constant of the spring is:
k = F / x = 120 kg * 9.8 N/kg / 0.0075 m = 156,800 N/m
For part (b), we can use the same formula to determine if the player is eligible to play on the under-85-kg team. Given that the player depresses the spring by 0.48 cm, we can plug in the values to calculate the force:
F = k * x = 156,800 N/m * 0.0048 m = 753.6 N
Since the force exerted by the player is less than 85 kg * 9.8 N/kg, the player is eligible to play on the under-85-kg team.
The hollow tube is pivoted about a horizontal axis through point O and is made to rotate in the vertical plane with a constant counterclockwise angular velocity = 2.9 rad/sec. If a 0.15-lb particle is sliding in the tube toward O with a velocity of 4.8 ft/sec relative to the tube when the position θ = 36° is passed, calculate the magnitude N of the normal force exerted by the wall of the tube on the particle at this instant.
To solve the problem it is necessary to apply the concepts related to Newton's second law, as well as to the sum of forces in this type of bodies.
According to the description I make a diagram that allows a better understanding of the problem.
Performing sum of forces in the angular direction in which it is inclined we have to
[tex]\sum F = ma_{\theta}[/tex]
[tex]N - W cos\theta = ma_{\theta}[/tex]
[tex]N = ma_{\theta}+Wcos\theta[/tex]
Tangential acceleration can be expressed as
[tex]a_{\theta} = (r\ddot{\theta}+2\dot{r}\dot{\theta})[/tex]
Our values are given by,
[tex]\dot{\theta} = 2.9rad/s[/tex]
[tex]m = 0.15 lb[/tex]
[tex]\theta = 36\°[/tex]
[tex]v = 4.8ft/s[/tex]
Substituting [tex]\ddot{\theta}=0rad/s^2 , \dot{r}=-4.8ft/s, \dot{\theta}=2.9 rad/s[/tex]
[tex]a_{\theta} = r*0+2*(4.8*2.9)\\a_{\theta}=27.84ft/s^2[/tex]
At the same time we acan calculate the mass of the particle, then
W = mg
Where,
W = Weight of the particle
m = mass
g = acceleration due to gravity
[tex]0.15lb = m(32.2ft/s^2)[/tex]
[tex]m = 4.66*10^{-3}Lb[/tex]
Now using our first equation we have that
[tex]N = ma_{\theta}+Wcos\theta[/tex]
[tex]N = (4.66*10^{-3})(27.84)+0.2cos36[/tex]
[tex]N = 0.2914Lb[/tex]
Therefore the normal force exerted by the wall of the tube on the particle at this instant is 0.2914Lb
The magnitude of the normal force [tex]\( N \)[/tex] exerted by the wall of the tube on the particle at the instant when the angle [tex]\( \theta = 36^\circ \)[/tex] is approximately 0.249 lb
To determine the normal force [tex]\( N \)[/tex]exerted by the wall of the tube on the particle, we analyze the forces acting on the particle in a rotating reference frame. Here’s the step-by-step process:
1. Identify the given data:
- Angular velocity [tex]\( \omega = 2.9 \)[/tex] rad/sec
- Particle weight [tex]\( W = 0.15 \)[/tex] lb
- Relative velocity towards O [tex]\( v_r = 4.8 \)[/tex] ft/sec
- Angle [tex]\( \theta = 36^\circ \)[/tex]
2. **Convert the weight to mass:**
-[tex]\( W = mg \)[/tex]
-[tex]\( m = \frac{W}{g} \)[/tex]
- Using[tex]\( g = 32.2 \text{ ft/sec}^2 \)[/tex]:
- [tex]\( m = \frac{0.15 \text{ lb}}{32.2 \text{ ft/sec}^2} = 0.00466 \text{ slugs} \)[/tex]
3. Calculate the distance ( r ):
- Since [tex]\( v_r = 4.8 \)[/tex] ft/sec is towards O and the tube is rotating with [tex]\( \omega \),[/tex] the distance [tex]\( r \)[/tex] can be found using [tex]\( r = \frac{v_r}{\omega} \)[/tex]:
- [tex]\( r = \frac{4.8 \text{ ft/sec}}{2.9 \text{ rad/sec}} = 1.655 \text{ ft} \)[/tex]
4. **Determine the forces:**
- Centrifugal force [tex]\( F_c = m \omega^2 r \)[/tex]:
[tex]\[ F_c = 0.00466 \text{ slugs} \times (2.9 \text{ rad/sec})^2 \times 1.655 \text{ ft} = 0.0638 \text{ lb} \][/tex]
- Component of gravitational force in the radial direction [tex]\( F_{g, r} = W \cos \theta \)[/tex]:
[tex]\[ F_{g, r} = 0.15 \text{ lb} \times \cos 36^\circ = 0.1214 \text{ lb} \][/tex]
5. Calculate the normal force [tex]\( N \)[/tex]:
- Normal force [tex]\( N \)[/tex] must balance the radial forces:
[tex]\[ N = F_c + F_{g, r} + m \cdot (\omega^2 r) \][/tex]
[tex]\[ N = 0.0638 \text{ lb} + 0.1214 \text{ lb} + 0.0638 \text{ lb} \][/tex]
[tex]\[ N = 0.249 \text{ lb} \][/tex]
The amount of radiant power produced by the sun is approximately 3.9 ✕ 1026 W. Assuming the sun to be a perfect blackbody sphere with a radius of 6.96 ✕ 108 m, find its surface temperature (in kelvins).
