Answer:
a. 3.86 m/s²
b. 332.64 N directed upwards towards the center of the ferris wheel.
c. 764.96 N directed downwards towards the center of the ferris wheel.
d. 589.84 N at 21.5° counterclockwise from the horizontal direction.
Explanation:
a. Since the ferris wheel rotates 4 times per minute, its period, T = 60 s/4 = 15 s.
We now find it angular speed ω = 2π/T = 2π/15 = 0.418 rad/s
We then calculate its centripetal acceleration from a = rω² where r = radius of ferris wheel = 22.0 m.
So, a = 22 m × (0.418 rad/s)² = 3.86 m/s²
b. At the lowest point, the normal force, N and the centripetal force, F both act in opposite directions to the weight, mg of the object. So,
N + F = mg
N = mg - F
N = mg - ma where a is the centripetal acceleration
N = m(g - a)
N = 56 kg(9.8 m/s² - 3.86 m/s²)
N = 56 kg × 5.94 m/s²
N = 332.64 m/s²
The normal force the seat exerts on the child is thus 332.64 N directed upwards towards the center of the ferris wheel.
c. At the highest point, the weight, mg of the object and the centripetal force, F both act in opposite directions to the normal force, N. So,
N = mg + F
N = mg + ma where a is the centripetal acceleration
N = m(g + a)
N = 56 kg(9.8 m/s² + 3.86 m/s²)
N = 56 kg × 13.66 m/s²
N = 764.96 N
The normal force the seat exerts on the child is thus 764.96 N directed downwards towards the center of the ferris wheel.
d. Half way between the top and the bottom of the ferris wheel, the normal force must balance the weight and the centripetal force so the child doesn't fall off. For this to happen, the normal force is thus the resultanf of the centripetal force and the weight of the child. Since these two forces are perpendicular at this instance,
N = √[(mg)² + (ma)²] = m√(g² + a²) = 56 kg√[(9.8 m/s²)² + (3.86 m/s²)²] = 56 kg√[(97.196 (m/s²)² + 14.8996(m/s²)²] = 56kg√110.9396 (m/s²)² = 56 kg × 10.53 m/s² = 589.84 N.
Since the centripetal force acts towards the center of the ferris wheel in the horizontal direction, it is equal to the horizontal component of the normal force, Also, the weight acts downwards and is equal to the vertical component of the normal force.
So, the direction of the normal force is gotten from
tanθ = ma/mg = a/g
θ = tan⁻¹(a/g) = tan⁻¹(3.86 m/s² / -9.8 m/s²) = tan⁻¹(-0.3939) = -21.5°. Since the angle is it shows a counter clockwise direction.
So, the normal force is 589.84 N at 21.5° counterclockwise from the horizontal direction.
(a) The centripetal acceleration of the child is 1.94 m/s².
(b) The force the seat exert on the child at the lowest point of the ride is 657.4 N.
(c) The force the seat exert on the child at the highest point of the ride is 440.2 N.
(d) The force the seat exert on the child when the child is halfway between the top and bottom is 108.64 N.
The given parameters;
mass of the child, m = 56 kgangular speed, ω = 4 rev/mindiameter of the wheel, d = 22 mradius of the wheel, r = 11 mThe centripetal acceleration of the child is calculated as follows;
[tex]a_c = \omega ^2 r\\\\[/tex]
where;
ω is angular speed of the wheel in rad/s
The angular speed of the wheel in rad/s is calculated as follows;
[tex]\omega = 4 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 0.419 \ rad/s[/tex]
The centripetal acceleration of the child is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (0.419)^2 \times 11\\\\a_c = 1.94 \ m/s^2[/tex]
The force the seat exert on the child at the lowest point of the ride is calculated as;
[tex]T_{bottom} = ma_c + mg\\\\T_{bottom} = m(a_c + g)\\\\T_{bottom} = 56(1.94 + 9.8)\\\\T_{bottom} = 657.4 \ N[/tex]
The force the seat exert on the child at the highest point of the ride is calculated as;
[tex]T_{top} = mg- ma_c\\\\T_{top} = m(g - a_c)\\\\T_{top} = 56(9.8 - 1.94)\\\\T_{top} = 440.2 \ N[/tex]
The force the seat exert on the child when the child is halfway between the top and bottom;
[tex]T = ma_c\\\\T = 56(1.94)\\\\T = 108.64 \ N[/tex]
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Imagine that an incident light wave encounters a thin film. Some of that light wave will be reflected off of the top surface of the thin film (call this Wave 1), while the rest will be transmitted through the film. Some of this transmitted light (Wave 2) will reflect off the back surface of the film and move back through the thin film and out into the original medium of the incident light wave. If the thin film has a thickness t, then what is the path length difference between Wave 1 and Wave 2?
a) t/4
b) t/2
c) t
d) 2t
Answer
D, 2t
Explanation:
See attached file
"2t" will be the path length difference between Wave 1 and Wave 2. A complete solution is below.
