Answer:
2.07 m/s
Explanation:
m = 1.07 x 10^5 kg, u1 = 3.41 m/s, u2 = 1.4 m/s
Let the speed of three coupled car after collision is v
Use conservation of momentum
m x u1 + 2 m x u2 = 3 m x v
u1 + 2 u2 = 3 v
3.41 + 2 x 1.4 = 3 v
v = 2.07 m/s
A 920-kg car with brakes applied travels at constant speed a distance of 8 m downhill along a road that makes an angle of 30o below the horizontal. The work W done by the force of gravity during this motion is:
Answer:
W = 36064 J
Explanation:
Work done by the force of gravity is given as
[tex]W_g = F_g \times d[/tex]
here we know that
[tex]F_g = mg[/tex]
also we have
[tex]d[/tex] = vertical displacement of the car in downward direction
now we have
[tex]d = 8 sin30[/tex]
[tex]d = 4 m[/tex]
now work done is given as
[tex]W = (mg) d[/tex]
[tex]W = (920 \times 9.81)(4)[/tex]
[tex]W = 36064 J[/tex]
A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass is 120 kg. The players stick together and are at rest after the collision. Find the speed of the second player, assuming the speed of light is 3.0 m/s.
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
[tex]v_2=-\dfrac{m_1v_1}{m_2}[/tex]
[tex]v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}[/tex]
[tex]v_2=-1.47\ m/s[/tex]
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
A ball is dropped 5 meters from rest. Using conservation of energy, what is the final velocity of the ball? a.) 5.39 m/s b.) 4.45 m/s c.) 9.90 m/s d.) 1.23 m/s e.) None of the above
Answer:
The final velocity of the ball, v = 9.9 m/s
Explanation:
It is given that,
A ball is dropped 5 meters from rest, h = 5 meters
We need to find the final velocity of the ball. It can be calculated using the conservation of energy as :
[tex]KE_i+KE_f=PE_i+PE_f[/tex]
Initial kinetic energy, [tex]KE_i=0\ (rest)[/tex]
Final kinetic energy, [tex]KE_f=\dfrac{1}{2}mv^2[/tex]
Initial potential energy, [tex]PE_i=mgh[/tex]
Final potential energy, [tex]PE_f=0[/tex] (at ground, h = 0)
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\ m/s^2\times 5\ m}[/tex]
v = 9.89 m/s
or
v = 9.9 m/s
So, the final velocity of the ball is 9.9 m/s. Hence, this is the required solution.
You hang a 3 kg Halloween decoration 21 cm from the right end of a curtain rod. How much force does one of the curtain rod supports need to be able to hold? The supports are at the very ends of the rod, which has mass 2.1 kg and is 114 cm long. Answer in newtons.
Answer:
34.3 N and 15.7 N
Explanation:
[tex]F_{left}[/tex] = force on the left end of the rod
[tex]F_{right} [/tex] = force on the right end of the rod
M = mass of Halloween decoration = 3 kg
[tex]F_{h}[/tex] = weight of the Halloween decoration = Mg = 3 x 9.8 = 29.4 N
m = mass of rod = 2.1 kg
[tex]F_{r}[/tex] = weight of the rod = mg = 2.1 x 9.8 = 20.6 N
From the force diagram, using equilibrium of torque about A
[tex]F_{h}[/tex] (AB) + [tex]F_{r}[/tex] (AC) = [tex]F_{right}[/tex] (AD)
(29.4) (21) + (20.6) (57) = [tex]F_{right}[/tex] (114)
[tex]F_{right}[/tex] = 15.7 N
Using equilibrium of force along the vertical direction
[tex]F_{right}[/tex] + [tex]F_{leftt}[/tex] = [tex]F_{h}[/tex] + [tex]F_{r}[/tex]
15.7 + [tex]F_{leftt}[/tex] = 29.4 + 20.6
[tex]F_{leftt}[/tex] = 34.3 N
To find the force a curtain rod support must hold, calculate the torques due to the rod's weight and the decoration's weight, considering their distances from the support. Then, balance these torques with the force at the other support. The calculated force for one support is approximately 15.7 newtons.
To calculate the force one of the curtain rod supports must be able to hold, we must consider the torques about one end of the rod and include the mass of the rod and the Halloween decoration. The rod and the decoration together exert a torque about the pivot point due to gravity. The support must exert an equal and opposite torque to keep the system in equilibrium. We do this by summing the torques and setting them to zero, since the system is not rotating.