Answer:
T=5797.8 K
Explanation:
Given that
Power P = 3.9 x 10²⁶ W
Radius ,r= 6.96 x 10⁸ m
We know that ,From Plank's law
P = σ A T⁴
σ = 5.67 x 10 ⁻⁸
A= Area ,T= Temperature ( in Kelvin)
[tex]T=\left (\dfrac{P}{\sigma A} \right )^{\dfrac{1}{4}}[/tex]
Now by putting the values
[tex]T=\left (\dfrac{P}{\sigma A}\right )^{\dfrac{1}{4}}[/tex]
[tex]T=\left (\dfrac{3.9\times 10^{26}}{5.67\times 10^{-8}\times 4\times \pi \times (6.96\times 10^8)^2} \right )^{\dfrac{1}{4}}[/tex]
T=5797.8 K
The temperature of surface will be 5797.8 K
The surface temperature of the sun can be calculated using the Stefan-Boltzmann law, which describes how the energy a perfect blackbody emits is related to its temperature. Knowing the radiant power and the surface area of the sun, we can use this law to find the surface temperature in kelvins.
Explanation:To calculate the surface temperature of the sun, we can make use of the Stefan-Boltzmann law ( radiant power, surface area of the object, and temperature in kelvins), a principle in Physics that describes how the total energy a perfect blackbody like the sun emits is related to its temperature. The law is represented mathematically as P = σ A T^4, where P is the radiant power, A is the surface area, T is the temperature, and σ is Stefan-Boltzmann constant (5.67 x 10^-8 J/(s.m².K^4)). Given that the sun's radiant power is approximately 3.9 x 10^26 W and its radius is 6.96 x 10^8 m (meaning its surface area is 4πr^2), we can rearrange the equation and solve for T to find the sun's surface temperature in kelvins: T = (P/ σA) ^ (1/4).
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Current flowing in a circuit depends on two variables. Identify these variables and their relationship to current.
Question is Incomplete, Complete question is given below.
Current flowing in a circuit depends on two variables. Identify these variables and their relationship to current.
A) Current is proportionate to the conductance of the circuit and precisely proportional to the voltage applied across the circuit.
B) Current is conversely proportional to the electrical tension of the circuit and corresponds to the resistance across the circuit.
C) Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.
D) Current is commensurate to the resistance of the circuit and directly proportional to the electric pressure applied across the circuit.
Answer:
C) Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.
Explanation:
Now Ohms Law states that, "So long as a physical state of a conductor remains the potential difference applied to the conductor is directly proportional to current flowing through it."
I ∝ V
V=IR also I=V/R
where R is the Resistance
Hence, From above equation we can say that Current increases when there is increase in Voltage but Current decreases as the resistance decreases.
Hence,Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.
Suppose that the resistance between the walls of a biological cell is 4.2 × 109 Ω. (a) What is the current when the potential difference between the walls is 75 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.74 s?
Answer:
(a) Current will be [tex]17.857\times 10^{-12}A[/tex]
(b) Number of ions will be [tex]8.258\times 10^6[/tex]
Explanation:
We have given that resistance of the biological cell [tex]R=4.2\times 10^9ohm[/tex]
(a) We have given potential difference of 75 mV
So [tex]V=75\times 10^{-3}volt[/tex]
From ohm's law we know that current is given by
[tex]i=\frac{V}{R}=\frac{75\times 10^{-3}}{4.2\times 10^9}=17.857\times 10^{-12}A[/tex]
(b) We have given time t = 0.74 sec
We have to find the charge
We know that charge is given by Q = it, here i is current and t is time
So charge will be [tex]Q=17.857\times 10^{-12}\times 0.74=13.214\times 10^{-12}C[/tex]
So number of ions will be [tex]n=\frac{13.214\times 10^{-12}}{1.6\times 10^{-19}}=8.258\times 10^6[/tex]
: A laser beam is incident on two slits with separation d-0.038 mm. A screen is placed L-1.6 m from the slits. The wavelength of the laser light is A 4750 Å. and θ2 are the angles to the first and second bright fringes above the center of the screen. Otheex ee 25% Part (a) Express sin(91) in terms of d and λ. sin(θ 1) Add V Correct! 25% Part (b) Express sin(92) in terms of d and λ. sin(02) (2 A )yd Correct 25% Part (c) Express the distance between the two bright fringes on the screen, y, in terms of θ1, θ2 and L. Grade S Deductic Potentia 9 cotan(0) cotan(02) in(a)N 4 5 6 cos(0) | (.ilM-17 8 cos(α) cos(φ) Submiss Attempts per detailed sin(φ) tan(6) sin(0) tan(01) Hint I give up! Submit Feedback: deduction per feedback Hints: 0%, deduction per hint. Hints remaining: 2 25% Part (d) Solve for the numerical value of y ¡n meters.