According to the question,
The path difference between Wave 1 and 2 will be:
When [tex]\Theta = 0^{\circ}[/tex],
→ [tex]P.d= \Delta x[/tex]
[tex]= AB+BC[/tex]
hence,
Path difference,
= [tex]t+t[/tex]
= [tex]2t[/tex]
Thus the above answer i.e., "option d" is appropriate.
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A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 10-11 m2/s, and the diffusion flux is found to be 1.2 10-7 kg/m2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3? Assume a linear concentration profile.
Answer:
The answer is 0.001 m
Explanation:
Solution
Recall that,
A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at = 1200°C
The diffusion coefficient for nitrogen in steel at this temperature is =6 * 10-11 m2/s
The diffusion flux is = 1.2 *10^-7 kg/m2s
Concentration of nitrogen in the steel at the high-pressure surface is= 4 kg/m3.
The high-pressure side will the concentration is estimated to be = 2.0 kg/m3
Now,
The flux = -D dC/dx
1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx
∫ˣ₀ dx = -5x^10-4 ∫²₄ dC
so,
x = (2-4) kg/m3 (-5x10-4 m4/kg)
where x = .001 m
Therefore x = 0.001 m
The distance from this high-pressure side will the concentration is 0.001 m
Calculation of the distance:Since the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, the diffusion flux is found to be 1.2x 10-7 kg/m2 -s.
We know that
The flux =[tex]-D\ dC\div dx[/tex]
So,
1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx
dC/dx = -2000
Now
-2000 = 4 - 2/0-x_B
x_B = 2/2000
x_B = 1*10^-3m
x_B = 0.001m
Hence, The distance from this high-pressure side will the concentration is 0.001 m
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"One possible explanation for a galaxy's type invokes the angular momentum of the protogalactic cloud from which it formed. Suppose a galaxy forms from a protogalactic cloud with a lot of angular momentum. Assuming its type has not changed as a result of other interactions, we'd expect this galaxy to be ______."
Answer:
Spiral galaxy
Explanation:
If a galaxy forms from a protogalactic cloud with a lot of angular momentum, we would expect this galaxy to be a spiral galaxy. This is due to the conservation of angular momentum, where greater angular momentum usually results in a flatter, disk-like shape seen in spiral galaxies. Other factors also affect galaxy formation.
Explanation:In the subject of astrophysics, the angular momentum of the protogalactic cloud has significant implications for the type of galaxy formed. Specifically, if a galaxy forms from a protogalactic cloud with a high angular momentum, we'd expect this galaxy to be a spiral galaxy.
Angular momentum is a property of rotating bodies, determined by the mass, shape, and spin rate of the object. In the context of galaxies, a protogalactic cloud with higher angular momentum is more likely to flatten out into a spinning disk shape, leading to the formation of a spiral galaxy. This is due to the conservation of angular momentum. In contrast, lower angular momentum could lead to the formation of elliptical galaxies, which are less flat and more randomly arranged.
In the realm of galaxy formation and evolution, these are widely accepted ideas, though it's important to note that other factors, like the gas content of the protogalactic cloud and interactions with other galaxies, can also play a major role.
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The sun continuously radiates energy into space in all directions. Some of the sun's energy is intercepted by the Earth. The average temperature of the surface of the Earth remains a little above 300 k. Why doesn't the Earth's temperature rise as it intercepts the sun's energy?
A. The Earth reflects the sun's light.
B. The Earth radiates an amount of energy into space equal to the amount it receives.
C. The energy only raises the temperature of the upper atmosphere and never reaches the surface.
D. The thermal conductivity of the Earth is low.
E. The heat is carried away from the Earth by convection currents.
B. The Earth radiates an amount of energy into space equal to the amount it receives.
Part of the solar energy is reflected by the Earth into space, this is known as albedo. The other part of the energy radiated by the Earth in the form of infrared radiation, is absorbed by the greenhouse gases, which cause most of this infrared radiation to be emitted into space. Therefore, the net flow of energy is zero.