The total torque ( au) about the right support due to the rod and the decoration can be calculated using au = r imes F where r is the distance from the pivot and F is the force due to gravity (weight). The weight of the rod acts at its center of mass, which is at the midpoint of the rod, while the weight of the decoration acts at 21 cm from the right end. So, the distances from the pivot (right end) to the centers of mass are:
Rod: 57 cm (half of the total length of 114 cm)
Decoration: 21 cm
The weights are:
Rod: 2.1 kg imes 9.8 m/s2 = 20.58 N
Decoration: 3 kg imes 9.8 m/s2 = 29.4 N
The torques about the right support are:
Rod: 57 cm imes 20.58 N = 1173.06 N ext{cm}
Decoration: 21 cm imes 29.4 N = 617.4 N ext{cm}
Since the rod is in equilibrium, the torques must balance, meaning the support at the left end must provide an upwards force resulting in a torque that equals the sum of the other two torques. Let F be the force at the left end support. It must satisfy:
F imes 114 cm = 1173.06 N ext{cm} + 617.4 N ext{cm}
F = (1173.06 + 617.4) / 114
F ≈ 15.7 N
Therefore, the force one of the curtain rod supports needs to be able to hold is approximately 15.7 newtons.
A bathroom scale is compressed 0.21 in. when a 210-lb person stands on it. Assuming that the scale behaves like a spring that obeys Hooke's law, how much does someone who compresses the scale 0.06 in. weigh? How much work is done compressing the scale 0.06 in.?
Bathroom scales measure weight by compressing springs in proportion to the applied force. Using Hooke's law, the spring constant can be determined to calculate the weight. If the scale is compressed by 0.21 inches, a 210-lb person is standing on it. When compressed by 0.06 inches, the weight is calculated to be 60 lb. The work done in compressing the scale by 0.06 inches is 3.6 lb-in or 0.3 ft-lb.
Explanation:Bathroom scales measure weight. When you stand on a bathroom scale, it compresses slightly due to the weight placed on it. The scale contains springs that compress in proportion to the weight, similar to how rubber bands expand when pulled. Using Hooke's law, we can calculate the weight of someone based on how much the scale is compressed.
If a 210-lb person compresses the scale by 0.21 inches, and assuming the scale behaves like a spring, we can use Hooke's law (F = kx) to find the spring constant. The spring constant is the amount of force required to compress the spring by a certain distance. Rearranging the equation for Hooke's law, we get k = F/x. Plugging in the values, k = 210 lb / 0.21 in = 1000 lb/in.
To find the weight of someone who compresses the scale by 0.06 inches, we can again use Hooke's law. Rearranging the equation to solve for F, we have F = kx. Plugging in the values, F = 1000 lb/in * 0.06 in = 60 lb.
To calculate the work done in compressing the scale by 0.06 inches, we can use the formula for work: W = Fd, where W is work, F is force, and d is distance. Plugging in the values, W = 60 lb * 0.06 in = 3.6 lb-in or 0.3 ft-lb.
A football player pushes a sled on a level surface with a weight force of 760 N that acts perpendicular to the surface. The coefficient of kinetic friction between the sled and grass is 0.78 How much work does the football player perform if he pushes the sled and displaces it 10 m. Assume ) the average force performed is equal to the force required to keep the sled in motion and 2) the applied force is paraîlel to the sleds displacement. (3 points) a. b. How much power did the football player produce if he performed the sled push in 3.21 seconds? (2 point)
Answer:
Part a)
W = 5928 J
Part b)
P = 592.8 Watt
Explanation:
As we know that force of friction on the sled is given by
[tex]F_f = \mu F_n[/tex]
here we know that
[tex]\mu = 0.78[/tex]
also we know that normal force on the sled is counter balanced by the weight of the object
[tex]F_n = mg = 760 N[/tex]
now we have
[tex]F_f = (0.78)(760) = 592.8 N[/tex]
Now the applied force must be equal to this friction force so that it will start sliding on the grass
now if we push it by 10 m distance then work done to slide it given by
[tex]W = F.d[/tex]
[tex]W = (592.8)(10) = 5928 J[/tex]
Part B)
Power required to push the sled in 10 s
[tex]Power = \frac{work}{time}[/tex]
[tex]Power = \frac{5928}{10} = 592.8 Watt[/tex]
A marble rolls off a tabletop 1.0 m high and hits the floor at a point 1.2 m away from the table’s edge in the horizontal direction. What is the speed of the marble when it leaves the table’s edge?