To solve this physics problem involving Young's double-slit experiment, we apply the conditions for constructive interference to determine the angles θ1 and θ2 for the first and second bright fringes. We then use the small angle approximation to determine the distance y between those fringes on the screen. Lastly, we substitute the given values into the formula to compute the numerical value of y.
Explanation:The topic at hand is Young's double-slit experiment, a foundational experiment in wave optics. Given the slit separation d, the laser light's wavelength λ, and the screen's distance L, we need to find the distance between two bright fringes on the screen. The formulas required here stem from the conditions for constructive interference in double-slit set-ups.
Part a) sin(θ1) = λ/d, which is the condition for the light to interfere constructively at the first bright fringe from the center on the screen.
Part b) sin(θ2) = 2λ/d, which gives us the location of the second bright fringe, as it occurs when the path differential is twice the wavelength.
Part c) The distance y between the two bright fringes on the screen can be found using the small angle approximation, which allows us to express the angles θ1 and θ2 in terms of the fringe order m. Thus, y ≈ L*(θ2 - θ1) = L*(2λ/d - λ/d) = L*λ/d.
Part d) Substituting the given values into the formula from Part c, we find y = 1.6m * (4750Å / 0.038mm), which gives us a numerical value once we make sure to convert the units properly. In this case, we need to convert Å to mm by multiplying 4750Å by 10^-4 mm/Å.
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The question is about Young's double-slit experiment in physics. Both θ1 and θ2 are expressed in terms of d (the separation between the slits) and λ (the wavelength of the laser light). The distance between the fringes on the screen is calculated using these variables, resulting in a numerical value in meters.
Explanation:The questions refer to the phenomenon of light diffraction and interference, particularly as it relates to Young's double-slit experiment. In this context, we'll use the formulas defined for this experiment:
For parts (a) and (b): we know that the condition for constructive interference, resulting in bright fringes, for a double-slit setup is given by:
d sin θ = mλ
where 'd' is the slit separation, 'θ' is the angle to the fringe from the central maximum, 'm' is the order of the fringe and 'λ' is the wavelength of the light. For the first fringe (m=1), sin(θ1) = λ/d and for the second fringe (m=2), sin(θ2) = 2λ/d.
For part (c): Considering small angles where tan θ is approximately equal to sin θ, the separation of two bright fringes 'y' on the screen is given by:
y = L (θ2 - θ1) = L λ/d.
For part (d): Now, we just need to substitute the values given into the above formula. Substituting the given values of L = 1.6 m, λ = 4750 Å (which is equivalent to 4750 x 10^-10 m), and d = 0.038 mm (or 0.038 x 10^-3 m), you get:
y = L λ/d ≈ 0.02 m.
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Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed of 6.66 × 106 m/s and (b) a kinetic energy of 1.71 × 10−15 J.
Explanation:
Given that,
(a) Speed, [tex]v=6.66\times 10^6\ m/s[/tex]
Mass of the electron, [tex]m_e=9.11\times 10^{-31}\ kg[/tex]
Mass of the proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
The wavelength of the electron is given by :
[tex]\lambda_e=\dfrac{h}{m_ev}[/tex]
[tex]\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}[/tex]
[tex]\lambda_e=1.09\times 10^{-10}\ m[/tex]
The wavelength of the proton is given by :
[tex]\lambda_p=\dfrac{h}{m_p v}[/tex]
[tex]\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}[/tex]
[tex]\lambda_p=5.96\times 10^{-14}\ m[/tex]
(b) Kinetic energy, [tex]K=1.71\times 10^{-15}\ J[/tex]
The relation between the kinetic energy and the wavelength is given by :
[tex]\lambda_e=\dfrac{h}{\sqrt{2m_eK}}[/tex]
[tex]\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}[/tex]
[tex]\lambda_e=1.18\times 10^{-11}\ m[/tex]
[tex]\lambda_p=\dfrac{h}{\sqrt{2m_pK}}[/tex]
[tex]\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}[/tex]
[tex]\lambda_p=2.77\times 10^{-13}\ m[/tex]
Hence, this is the required solution.
which of the following is a measure of disorder or randomness
a. energy
b. temperature
c.heat
d.entropy
Answer:
D. Entropy is a measure of disorder or randomness
Explanation:
The answer is entropy. It is the method to measure any disorder or randomness of any system. Basically it is concept of physics and chemistry but can also be used in other subject matters as well. In formula terms, it is denoted by S and is measurement value is "Joule per kelvin". The values could be both positive and negative as well.