A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating and drops onto the first flywheel and sticks to it. Both flywheels have radius R. What is the final angular velocity in terms of the initial angular velocity ω0? Treat each flywheel as a disk (I = (1/2) MR2).
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
Consider two blocks of copper. Block A contains 800 atoms and initially has a total of 20 quanta of energy. Block B contains 200 atoms and initially has 80 quanta of energy. The two blocks are placed in contact with each other, inside an insulated container (so no thermal energy can be exchanged with the surroundings). After waiting for a long time (for example, an hour), which of the following would you expect to be true?
a. Approximately 50 quanta of energy are in block A, and approximately 50 quanta of energy are in block B.
b. Approximately 80 quanta of energy are in block A, and approximately 20 quanta of energy are in block B.
c. The entropy of block A is equal to the entropy of block B.
d. The temperature of block A and the temperature of block B are equal.
Answer:
Option B and Option D are true
Explanation:
We are given;
Number of atoms in block A = 800
Energy content in block A = 20 quanta
Number of atoms in block B = 200
Energy content in block B = 80 quanta
The energy of a system which is an extensive quantity,depends on the mass or number of moles of the system. However, at equilibrium, the energy density of the two copper blocks will be equal. That is, each atom of Cu in the two blocks will, on average, have the same energy. Because block A has 4 times more atoms than block B, it will have 4 times more quanta of energy. Thus, option B is therefore true while option A is false.
Temperature is a measure of the average kinetic energy of the atoms in a material. Now, if each atom in blocks A and B have the same average energy, then the temperatures of blocks A and B will be equal at equilibrium. Thus, option D is true.
Entropy of a system is an extensive quantity that depends on the the mass or number of atoms in the system. Because block A is bigger than block B, it will have higher entropy. However, that the specific entropy (the entropy per mole or per unit mass) is an intensive quantity -- it is independent of the size of a system. The molar entropy of blocks A and B are equal at equilibrium. Thus option C is false.
For the given condition in the paragraph, the statements (a) and (c) are false. And the statements given in option (b) and (d) are true.
Given data:
Number of atoms in Block A is, 800 atoms.
Initial amount of energy in Block A is, 20 quanta.
Number of atoms in Block B is, 200 atoms.
Initial amount of energy in Block B is, 80 quanta.
We will use the concept of Energy to relate it with the temperature, to solve the given problem. Also, there is some glance about the equilibrium condition as discussed:
The energy of a system is an extensive quantity because it depends on the mass or number of moles of the system. However, at equilibrium, the energy density of the two copper blocks will be equal. That is, each atom of Cu in the two blocks will, on average, have the same energy. Because block A has 4 times more atoms than block B, it will have 4 times more quanta of energy. Thus, option B is true while option A is false.Entropy of a system is an extensive quantity that depends on the the mass or number of atoms in the system. Because block A is bigger than block B, it will have higher entropy. However, that the specific entropy (the entropy per mole or per unit mass) is an intensive quantity because it is independent of the size of a system. The molar entropy of blocks A and B are equal at equilibrium. Thus option C is false.Temperature is a measure of the average kinetic energy of the atoms in a material. If each atom in blocks A and B have the same average energy, then the temperatures of blocks A and B will be equal at equilibrium. Thus, option D is true.
Thus, we conclude that for the given condition, the statements (a) and (c) are false. And the statements given in option (b) and (d) are true.
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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum that has a wavelength of 700 nm and a peak electric field magnitude of 3.5 V/m. (Use the following as necessary: t and x. Assume that E is in volts per meter, B is in teslas, t is in seconds, and x is in meters. Do not include units in your answer. Assume that E = 0 and B = 0 when x = 0 and t = 0.) E(x, t) = B(x, t) =
Answer:
[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex]
[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex]
Explanation:
The electric field equation of a electromagnetic wave is given by:
[tex]E=E_{max}(kx-\omega t)[/tex] (1)
E(max) is the maximun value of E, it means the amplitude of the wave.k is the wave numberω is the angular frequencyWe know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).