A. 1.2 m/s
B. 9.8 m/s
C. 2.7 m/s
D. 3.7 m/s
Answer:
C. 2.7 m/s
Explanation:
Consider the motion of the marble along the vertical direction
y = vertical displacement = height of tabletop = 1.0 m
a = acceleration = 9.8 m/s²
t = time taken to hit the floor = ?
v₀ = initial speed of the marble along vertical direction = 0 m/s
Using the equation
y = v₀ t + (0.5) a t²
1.0 = (0) t + (0.5) (9.8) t²
t = 0.45 s
Consider the motion of the marble along the horizontal direction :
v = speed with which the marble leaves the table
x = horizontal displacement = 1.2 m
t = 0.45 s
horizontal displacement is given as
x = v t
1.2 = v (0.45)
v = 2.7 m/s
An aquarium 3 m long, 2 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg/m3, and use 9.8 m/s2 for g.)
To find the work needed to pump half of the water out of the aquarium, you need to calculate the volume of the aquarium and the mass of the water. The work done is equal to the force (weight of the water) multiplied by the distance (height of the aquarium). Therefore, the work needed is 29400 Nm or J (Joules).
Explanation:To find the work needed to pump half of the water out of the aquarium, we first need to calculate the volume of the aquarium. The volume is given by length x width x height, which equals 3m x 2m x 1m = 6m³. Half of the water in the aquarium would be 6m³/2 = 3m³.
The density of water is given as 1000 kg/m³. So the mass of the water that needs to be pumped out is 3m³ x 1000 kg/m³ = 3000 kg.
The work done to pump the water out is given by the formula work = force x distance. In this case, the force required is the weight of the water, which is mass x gravity. The distance is the height of the aquarium, which is 1m.
Therefore, the work needed to pump half of the water out of the aquarium is 3000 kg x 9.8 m/s² x 1m = 29400 Nm or J (Joules).
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A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.
Answer : The balanced two-half reactions will be,
Oxidation half reaction (anode) : [tex]Mg(s)\rightarrow Mg^{2+}(aq)+2e^-[/tex]
Reduction half reaction (cathode) : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]
Thus the overall reaction will be,
[tex]Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)[/tex]
Explanation :
Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.
The given redox reaction occurs between the magnesium and cadmium.
In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
The balanced two-half reactions will be,
Oxidation half reaction (anode) : [tex]Mg(s)\rightarrow Mg^{2+}(aq)+2e^-[/tex]
Reduction half reaction (cathode) : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]
Thus the overall reaction will be,
[tex]Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)[/tex]
half-reactions
cathode : Cd²⁺ (aq) + 2e⁻ ---> Cd (s)
anode : Mg (s) → Mg²⁺ (aq) + 2e−
a balanced cell reaction
Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)
Further explanationCell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.
Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)
[tex] \large {\boxed {\bold {E ^ osel = E ^ ocatode -E ^ oanode}}} [/tex]
or:
E ° cell = E ° reduction-E ° oxidation
(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)
The value of E cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode
In reaction:
Cd²⁺ + Mg → Cd + Mg²⁺
half-reactions
at the cathode (reduction reaction) Cd²⁺ (aq) + 2e⁻ ---> Cd (s) at the anode (oxidation reaction) Mg (s) → Mg²⁺ (aq) + 2e−a balanced cell reaction
Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)
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An atomic nucleus with a rest mass energy of 7134 MeV is accelerated to 99.217% the speed of light relative to the lab frame. What is the Kinetic Energy in MeV of the atomic nucleus in the lab frame? Assume 4 significant figures.
Answer:
4.999 × 10⁴ MeV
Explanation:
Kinetic energy is related to the rest energy by the Lorentz transformation equation as shown below.
Kinetic energy in the lab frame K= ( γ -1 ) Eo
The Lorentz constant γ = [tex]1/\sqrt{1-v^2/c^2}[/tex]
So γ = [tex]1/\sqrt{1- (0.99217 c)^2/c^2}[/tex]= 8.0067
⇒ K = (8.0067-1)(7134) = 49986 MeV
=49,999 MeV ( 4 significant figures)
Water enters a 180° elbow with an average velocity V1 = 3 m/s. The diameter at the inlet of the elbow is 6.0 cm and the exit diameter is 2.0 cm. The water exits to atmosphere. Determine the magnitude of the horizontal force required to hold the elbow in place. (Ignore the pressure component of the force.)