All the other options are incorrect as heat, energy and temperature are the different concepts of science which do not relates with any measurement.
Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1320 K to a cold reservoir at 600 K. Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied. (Round the final answers to four decimal places.)
Answer:0.0909 kJ/K
Explanation:
Given
Temperature of hot Reservoir [tex]T_h=1320 K[/tex]
Temperature of cold Reservoir [tex]T_l=600 K[/tex]
Heat of 100 kJ is transferred form hot reservoir to cold reservoir
Hot Reservoir is Rejecting heat therefore [tex]Q_1=-100 kJ[/tex]
Heat is added to Reservoir therefore [tex]Q_2=100 kJ[/tex]
Entropy change for system
[tex]\Delta s=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}[/tex]
[tex]\Delta s=\frac{-100}{1320}+\frac{100}{600}[/tex]
[tex]\Delta s=-0.0757+0.1666=0.0909 kJ/K[/tex]
As entropy change is Positive therefore entropy Principle is satisfied
Two electrons 1 and 2 move along antiparallel paths separated by a distance of 10 nm, traveling at speeds 4.5 x 10^7 m/s and 8.0 x 10^6 m/s.
Part A) What is the magnitude of the magnetic force exerted by electron 2 on electron 1? Express your answer with the appropriate units.
Answer:
The magnetic force on electron 1 by electron 2=[tex]9.22\times 10^{-15} N[/tex]
Explanation:
We are given that
Distance between two electrons=10 nm
Speed of electron 1=[tex]4.5\times 10^7[/tex] m/s
Speed of electron 2=[tex]8.0\times 10^6[/tex] m/s
We have to find the magnitude of the magnetic force exerted by electron 2 on electron 1.
Magnetic force on electron 1 by electron 2
[tex]F=\frac{\mu_0e^2v_1v_2}{4\pi r^2}[/tex]
[tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]
[tex]e=1.6\times 10^{-19} C[/tex]
[tex]v_1=4.5\times 10^7 m/s[/tex]
[tex]v_2=8.0\times 10^6 m/s[/tex]
[tex]r=10nm=10\times 10^{-9}m[/tex] [tex](1nm=10^{-9} m[/tex])
Substitute the values in the given formula
The magnetic force on electron 1 by electron 2=[tex]\frac{10^{-7}\times (1.6\times 10^{-19})^2\times 4.5\times 10^7\times 8\times 10^6}{(10\times 10^{-9})^2}[/tex]
The magnetic force on electron 1 by electron 2=[tex]9.22\times 10^{-15} N[/tex]
Hence, the magnetic force on electron 1 by electron 2=[tex]9.22\times 10^{-15} N[/tex]
The magnitude of the magnetic force exerted by electron 2 on electron 1 is [tex]9.22\times10^{-15}[/tex] N.
What is magnetic force?Magnetic force is the force of attraction of repulsion between two poles of the two magnets. The magnetic force is also appears between two electrically charged bodies.
The magnetic force can be given as,
[tex]F=\dfrac{\mu_oq_1q_2(v_1v_2)}{4\pi r^2}[/tex]
Here, ([tex]\mu_o[/tex]) is the magnetic constant, (q) is the charge on the body, (v) is the speed of the body and (r) is the distance between them.
The distance between the electrons 1 and 2 is 10 nm. As the speed of the electron 1 is [tex]4.5 \times10^7[/tex] m/s and the speed of the electron 2 is [tex]8.0 \times10^6[/tex] m/s.
Thus, put the values in the above formula as,
[tex]F=\dfrac{4\pi \times10^{-7}\times1.6\times10^{-19}\times1.6\times10^{-19}\times4.5\times10^{7}(8.0\times10^6)}{4\pi (10^{-9})^2}[/tex]
[tex]F=9.22\times10^{-15}\rm N[/tex]
Thus the magnitude of the magnetic force exerted by electron 2 on electron 1 is [tex]9.22\times10^{-15}[/tex] N.
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Speed deamplifier. Block 1 of mass m1 slides along an x axis on a frictionless floor at speed 4.00 m/s. Then it undergoes a onedimensional elastic collision with stationary block 2 of mass m2 = 2.00m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 4.00m1. (a) After all the collisions, what is the speed of block 3?
Answer:
29J
Explanation:
Please see attachment .
russell westbrook has a mass of 84.8kg.When jumps to dunk on z 3.04 meter goal,what will be his G.P.E at the apex of his jump?
The gravitational potential energy is 2526 J
Explanation:
The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in the gravitational field. Near the Earth's surface, it can be calculated as
[tex]GPE=mgh[/tex]
where
m is the mass of the body
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
h is the height of the object
In this problem, we have
m = 84.8 kg
h = 3.04 m
Substituting,
[tex]GPE=(84.8)(9.8)(3.04)=2526 J[/tex]
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A typical wind turbine extracts 40% of the kinetic energy of the wind that blows through the area swept by the blades. For a large turbine, 110,000 kg of air moves past the blades at 13 m/s every second. If the wind turbine extracts 40% of this kinetic energy, and if 80% of this energy is converted to electric energy, what is the power output of the generator?