By definition:
[tex]k=\frac{2\pi}{\lambda}[/tex]
[tex]k=8.98*10^{6} [rad/m][/tex]
And the relation between λ and f is:
[tex]c=\lambda f[/tex]
[tex]f=\frac{c}{\lambda}[/tex]
[tex]f=\frac{3*10^{8}}{700*10^{-9}}[/tex]
[tex]f=4.28*10^{14}[/tex]
The angular frequency equation is:
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\pi*4.28*10^{14}[/tex]
[tex]\omega=2.69*10^{15} [rad/s][/tex]
Therefore, the E equation, suing (1), will be:
[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex] (2)
For the magnetic field we have the next equation:
[tex]B=B_{max}(kx-\omega t)[/tex] (3)
It is the same as E. Here we just need to find B(max).
We can use this equation:
[tex]E_{max}=cB_{max}[/tex]
[tex]B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}[/tex]
[tex]B_{max}=1.17*10^{-8}T[/tex]
Putting this in (3), finally we will have:
[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex] (4)
I hope it helps you!
Final answer:
To find the equations of the fields for an electromagnetic wave in the red spectrum with given parameters, calculate the frequency using the speed of light and wavelength, then apply it to the sinusoidal equations representing the electric and magnetic fields.
Explanation:
To determine the equations for the electric and magnetic fields (E and B, respectively) of an electromagnetic wave in the visible red spectrum with a wavelength (λ) of 700 nm and a peak electric field magnitude (Ē) of 3.5 V/m, we first need to calculate the frequency (f) of the wave.
Since the speed of light (c) is approximately 3 × 108 m/s, the frequency can be calculated by using the relationship c = f × λ. After finding the frequency, we can then write down the equation for the electric field E(x, t). Assuming that the wave is propagating in the +x direction and that the electric field oscillates in the y-direction, with no initial phase change, the equation for the electric field is:
E(x, t) = Ē × sin(2π(ft - x/λ))
To find the associated magnetic field B, we use the fact that the magnitudes of E and B are related by the speed of light c, such that B = E/c. Since electromagnetic waves have their electric and magnetic fields perpendicular to each other, if E is in the y-direction, B will be in the z-direction. The equation for the magnetic field is:
B(x, t) = B × sin(2π(ft - x/λ))
Where B is the peak magnetic field strength.
The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from the black hole. That is, no information about the interior of the black hole may escape to any observer located outside of the black hole. According to general relativity, RH = (2GM / c^2), where M is the mass of the black hole and c is the speed of light. you want to study a black hole by getting near it ith a radial distance of 50 RH. However, you don't want the difference in gravitational acceleration between your head and your feet to exceed 10 m/s^2.
a) As a multiple of the Sun's mass, approximate what is the limit to the mass of the black hole you can tolerate at the given radial distance.
b) Is the limit an upper limit(you can tolerate smaller masses) or a lower limit (you can tolerate larger masses)?
Solution:
a) We know acceleration due to gravity, g = GM/r²
Differential change, dg/dr = -2GM/r³
Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²
My height, h=dr = 1.7 m
Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²
or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³
or, 5.9*100³*G²*M² = 2c⁶
or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms
Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.
b) This limit an upper limit ,we can tolerate smaller masses only.
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 Joules. What is the maximum height h reached by the ball? Consider air friction to be negligible.
A. h ≈ 30 m
B. h ≈ 40 m
C. h ≈ 50 m
D. h ≈ 10 m
E. h ≈ 20 m
Given that,
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules. It is moving upward with a kinetic energy of 50 Joules.
We need to find the maximum height h reached by the ball. Let at a height of 10 meters, it has a potential energy of 50 Joules. So,
[tex]mgH=50\\\\mg=\dfrac{50}{h}\\\\mg=\dfrac{50}{10}\\\\mg=5\ N[/tex] ........(1)
Let at a height of h m, it reaches to a maximum height. at this point, it has a total of 100 J of energy. So,
[tex]mgh=50+50\\\\mgh=100\\\\h=\dfrac{100}{5}\\\\h=20\ m[/tex]
So, the correct option is (E) "h = 20 m".
The maximum height reached by the ball is about 20 meters, as determined by using the conservation of mechanical energy principle, considering that the total mechanical energy at the height of 10 meters was 100 Joules.
Explanation:To solve for the maximum height reached by the ball, we can use the conservation of mechanical energy principle, which states that the total mechanical energy (potential energy + kinetic energy) of the ball remains constant in the absence of air friction.