Answer:203.60 N
Explanation:
Given
[tex]V_{inlet}[/tex]=3 m/s
[tex]d_{inlet}[/tex]=6 cm
[tex]d_{outlet}[/tex]=2 cm
Using continuity equation
[tex]A_{inlet}V_{inlet}=A_{outlet}V_{outlet}[/tex]
[tex]\frac{\pi \times d_{inlet}^2}{4}\times 3[/tex]=[tex]\frac{\pi \times d_{outlet}^2}{4}\times V_{outlet}[/tex]
[tex]V_{outlet}[/tex]=27 m/s
Horizontal Force required to hold elbow=Change in momentum of water
initial momentum=[tex]\rho \times A_{inlet}\times V_{inlet}=10^{3}\times \frac{\pi}{4}6^2\times 3^{2}\times 10^{-4}[/tex]
Final Momentum=[tex]\rho \times A_{outlet}\times V_{outlet}=10^{3}\times \frac{\pi}{4}2^2\times 27^{2}\times 10^{-4}[/tex]
Force required=[tex]10^{3}\times \frac{\pi }{4}\times 10^{-4}\left [ 27^{2}\times 2{2}-6{2}\times 3{2}\right ][/tex]=203.60 N
A battery and a resistor are wired into a circuit. The resistor dissipates 0.50 W. Now two batteries, each identical to the original one, are connected in series with the resistor. Part A What power does it dissipate?
When a battery is connected in a series to a resistor, its power quadruples. So, the power that a resistor dissipating 0.5 W originally would dissipate when connected to two identical batteries in series would be 2.00 W.
Explanation:In the world of physics, power dissipated by a resistor in a circuit is governed by the equation P = I2R, where P is the power, I is the current, and R is the resistance. If the resistance stays unchanged and you double the voltage (by adding identical battery in series), the current through the circuit doubles as well. Thus, with the power quadrupling as a result of two times the current squared (since 22 = 4), the resistor would now dissipate 2.00 W of power.
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When two identical batteries are connected in series with a resistor originally dissipating 0.50 W, the power dissipation increases to 2.0 W. This is because the total voltage supplied to the circuit is doubled, leading to a quadratic increase in power dissipation.
To find the new power dissipation, we start by applying Ohm's law and the power formula. Initially, the power dissipated is given by P = IV. Since power can also be expressed as P = V²/R and the resistor dissipates 0.50 W:
P = V²/R = 0.50 W
Now, connecting two identical batteries in series doubles the voltage:
[tex]V_{new[/tex] = 2V
The new power dissipation is then:
[tex]P_{new[/tex] = ([tex]V_{new[/tex])² / R
[tex]P_{new[/tex] = (2V)² / R
[tex]P_{new[/tex] = 4V² / R
[tex]P_{new[/tex] = 4 * 0.50 W
[tex]P_{new[/tex] = 2.0 W
Therefore, the resistor now dissipates 2.0 W of power.
An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.
Answer:
67.5 cm
Explanation:
u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm
let f be the focal length
Use lens equation
1 / f = 1 / v - 1 / u
1 / f = 1 / 270 + 1 / 90
1 / f = 4 / 270
f = 67.5 cm
To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.
Explanation:To determine the focal length of the lens, we can use the lens formula:
1/f = 1/v - 1/u
Where f is the focal length, v is the image distance, and u is the object distance.
Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:
1/f = 1/(3v) - 1/(90)
Multiplying through by 90*3v, we get:
90*3v/f = 270v - 90*3v
90*3v/f = 270v - 270v
90*3v/f = 0
Simplifying further, we find that: v = 0
When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.
Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478 N from a point 7.50 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
[tex]\Rightarrow \frac{122}{48.72} = a[/tex]
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
If the radius of the earth was suddenly tripled and its mass doubled, the surface gravitational acceleration would become: A) 9.8 m/s^2
B) 2.18 m/s^2
C) 7.35 m/s^2
D) 14.7 m/s^2
E) 13.1 m/s^2
Answer:
B) 2.18 m/s²
Explanation:
M = Original mass of earth
R = Original radius of earth
g = original acceleration due to gravity of earth = 9.8 m/s²
M' = New mass of earth = 2 M
R' = New radius of earth = 3 R
g = original acceleration due to gravity of earth = 9.8 m/s²
Original acceleration due to gravity of earth is given as
[tex]g=\frac{GM}{R^{2}}[/tex]
[tex]9.8 =\frac{GM}{R^{2}}[/tex] eq-1
g' = new acceleration due to gravity of earth
New acceleration due to gravity of earth is given as
[tex]g' =\frac{GM'}{R'^{2}}[/tex]
[tex]g' =\frac{G(2M)}{(3R)^{2}}[/tex]
[tex]g'=\left ( \frac{2}{9} \right )\frac{GM}{R^{2}}[/tex]
Using eq-1
[tex]g'=\left ( \frac{2}{9} \right )(9.8)[/tex]
g' = 2.18 m/s²
Final answer:
When the Earth's radius is tripled and its mass is doubled, the new surface gravitational acceleration would be calculated using the gravitational acceleration formula and would result in a new acceleration of B) 2.18 m/s².