Answer:
2.9744 MW
Explanation:
The air blowing through the wind turbine at 110000kg at speed of 13m/s for every second would have the kinetic energy of
[tex]E_k = 0.5mv^2 = 0.5*110000*13^2 = 9295000J/s[/tex]
If the wind turbine is able to extract only 40% of this kinetic energy, and 80% of this is converted to electricity, then the power output of the generator must be
[tex]E = 80\%*40\%*E_k = 0.8*0.4*9295000= 2974400J/s[/tex]
or 2.9744 MegaWatt
The total power output of the wind turbine is calculated by first finding the kinetic energy of the wind, and then determining the amount of that energy that is harvested and converted into electricity. The wind turbine extracts 40% of the kinetic energy, and 80% of this energy is converted into electricity, resulting in a power output of 2.974 gigawatts.
Explanation:To calculate the power output of the generator, we'll first need to compute the kinetic energy of the air moving through the turbine. The equation for kinetic energy is given by: KE = 0.5*mass*velocity^2.
So, the kinetic energy of the air per second is KE = 0.5*110,000 kg*(13 m/s)^2 = 9.295 gigajoules per second. Since the wind turbine extracts 40% of this energy, the energy extracted would be 3.718 gigajoules per second.
Of this extracted energy, 80% is converted to electric energy, which corresponds to 2.974 gigajoules per second. Because power is defined as the energy transferred or converted per unit of time, and here the energy is given per second, the power output is therefore 2.974 gigawatts.
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A soccer ball has a mass of 0.40 kg. Initially, it is moving to the left at 20m/s, but then it is kicked and given a velocity of 45 degrees upward and to the right, with a magnitude of 30m/s. Find the impulse of the net force and the average net force, assuming a collision time of .010 seconds.
Answer:
Impulse = 18.53 kg m / s
Net force = 1853N
Explanation:
Mass of the ball, m = 0.4 kg
Initial velocity = 20 m/s horizontally left
Final velocity = 30 m/s at 45 degrees from horizontal
Time of collision = 0.010 s
Check attachment to see full explanation
The impulse and the average net force is 16.484 Ns and 1648.4 N respectively.
To calculate the impulse of the ball, we use the formula below.
Formula:
I = m(v-u).............. Equation 1Where:
I = impulse of the net forcem = mass of the ballu = initial velocity of the ballv = final velocity of the ballFrom the question,
Given:
m = 0.40 kgu = -20 m/s (to the left)v = 30cos45° (to the right) = 21.21 m/s.Substitute these values into equation 1
I = 0.4[21.21-(20)]I = 0.4(21.21+20)I = 0.4(41.21)I = 16.484 Ns.To calculate the average force, we use the formula below.
F = I/t............. Equation 2Where:
F = Average forcet = time.From the question,
Given:
I = 16.484 Nst = 0.01 sSubstitute these values into equation 1
F = 16.484/0.01F = 1648.4 NHence, The impulse and the average net force is 16.484 Ns and 1648.4 N respectively.
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A transmission line that has a resistance per unit length of 2.40 10-4 Ω/m is to be used to transmit 5.00 MW across 400 mi (6.44 105 m). The output voltage of the source is 4.50 kV. (a) What is the line loss if a transformer is used to step up the voltage to 506 kV? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered in attempting to transmit the 5.00 MW at the source voltage of 4.50 kV?
Answer:
(a) 30.174 kW
(b) [tex]6.035\times 10^{- 3}\ N[/tex]
(c) I = 1.112 kA
Solution:
As per the question:
Resistance per unit length, [tex]\sigma = 2.40\times 10^{- 4}\Omega/ m[/tex]
Power transmitted, P = 5.00 MW = [tex]5\times 10^{6}\ W[/tex]
Length, L = [tex]6.44\times 10^5\ m[/tex]
Output voltage, V = 4.50 kV
Stepped-up Voltage, V' = 506 kV
Now,
(a) The line loss can be calculated as:
Current, I = [tex]\frac{P}{V'} = \frac{5\times 10^{6}}{506\times 10^{3}} = 9.88\ A[/tex]
To calculate the power loss:
[tex]P_{L} = I^{2}R[/tex]
where
R = Resistance in the transmission line
R = [tex]2\sigma L = 2\times 2.40\times 10^{- 4}\times 6.44\times 10^5 = 309.12\Omega [/tex]
Thus
[tex]P_{L} = 9.88^{2}\times 309.12 = 30.174\ kW[/tex]
(b) Fraction of the input power lost is given by:
[tex]f = \frac{P_{L}}{P} = \frac{30.174\times 10^{3}}{5\times 10^{6}} = 6.035\times 10^{- 3}\ W[/tex]
(c) To calculate the current at the sending end at ther source voltage of 4.50 kV:
[tex]I = \frac{P}{V} = \frac{5\times 10^{6}}{4.50\times 10^{3}} = 1.112\ kA[/tex]
This value of current is very high and sending such a high watt power is impossible, even if the circuit at the other end is open then also this current can be destructive.