At 10 meters above the ground, the ball has a potential energy (PE) of 50 Joules and a kinetic energy (KE) of 50 Joules. Therefore, the total mechanical energy at that height is:
PE + KE = 50 J + 50 J = 100 J
As the ball rises, its kinetic energy is converted into potential energy until the kinetic energy becomes zero at the maximum height. The total mechanical energy at maximum height will be equal to the potential energy:
PE at maximum height = total mechanical energy = 100 J
Using the formula for gravitational potential energy, PE = mgh (where m is mass, g is the acceleration due to gravity (9.81 m/s²), and h is the height), and knowing that the PE at 10 meters is 50 J, we can find the mass of the ball:
50 J = m * 9.81 m/s² * 10 m
m = 50 J / (9.81 m/s² * 10 m) = 0.51 kg
With the mass of the ball, we can now calculate the maximum height using the total mechanical energy:
100 J = 0.51 kg * 9.81 m/s² * h
h = 100 J / (0.51 kg * 9.81 m/s²) ≈ 20 meters
Therefore, the maximum height h reached by the ball is about 20 meters.
Two boxes, mA = 18 kg and mB = 14 kg, are attached by a string under tension T1 . The rightmost box is being pulled horizontally across the floor by a different string under tension T2 . The coefficient of kinetic friction between the boxes and the floor is µK = 0.240. If the boxes are accelerating at 3.5 m/s2 to the right, what is the tensions T1 and T2 ?
Answer:
[tex]T_{1}=105.38 N[/tex]
[tex]T_{2}=187.34 N[/tex]
Explanation:
Applying the second Newton's law for the first box we have.
[tex]-f_{1f}+T_{1}=m_{A}a[/tex]
[tex]-\mu_{k}N_{1}+T_{1}=m_{A}a[/tex]
We know that the normal force is the product between the weight and the kinetic friction, so we have:
[tex]-\mu_{k}m_{A}g+T_{1}=m_{A}a[/tex]
Now we can find T₁:
[tex]\mu_{k}m_{A}g+m_{A}a=T_{1}[/tex]
The acceleration is the same for both boxes.
[tex]T_{1}=m_{1}(\mu_{k}g+a)[/tex]
[tex]T_{1}=18*(0.240*9.81+3.5)[/tex]
[tex]T_{1}=105.38 N[/tex]
Now let's analyze the forces of the second box.
[tex]-f_{2f}-T_{1}+T_{2}=m_{B}a[/tex]
[tex]-\mu_{k}m_{B}g-T_{1}+T_{2}=m_{B}a[/tex]
Let's solve it for T₂.
[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]
[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]
[tex]T_{2}=m_{B}(a+\mu_{k}g)+T_{1}[/tex]
[tex]T_{2}=14(3.5+0.240*9.81)+105.38[/tex]
[tex]T_{2}=187.34 N[/tex]
I hope it helps you!
A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initially, the fluid level is such that the block is at a depth d and the tension in the thread is T. Then, the fluid level is decreased such that the depth is 0.5d. What is the tension in the thread when the block is at the new depth?
Answer:
(C) T
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence the tension.
Note: The figure of question is added in the attachment
The tension in the thread when the block is at the new depth is :
(C) T
Buoyant forceThe definition of buoyancy refers to whether something can drift in water or discuss, or the control of water or other fluids to keep water above water, or an idealistic disposition.
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence, the tension.
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A 0.0140-kg bullet is fired straight up at a falling wooden block that has a mass of 2.42 kg. The bullet has a speed of 555 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.
Answer:
0.16 s
Explanation:
• Falling from rest (V_block= 0 m/s) the block attains a final velocity V_block before colliding with the bullet. This velocity is given by Equation 2.4 as
V_block(final velocity of block just before hitting) =V_0,block +at
where a is the acceleration due to gravity (a = —9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is
V_block = at
• During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that
(M_bullet+M_block)V_f = M_bullet*V_bullet+ M_block*V_block
Total linear momentum after collision = Total linear momentum before collision
Here V_f is the final velocity of the bullet/block system after the collision, and V_bullet and V_block are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that V_f, the final velocity of the bullet/block system after the collision must have the same magnitude as V_block, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that of V_f =-V_block, Substituting this relation and Equation (1) into Equation (2) gives
(M_bullet + M_block)(-at) = M_bullet*V_bullet + M_block(at)
t = -M_bullet*V_bullet/a(M_bullet +2M_block)
=-(0.0140-kg)*555 m/s/-9.8(0.0140-kg+2(2.42 kg)
=0.16 s
Final answer:
The student's physics question involves applying the conservation of momentum and kinematic equations to determine the time the block was falling before being struck by a bullet. By considering the bullet's speed, the mass of both the bullet and block, and the motions involved, one can calculate the time of fall.