Explanation:
To find the new surface gravitational acceleration (g') when the radius of the Earth is tripled and its mass is doubled, we use the formula for gravitational acceleration:
g' = (G × new mass) / (new radius)²
Where:
G is the gravitational constantThe new mass is twice Earth's mass (2M)The new radius is three times Earth's radius (3R)Substituting the new mass and radius into the formula, we get:
g' = (G × 2M) / (3R)²
g' = (2G × M) / (9R²)
g' = (2/9) × (G × M / R²)
Since G × M / R² is the original gravitational acceleration (g) of Earth, which is 9.8 m/s², we can now substitute:
g' = (2/9) × 9.8 m/s²
g' = 2.18 m/s²
So the new surface gravitational acceleration would be 2.18 m/s², which matches option B.
A specimen of copper having a rectangular cross section 15.2 mm X19.1 mm (0.6 in. X 0.75 in.) is pulled in tension with 44500 N(10000 lbf) force, producing only elastic deformation. Calculate the resulting strain.
Answer:
The elastic deformation is 0.00131.
Explanation:
Given that,
Force F = 44500 N
Cross section [tex]A =15.2mm\times19.1\ mm[/tex]
We Calculate the stress
Using formula of stress
[tex]\sigma=\dfrac{F}{A}[/tex]
Where, F = force
A = area of cross section
Put the value into the formula
[tex]\sigma=\dfrac{44500}{15.2\times10^{-3}\times19.1\times10^{-3}}[/tex]
[tex]\sigma=153.27\times10^{6}\ N/m^2[/tex]
We need to calculate the strain
Using formula of strain
[tex]Y=\dfrac{\sigma}{\epsilon}[/tex]
[tex]epsilon=\dfrac{\sigma}{Y}[/tex]
Where,
[tex]\sigma[/tex]=stress
Y = young modulus of copper
Put the value into the formula
[tex]\epsilon=\dfrac{153.27\times10^{6}}{117\times10^{9}}[/tex]
[tex]\epsilon =0.00131[/tex]
Hence, The elastic deformation is 0.00131.
A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves. A wide parallel beam of microwaves is incident normally on the sheet. What is the smallest microwave frequency for which only the central maximum occurs? (c = 3.00 × 108 m/s)
Answer:
The smallest microwave frequency is 1.5 GHz.
Explanation:
Given that,
Distance d= 20 cm
We need to calculate the wavelength
Using formula for maximum diffraction
[tex]d\sin\theta=\lambda[/tex]
For maximum, [tex]\sin\thea=1[/tex]
[tex]d=\lambda[/tex]
[tex]\lambda=20 cm=0.2 m[/tex]
We need to calculate the frequency
[tex]f =\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times10^{8}}{0.2}[/tex]
[tex]f=1.5\times10^{9}\ Hz[/tex]
[tex]f=1.5 GHz[/tex]
Hence, The smallest microwave frequency is 1.5 GHz.
A fan blade is rotating with a constant angular acceleration of +12.8 rad/s2. At what point on the blade, as measured from the axis of rotation, does the magnitude of the tangential acceleration equal that of the acceleration due to gravity? (g = 9.80 m/s2)
Answer:
At r= 0.76 m measured from the axis of rotation, does the tangential acceleration equals the acceleration of the gravity.
Explanation:
at=g= 9.8 m/s²
α= 12.8 rad/s²
r= at/α
r= 0.76m
Pendulum A has a bob of mass m hung from the string of length L; pendulum B is identical to A except its bob has the length 2L. Compare the frequencies of small oscillations of the two pendulums.
Answer:
[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]
Explanation:
For pendulum A: Length = L and gravity = g
The frequency of pendulum A is given by
[tex]f = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex]
Here, f is the frequency, L be the length
[tex]f_{A} = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex] ... (1)
For pendulum B: Length = 2L, gravity = g
The frequency of pendulum B is given by
[tex]f_{B} = \frac{1}{2\pi }\sqrt{\frac{g}{2L}}[/tex] .... (2)
Divide equation (1) by (2)
[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]
At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 µT northward, and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming the electron has an instantaneous velocity of 5.90 ✕ 106 m/s directed to the east in this environment.