To calculate the line loss, we first need to calculate the value of current flowing through the transmission line. The line loss is approximately 2.08 * 10^5 A and the fraction of input power lost to the line is approximately 4.33 * 10^4 MW. Transmitting the power at the source voltage of 4.50 kV would result in high power losses.
Explanation:To calculate the line loss, we first need to calculate the value of current flowing through the transmission line. Using Ohm's Law, we have:V = IR
where V is the voltage, I is the current, and R is the resistance. Rearranging the equation, we get:I = V/R
Substituting the given values, we have:I = (506,000 V)/((2.40 * 10^-4) Ω/m * 6.44 * 10^5 m)
Simplifying, we find that the current is approximately 1.23 * 10^3 A. To find the line loss, we use the formula:P_loss = I^2 * R * L
where P_loss is the line loss, I is the current, R is the resistance per unit length, and L is the length of the transmission line. Substituting the given values, we find that the line loss is approximately 2.08 * 10^5 A.
The fraction of input power lost to the line can be found using the formula:
Fraction = P_loss / P_input
where P_input is the input power. Substituting the given values, we find that the fraction of input power lost to the line is approximately 4.33 * 10^4 MW.
Lastly, if we were to transmit the 5.00 MW at the source voltage of 4.50 kV, we would encounter difficulties due to the high power losses in the transmission lines. The power losses increase with higher currents, so the voltage is stepped up to reduce the current and subsequently decrease the power losses.
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Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius 2 times that of sphere B. Let QA and QB be the charges on the two spheres, and let EA and EB be the electric-field magnitudes at the surfaces of the two spheres.What is the ratio QB/QA?What is the ratio EB/EA?
Answer:
Explanation:
Given
Radius of A is twice of B i.e.
[tex]R_A=2R_B[/tex]
Also Potential of both sphere is same
[tex]V_A=V_B[/tex]
[tex]V=\frac{kQ}{R}[/tex]
thus
[tex]k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}[/tex]
[tex]\frac{Q_A}{Q_B}=\frac{R_A}{R_B}[/tex]
[tex]\frac{Q_A}{Q_B}=\frac{2}{1}=2[/tex]
[tex]\frac{Q_B}{Q_A}=\frac{1}{2}[/tex]
(b)Ratio of [tex]\frac{E_B}{E_A}[/tex]
Electric Field is given by [tex]E=\frac{kQ}{R^2}[/tex]
thus [tex]E_A=\frac{kQ_A}{R_A^2}[/tex]----1
[tex]E_B=\frac{kQ_B}{R_B^2}[/tex]----2
Divide 2 by 1
[tex]\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}[/tex]
[tex]\frac{E_B}{E_A}=\frac{1}{2}\times 4=2[/tex]
The ratio QB/QA for the charges on the spheres is 2, and the ratio EB/EA for the magnitudes of the electric field at the surfaces of the spheres is 1/2.
Explanation:The question revolves around the principles of electric fields and electric charges on two spherical objects. The spheres in question here are metal and have different sizes, but they hold the same electric potential at their surfaces. This situation is governed by the principles of physics, specifically electrostatics.
To solve this, we begin by noting that the electric potential (V) at the surface of a charged sphere is given by the equation V = KQ/r, where K is Coulomb's constant, Q is the charge on the sphere, and r is the radius of the sphere. Since the potential is the same for both spheres, we get that QA/2R = QB/R. Simplifying, we get the ratio QB/QA = 2.
Next, we calculate the electric field. The electric field (E) at the surface of a charged sphere is given by E = KQ/r^2. Therefore, EA = K×QA/(2R)^2 and EB = K×QB/R^2. Inserting the earlier found ratio of QB/QA = 2, the ratio EB/EA simplifies to 1/2.
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Suppose that instead of a long straight wire, a shortstraight wire was used. The distance from the wire to thepoint that the magnetic field is measured would then be comparableto the length of the short wire. How would you expect themeasured value of the magnetic field to change if the field weresampled at the same distance from each of the two wires?
Answer:
Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.
Explanation:
The magnetic field, B of long straight wire can be obtained by applying ampere's law
[tex]B= \frac{\mu_0 I}{2\pi r}[/tex]
I is here current, and r's the distance from the wire to the field of measurement.
The magnetic field is obviously directly proportional to the current wire. From this expression.
As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.
which of the following tends to increase in a system?
A. temperature
B. Heat
C. Entropy
D. Energy
Answer: entropy
Explanation: entropy is the degree of disorderliness or randomness in a system
A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where t is in seconds and I is in amperes. The capacitor initially has no charge.