Explanation:
The student's question involves a physics problem related to conservation of momentum and projectile motion. The bullet's speed when it strikes the block, and the subsequent motion of the block and bullet system provide key information. To solve for the time t that the block was falling, we need to use the principles of physics that dictate how objects move under the influence of gravity and how they interact in collisions.
To find the time t, we'll take the following steps:
Use the conservation of momentum to find the velocity of the block and bullet immediately after the collision.
Use kinematic equations to relate this velocity to the maximum height reached by the block and bullet, taking into account the direction reversal.
Use the kinematic equations again to find the time t during which the block was falling before the collision.
Assuming the collision is perfectly inelastic, the bullet embeds itself in the block, so we have a combined mass moving upward after the collision. This combined mass will move up to the original height due to the conservation of energy principle, as the initial kinetic energy is converted entirely to gravitational potential energy at the peak of the ascent after which it momentarily comes to a halt.
A man is traveling on a bicycle at 14 m/s along a straight road that runs parallel to some railroad tracks. He hears the whistle of a train that is behind him. The frequency emitted by the train is 840 Hz, but the frequency the man hears is 778 Hz. Take the velocity of sound to be 340 m/s.
a) What frequency is heard by a stationary observer located between the train and the bicycle?
b) What is the speed of the train, and is the train traveling away from or towardthe bicycle?
Answer:
Explanation:
b ) The problem is based on Doppler's effect of sound
f = f₀ x (V - v₀) /( [tex]V+v_s[/tex])
f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away , [tex]v_s[/tex] is velocity of source going away
778 = 840 x (340 - 14)/ (340 + [tex]v_s[/tex])
340 + [tex]v_s[/tex] = 341.18
[tex]v_s[/tex] = 1.18 m /s
it will go away from the observer or the cyclist.
speed of train = 1.18 m /s
a )
For a stationary observer v₀ = 0
f = f₀ x V /( [tex]V+v_s[/tex])
= 840 x 340 / (340 + 1.180)
= 837 Hz
Final answer:
The stationary observer between the train and the bicycle hears a frequency of 809 Hz, and the train is moving away from the bicycle at 42 m/s.
Explanation:
A stationary observer located between the train and the bicycle would hear a frequency of 809 Hz as the train approaches and passes by.
The speed of the train can be calculated to be 42 m/s moving away from the bicycle.
Suppose that a pendulum clock keeps time exactly correctly on Earth. Now suppose that you take this clock to planet X and find that the clock’s hour hand makes α revolutions every hour. Find a formula for the magnitude of the acceleration due to gravity near the surface of planet X. What would α be on the moon?
Answer:
The gravity at planet X is [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
The value of [tex]\alpha[/tex] on the moon is [tex]\alpha = 730.38 \ revolutions[/tex]
Explanation:
From the question we are told that
The clocks hour hand makes [tex]\alpha[/tex] revolution every 1 hour which is 3600 sec
this implies that the time peroid for 1 revolution would be [tex]= \frac{3600}{\alpha }sec[/tex]
The peroid for a pendulum is mathematically represented as
[tex]T = 2 \pi\sqrt{\frac{L}{g} }[/tex]
Where L is the pendulum length
g is the acceleration due to gravity
Let assume that we have a pendulum that counts in second on earth
This implies that its peroid would be = 2 second
i.e one second to swimg forward and one second to swing back to its original position
Now the length of this pendulum on earth is
[tex]L = \frac{gT^2}{4 \pi^2}[/tex] [Making L the subject in above equation]
Substituting values
[tex]L = \frac{9.8 * (2)^2}{4 * (3.142)^2}[/tex]
[tex]= 1[/tex]
When the same pendulum is taken to planet X the peroid would be
[tex]T = \frac{3600}{\alpha }[/tex]
Recall this value was obtained above for 1 revolution (from start point to end point back to start point)
So the acceleration due to gravity on this planet would be mathematically represented as
[tex]g = \frac{4 \pi L }{T^2}[/tex] [making g the subject in the above equation]
substituting values
[tex]g = \frac{4 * 3.142^2 * 1}{[ \frac{3600}{\alpha } ]^2}[/tex]
[tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
On moon the acceleration due to gravity has a constant value of
[tex]g = 1.625 m/s^2[/tex]
The period of this pendulum on the moon can be mathematically evaluated as
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
substituting value
[tex]T = 2 *3.142 \sqrt{\frac{1}{1.625} }[/tex]
[tex]= 4.929s[/tex]
given that
[tex]1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s[/tex] {Note 1 revolution takes a peroid }
Making [tex]\alpha[/tex] the subject of the formula
[tex]\alpha =\frac{3600}{4.929}[/tex]
[tex]\alpha = 730.38 \ revolutions[/tex]
A glass lens, nglass=1.52, has a 127 nm thick antireflective film coating one side, nfilm=1.39. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating? Assume that nair=1.00.