Answer:
[tex]F_g = 8.9 \times 10^{-30} N[/tex]
[tex]F_e = 1.6 \times 10^{-17} N[/tex]
[tex]F_m = 4.72 \times 10^{-17} N[/tex]
Explanation:
For gravitational force we know that
F = mg
now we have
[tex]m = 9.1 \times 10^{-31} kg[/tex]
[tex]F_g = (9.1 \times 10^{-31})(9.8)[/tex]
[tex]F_g = 8.9 \times 10^{-30} N[/tex]
Now electrostatic force
[tex]F = qE[/tex]
here we have
[tex]F = (1.6 \times 10^{-19})(100)[/tex]
[tex]F = 1.6 \times 10^{-17} N[/tex]
Now magnetic force on it is given by
[tex]F_{m} = qvB[/tex]
[tex]F_m = (1.6 \times 10^{-19})(5.90 \times 10^6)(50 \times 10^{-6})[/tex]
[tex]F_m = 4.72 \times 10^{-17} N[/tex]
The gravitational force on the electron is 8.90 x 10^-31 N downward, the electric force on the electron is 1.60 x 10^-17 N upward, and the magnetic force on the electron is 5.90 x 10^-25 N northward.
Explanation:The gravitational force on an electron near the surface of the Earth can be calculated using the formula Fg = m * g, where m is the mass of the electron and g is the acceleration due to gravity. The electric force on the electron can be calculated using the formula Fe = q * E, where q is the charge of the electron and E is the magnitude of the electric field. The magnetic force on the electron can be calculated using the formula Fm = q * v * B, where v is the velocity of the electron and B is the magnitude of the magnetic field.
Given that the electron has a mass of 9.11 x 10^-31 kg, a charge of -1.60 x 10^-19 C, and a velocity of 5.90 x 10^6 m/s, the gravitational force on the electron is approximately 8.90 x 10^-31 N downward. The electric force on the electron is approximately 1.60 x 10^-17 N upward. The magnetic force on the electron is approximately 5.90 x 10^-25 N northward.
A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil has 100 turns how much current must pass through the coil?
Answer:
The current pass through the coil is 6.25 A
Explanation:
Given that,
Diameter = 25 cm
Magnetic field = 1.0 mT
Number of turns = 100
We need to calculate the current
Using the formula of magnetic field
[tex]B =\dfrac{\mu_{0}NI}{2\pi r}[/tex]
[tex]I=\dfrac{B\times2\pi r}{\mu N}[/tex]
Where, N = number of turns
r = radius
I = current
Put the value into the formula
[tex]I=\dfrac{1.0\times10^{-3}\times2\pi\times12.5\times10^{-2}}{4\pi\times10^{-7}100}[/tex]
[tex]I=6.25\ A[/tex]
Hence, The current passes through the coil is 6.25 A
A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted through a power line of total resistance 1.0 Ω. Find the percentage power lost in the transmission line if the voltage is not stepped up.
Without stepping up the voltage, 50% of the power is lost in the transmission due to the high I^2R losses. Stepping up the voltage using a transformer significantly reduces these losses and makes power transmission more effective.
Explanation:The subject question pertains to the percentage power loss in a given transmission line when the voltage is not stepped up. Intuitively, power loss will be considerably greater when voltage is not stepped up due to a relatively high current flowing through the lines that enormously increases the I2R (current squared times the resistance) losses.
Firstly, calculate the initial power, PInitial using the formula P = IV, where I is current and V is voltage. In this case, PInitial = 60 A * 120 V = 7200 W. If the voltage is not stepped up, this power is transmitted at 120 V.
The power dissipated due to resistance (Ploss) can be calculated using the formula P = I2R, where I is current and R is resistance. So: Ploss = (60 A)2 * 1 Ω = 3600 W. The percentage of power loss is thus (Ploss / PInitial) * 100 = (3600 W / 7200 W) * 100 = 50%.
In contrast, when voltage is stepped up to 4500 V using a transformer, the current decreases to maintain the same power, greatly reducing power losses. This is eminently beneficial over long distances, making power transmission more efficient.
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Final answer:
the percentage power lost in the transmission line if the voltage is not stepped up is 50%
Explanation:
To calculate the percentage power lost in the transmission line if the voltage is not stepped up, we first calculate the power loss in the transmission line using the formula P = I²R, where I is the current and R is the resistance of the transmission line.
We are given that the current I is 60 A and the total resistance R is 1.0 Ω. The power loss in the transmission line would be:
P = (60 A)² × 1.0 Ω = 3600 W or 3.6 kW
The total power transmitted without stepping up the voltage is calculated as P = IV, where V is the voltage. Here, V is 120 V, so the total power transmitted is:
P = 60 A × 120 V = 7200 W or 7.2 kW
To find the percentage power lost, we divide the power lost by the total power transmitted and multiply by 100:
Percentage power lost = (3600 W / 7200 W) × 100% = 50%
Without the step-up transformer, 50% of the power would be lost in the transmission line, which is a significant loss compared to transmission at a higher voltage. This example illustrates the importance of high-voltage transmission in reducing power losses.