(a) Determine the voltage across the inductor as a function of time. mV
(b) Determine the voltage across the capacitor as a function of time. (V/s2) t2
(c) Determine the time when the energy stored in the capacitor first exceeds that in the inductor.
Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
The speed of sound in air is around 345 m/s. A tuning fork vibrates at 850 Hz above the open end of the sound resonance tube. What is the wavelength (in cm) of the sound waves in the tube? Never include units with a numerical answer.
For the development of this problem it is necessary to apply the concepts related to the wavelength depending on the frequency and speed of light.
By definition we know that frequency can be expressed as
[tex]f = \frac{v}{\lambda}[/tex]
Where,
v = Velocity
[tex]\lambda =[/tex] Wavelength
Our values are
[tex]v=345m/sec[/tex]
[tex]f=850Hz[/tex]
Re-arrange to find Wavelength
[tex]f = \frac{v}{\lambda}[/tex]
[tex]\lambda = \frac{v}{f}[/tex]
[tex]\lambda = \frac{345}{850}[/tex]
[tex]\lambda = 0.4058m[/tex]
Converting to centimeters,
[tex]\lambda = 0.4058m(\frac{100cm}{1m})[/tex]
[tex]\lambda = 40.58cm[/tex]
Therefore the wavelength of the sound waves in the tube is 40.58cm
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?
Answer:[tex]\alpha =10.66 rad/s^2[/tex]
Explanation:
Given
mass of disk [tex]m=5 kg[/tex]
diameter of disc [tex]d=30 cm[/tex]
Force applied [tex]F=4 N[/tex]
Now this force will Produce a torque of magnitude
[tex]T=F\cdot r[/tex]
[tex]T=4\dot 0.15[/tex]
[tex]T=0.6 N-m[/tex]
And Torque is given Product of moment of inertia and angular acceleration [tex](\alpha )[/tex]
[tex]T=I\cdot \alpha [/tex]
Moment of inertia for Disc [tex]I= \frac{Mr^2}{2}[/tex]
[tex]I=0.05625 kg-m^2[/tex]
[tex]0.6=0.05625\cdot \alpha [/tex]
[tex]\alpha =10.66 rad/s^2[/tex]
A student needs to prepare 100.0 mL of vitamin C solution according to the directions in Part 1 of the experiment. S/he weighs out 0.0499 g vitamin C (molar mass = 176.124 g/mol) and dissolves it in water to give 100.0 mL solution. S/he pipets 10.00 mL of the solution into a beaker and adds 20 mL of water and 10 mL of buffer. (a) How many mg of vitamin C are in the beaker?
Answer:
4.99 mg of vitamin C are in the beaker.
Explanation:
Given that,
Weight of vitamin = 0.0499 g
Molar mass = 176.124 g/mol
Weight of water = 100.0 ml
We need to calculate the mg of vitamin C in the beaker
We dissolve 0.0499 g vitamin C in water to from 100.0 ml solution.
100 ml solution contain 49.9 mg vitamin C
Now, we take 10 ml of this vitamin C solution in breaker
Since, 100 ml solution =49.9 mg vitamin C
Therefore,
[tex]10\ ml\ solution =\dfrac{10\times49.9}{100}[/tex]
[tex]10\ ml\ solution =4.99\ mg[/tex]
Hence, 4.99 mg of vitamin C are in the beaker.
A small mass m on a string is rotating without friction in a circle. The string is shortened by pulling it through the axis of rotation without any external torque. What happens to the angular velocity of the object?
a. It decreases.
b. It increases.
c. It remains the same.
A small mass m on a string is rotating without friction in a circle. When the string is shortened by pulling it through the axis of rotation without any external torque, angular velocity of the object increases.
What is angular momentum?The characteristic of any rotating object determined by the product of the moment of inertia and the angular velocity is known as angular momentum.
It is a characteristic of rotating bodies determined by the product of their moment of inertia and angular velocity. Since it is a vector quantity, the direction must also be taken into account in addition to the magnitude.
As no external torque applied; total angular momentum of the object remains conserved.
Angular momentum of an object of mass m moving with angular velocity ω in a circular path of radius r is given by: L = mω²r.
As the string is shortened by pulling it through the axis of rotation without any external torque, the angular momentum of the object remains same and its angular velocity increases.
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A wheel is spinning about its axle such that the wheel rotates by 2π radians (1 rev) every 0.43 s. While the wheel spins, the axle is tilted through an angle of π/2 radians (90°) in a time of 0.30 s. The magnitude of the angular momentum changes by 50.00 kg m2/s when the axle is tilted. What is the magnitude of the torque applied to the wheel in order to tilt the axle by 90°?
Answer:
τ = 166.67 N m
Explanation:
The torque is related to the angular momentum
τ = dL / dt
Bold indicates vectors, where tau is the torque and L is the angular momentum
In this case we put the variations, It's like looking for middle values
τ = ΔL / Δt
Let's calculate
τ = 50.00 / 0.30
τ = 166.67 N m
Applied in the direction of rotation at the end of the wheel
If a metal has a specific heat, then the metal could make an excellent material for cookware because
A. it can absorb much more heat before the temperature increases.
B. it maintains a constant temperature.
C. it takes less heat to raise the temperature of the metal.
D. it will not burn food.
Answer:
C) Because it takes less heat to raise the temperature of the metal
Explanation:
A) No. Metals have low specific heats compared to other solids thus they would absorb less heat in order to increase the temperature
B) No. Since they have lower specific heats they are more vulnerable to temperature changes due to variations in the heat rate
C) Yes. It takes less heat to raise the temperature of water due to their lower specific heats
D) No. Because the temperature can be chosen regulating the heat rate to the cookware. Also food burning is related with thermal conduction of cookware and from the cookware to food.
Answer:
C is correct
Explanation:
The magnitude of a component of a vector must be:
a. equal to the magnitude of the vector.
b. greater than or equal to the magnitude of the vector.
c. less than, equal to, or greater than the magnitude of the vector.
d. less than or equal to the magnitude of the vector.
Answer:
(d) Less than or equal to the magnitude of vector
Explanation:
The magnitude of any vector is vector sum of magnitude of its component
As magnitude of vector is sum of its component so magnitude of its component never be greater than the magnitude of vector
It can be equal to the magnitude of vector in one case when the magnitude of other component of the vector is zero
So it can be less or equal to the magnitude of vector
The magnitude of a component of a vector must be less than or equal to the magnitude of the vector.
Explanation:The magnitude of a component of a vector must be less than or equal to the magnitude of the vector.
When a vector is resolved into its components, the magnitude of each component represents the projection of the vector onto that specific axis. The maximum magnitude of any component can be equal to the magnitude of the vector itself, but it cannot exceed it.
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A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At what radius inside and outside the capacitor gap is the magnitude of the induced magnetic field equal to half of its maximum value? (Use any variable or symbol stated above as necessary.)
To solve this problem it is necessary to apply the concepts related to the magnetic field.
According to the information, the magnetic field INSIDE the plates is,
[tex]B=\frac{1}{2} \mu \epsilon_0 r[/tex]
Where,
[tex]\mu =[/tex]Permeability constant
[tex]\epsilon_0 =[/tex]Electromotive force
r = Radius
From this deduction we can verify that the distance is proportional to the field
[tex]B \propto r[/tex]
Then the distance relationship would be given by
[tex]\frac{r}{R} = \frac{B}{B_{max}}[/tex]
[tex]r =\frac{B}{B_{max}} R[/tex]
[tex]r = \frac{0.5B_{max}}{B_{max}}R[/tex]
[tex]r = 0.5R[/tex]
On the outside, however, it is defined by
[tex]B = \frac{\mu_0 i_d}{2\pi r}[/tex]
Here the magnetic field is inversely proportional to the distance, that is
[tex]B \not\propto r[/tex]
Then,
[tex]\frac{r}{R} = \frac{B_{max}{B}}[/tex]
[tex]r = \frac{B_{max}{B}}R[/tex]
[tex]r = \frac{B_{max}{0.5B_{max}}}R[/tex]
[tex]r = 2R[/tex]
(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are 3.90x10—25 kg and 3.95x10—25 kg , respectively, and they travel at 3.00x105 m/s in a 0.250-T field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238.
Answer:
(a) 2.5 cm
(b) Yes
Solution:
As per the question:
Mass of Uranium-235 ion, m = [tex]3.95\times 10^{- 25}\ kg[/tex]
Mass of Uranium- 238, m' = [tex]3.90\times 10^{- 25}\ kg[/tex]
Velocity, v = [tex]3.00\times 10^{5}\ m/s[/tex]
Magnetic field, B = 0.250 T
q = 3e
Now,
To calculate the path separation while traversing a semi-circle:
[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]
The radius of the ion in a magnetic field is given by:
R = [tex]\frac{mv}{qB}[/tex]
[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]
[tex]\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})[/tex]
[tex]\Delta x = 2(\frac{m - m'}{qB}v)[/tex]
Now,
By putting suitable values in the above eqn:
[tex]\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm[/tex]
[tex]\Delta x = 1.25\ cm[/tex]
(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.
In a mass spectrometer, uranium-235 and uranium-238 ions can be separated based on their masses and velocities. The separation distance between their paths can be determined using the equation: d = mv/(qB). The distance between their paths is practical for the separation of uranium-235 from uranium-238.
Explanation:In a mass spectrometer, triply charged uranium-235 and uranium-238 ions are separated based on their masses and velocities.
The separation of their paths when they hit a target after traversing a semicircle can be calculated using the equation:
d = mv/(qB)
where d is the separation, m is the mass of the ion, v is the velocity, q is the charge, and B is the magnetic field.
The distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238, as even a small separation can result in significant enrichment over multiple passes through the mass spectrometer.
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