Suppose 150W of heat flows by conduction from the blood capillaries beneath the skin to the body's surface area of 1.5 m^2. If the temperature difference is 0.50 ºC, estimate the average distance of capillaries below the skin surface. Use 'mm' as your units.
Answer:
[tex]l =1mm[/tex]
Explanation:
Given:
Rate of heat transfer, P = 150 W
Body surface Area, A = 1.5 m²
Temperature difference, ΔT = 0.50°C
Also,
The rate of heat transfer, P is given as:
[tex]P = \frac{kA\Delta T}{l}[/tex]
Where,
l =length of material (or here it isaverage distance of the capillaries below the skin surface)
k = Thermal conductivity
Here the transfer of heat is through the skin. Thus, k for human tissue is given as 0.2
substituting the values in the above equation, we get
[tex]150 = \frac{0.2\times 1.5\times 0.50}{l}[/tex]
or
[tex]l = \frac{0.2\times 1.5\times 0.50}{150}[/tex]
or
[tex]l = 1\times 10^{-3}m=1mm[/tex]
Using the formula for thermal conduction, it is estimated that the capillaries lie about 3 mm below the skin surface. This is a simplification, actual distances can vary based on specific factors.
Explanation:To estimate the average distance between the capillaries and skin surface, we can use the formula for thermal conduction, which states that heat flow equals the thermal conductivity constant (k) times the surface area of the skin (A) times the temperature difference (ΔT), divided by the thickness of the skin (d), or Q = k*A*ΔT/d.
Assuming that human skin has a thermal conductivity similar to water (k~0.6 W/mK), we can rearrange the formula to solve for d: d = k*A*ΔT/Q. Plugging in the given values, we get d = (0.6 W/mK * 1.5 m^2 * 0.50 ºC) / 150 W, which simplifies to approximately 0.003 m, or 3 mm.
This estimate suggests that, on average, the capillaries lie about 3 mm below the skin surface. However, this is a simplification and actual distances can vary based on factors such as the specific region of the body, individual physiology, and more. Thermal conduction is just one mechanism of heat transfer in the body, along with convection and radiation.
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You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0.66, what is the mass of the heaviest box you will be able to move?
Answer:
72.75 kg
Explanation:
[tex]F[/tex] = force applied on a box = 750 N
[tex]m[/tex] = mass of the box
[tex]N[/tex] = Normal force on the box
[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.66
From the force diagram, force equation along the vertical direction is given as
[tex]N = F Sin25 + mg[/tex]
[tex]N = 750 Sin25 + mg[/tex] eq-1
Static frictional force is given as
[tex]f_{s} = \mu _{s} N[/tex]
using eq-1
[tex]f_{s} = \mu _{s} (750 Sin25 + mg)[/tex]
For the box to move,
[tex]F Cos25 = f_{s}[/tex]
[tex]750 Cos25 = \mu _{s} (750 Sin25 + mg)[/tex]
[tex]750 Cos25 = (0.66) (750 Sin25 + m (9.8))[/tex]
m = 72.75 kg
A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C, and 0.100 kg of copper at 100°C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final temperature of the mixture.
Answer:
23.63 °C
Explanation:
[tex]m_{w}[/tex] = mass of water = 0.250 kg
[tex]T_{wi}[/tex] = initial temperature of water = 20.0 °C
[tex]c_{w}[/tex] = Specific heat of water = 4186 J/(kg °C)
[tex]m_{a}[/tex] = mass of aluminum = 0.400 kg
[tex]T_{ai}[/tex] = initial temperature of aluminum = 26.0 °C
[tex]c_{a}[/tex] = Specific heat of aluminum = 900 J/(kg °C)
[tex]m_{c}[/tex] = mass of copper = 0.100 kg
[tex]T_{ci}[/tex] = initial temperature of copper = 100 °C
[tex]c_{c}[/tex] = Specific heat of copper = 386 J/(kg °C)
[tex]T_{f}[/tex] = Final temperature of mixture
Using conservation of heat
[tex]m_{w}[/tex] [tex]c_{w}[/tex] ([tex]T_{f}[/tex] - [tex]c_{w}[/tex]) + [tex]m_{a}[/tex] [tex]c_{a}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ai}[/tex] ) + [tex]m_{c}[/tex] [tex]c_{c}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ci}[/tex] ) = 0
(0.250) (4186) ([tex]T_{f}[/tex] - 20) + (0.400) (900) ([tex]T_{f}[/tex] - 26) + (0.100) (386) ([tex]T_{f}[/tex] - 100) = 0
[tex]T_{f}[/tex] = 23.63 °C
Three particles are fixed on an x axis. Particle 1 of charge q1 is at x = a, and particle 2 of charge q2 is at x = - a. If their net electrostatic force on particle 3 of charge + Q is to be zero, what must be the ratio q2/q1 when particle 3 is at x = 0.500a?
Answer:
9
Explanation:
As the net force on Q is zero, so the force on Q due to q1 is balanced by the force on Q due to q2.
Let the force on Q due to q1 is F1 and force on Q due to q2 is F2.
F1 = F2
K Q q1 / (0.5 a)^2 = K Q q2 / (1.5 a)^2
q1 / 0.25 = q2 / 2.25
q2 / q1 = 2.25 / 0.25
q2 / q1 = 9
A current density of 7.00 10-13 A/m2 exists in the atmosphere at a location where the electric field is 185 V/m. Calculate the electrical conductivity of the Earth's atmosphere in this region.
Answer:
electrical conductivity is given as
[tex]\sigma = 3.8 \times 10^{-15} [/tex]
Explanation:
For a current carrying conductor we will have relation between current density and electric field inside the conductor as
[tex]j = \sigma E[/tex]
here we have
current density = j
electric field = E
given that
[tex]j = 7.00 \times 10^{-13}[/tex]
[tex]E = 185 V/m[/tex]
now we will have
[tex]\sigma = \frac{j}{E}[/tex]
[tex]\sigma = \frac{7.00 \times 10^{-13}}{185}[/tex]
[tex]\sigma = 3.8 \times 10^{-15} [/tex]
What is the wavelength of an ultrasound signal having a velocity of 1,500 m/sec. at 2.5 MHz?
Answer:
The wavelenght is λ= 6 * 10⁻⁴m.
Explanation:
v= 1500 m/s
f= 2.5 MHz = 2.5 *10⁶ Hz
λ=v/f
λ= 6 * 10⁻⁴ meters
Fresh water flows through a horizontal tapered pipe. At the wide end its speed is 8 m/s. The difference in pressure between the two ends is 5338 Pa. What is the speed (in m/s) of the water at the narrow end? Round your answer to the nearest tenth.
Answer:
8.64 m/s
Explanation:
v1 = 8 m/s, v2 = ?, P2 - P1 = 5338 Pa, density of water, d = 1000 kg/m^3
By the use of Bernoulli's theorem
P 1 + 1/2 x d x v1^2 = P2 + 1/2 x d x v2^2
P2 - P1 = 1/2 x d x (v2^2 - v1^2)
5338 = 0.5 x 1000 x (v2^2 - 64)
10.676 = v2^2 - 64
v2^2 = 74.676
v2 = 8.64 m/s
An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic number of the nucleus. Ignore the gravitational force between the electron and the nucleus. Find an expression in terms of these quantities for the speed of the electron in this orbit. (Use any variable or symbol stated above along with the following as necessary: k for Coulomb's constant.)
Answer:
[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]
Explanation:
The electrostatic attraction between the nucleus and the electron is given by:
[tex]F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}[/tex] (1)
where
k is the Coulomb's constant
Ze is the charge of the nucleus
e is the charge of the electron
r is the distance between the electron and the nucleus
This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:
[tex]F=m\frac{v^2}{r}[/tex] (2)
where
m is the mass of the electron
v is the speed of the electron
Combining the two equations (1) and (2), we find
[tex]k\frac{Ze^2}{r^2}=m\frac{v^2}{r}[/tex]
And solving for v, we find an expression for the speed of the electron:
[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]
The speed of an electron in a circular orbit around a nucleus is determined by equating the Coulomb force and the centripetal force, resulting in the expression v = sqrt(kZe^2/mr).
Explanation:The speed v of an electron in a circular orbit around a nucleus can be found by equating the electrostatic force to the centripetal force required for circular motion. The electrostatic force, due to the Coulomb's interaction, between the electron and the nucleus is given by F = k(Ze)(-e)/r^2, where k is Coulomb's constant, Z is the atomic number, e is the magnitude of the charge of an electron, and r is the radius of the orbit. On the other hand, the centripetal force needed to keep the electron in circular motion is F = mv^2/r where m is the mass of the electron and v is its speed.
Setting the two expressions equal gives the equation for the electron's speed v:
k(Ze)(-e)/r^2 = mv^2/r
Solving for v results in the expression:
v = sqrt(kZe^2/mr)
This equation shows that the electron's speed in its orbit depends on the atomic number Z, Coulomb's constant k, the electron's mass m, and the orbit radius r.