Answer:
The wavelength is [tex]\lambda = 706nm[/tex]
Explanation:
From the question we are told that
The refractive index of the glass is [tex]n__{glass}} = 1.52[/tex]
The thickness of film is [tex]D = 127nm = 127*10^{-9}m[/tex]
The refractive index of film [tex]n__{film}} = 1.39[/tex]
The refractive index of air is [tex]n__{air}} = 1.00[/tex]
Generally the thickness of the film can be obtained mathematically from this expression
[tex]D = \frac{\lambda}{4 * n__{film}} }[/tex]
Where [tex]\lambda[/tex] is the wavelength
Making the wavelength the subject of the formula
[tex]\lambda = 4 * n__{film}} * D[/tex]
Substituting values
[tex]\lambda = 4 *1.39 * 127 *10^{-9}[/tex]
[tex]\lambda =7.06 *10^{-7}m = 706nm[/tex]
Answer:
[tex]\lambda = 706.12 nm[/tex]
Explanation:
The optical path length for the reflection of light = [tex]2 n_{film} t[/tex]
For destructive interference, [tex]2 n_{film} t = \frac{\lambda}{2}[/tex]
The thickness of the anti-reflective film = 127 nm
The largest wavelength of the reflected light, [tex]\lambda = 4n_{film} t[/tex]
[tex]\lambda = 4 * 1.39 *127 * 10^{-9}[/tex]
[tex]\lambda = 706.12 * 10^{-9} m\\\lambda = 706.12 nm[/tex]
Suppose your telescope has perfect mirrors and is limited only by diffraction. If you are viewing with light of 400 nm wavelength, what diameter mirror in a space telescope would have 1.0-km resolution of objects on Mars when viewed from above the earth at a time when the Earth-Mars distance of 80 million kilometers
Answer:
39 m
Explanation:
We are given that
Wavelength=[tex]\lambda=400 nm=400\times 10^{-9} m[/tex]
1 nm=[tex]10^{-9} m[/tex]
y=1 km=[tex]1000 m[/tex]
1 km=1000 m
Earth mars distance =x=80 million Km=[tex]80\times 10^9 m[/tex]
1million km=[tex]10^9 m[/tex]
[tex]sin\theta=\frac{1.22\lambda}{d}[/tex]
[tex]sin\theta=\frac{y}{x}[/tex]
[tex]\frac{y}{x}=\frac{1.22\lambda}{d}[/tex]
[tex]\frac{1000}{80\times 10^9}=\frac{1.22\times 400\times 10^{-9}}{d}[/tex]
[tex]d=\frac{1.22\times 400\times 10^{-9}\times 80\times 10^9}{1000}[/tex]
[tex]d=39.04 m\approx 39 m[/tex]
Diameter of mirror =39 m
To achieve a 1.0-km resolution on Mars with light of a 400 nm wavelength, the size of the space telescope's mirror would need to be approximately 6 mm. This calculation is based on diffraction limit theory and may not be precisely accurate in practical application due to other potential limiting factors.
Explanation:This question is related to the resolution of a telescope which is primarily influenced by the aperture size and wavelength of light being observed. The bigger the telescope aperture (the diameter of the primary mirror), the greater the resolution as more light can be collected. The formula for the resolution limit due to diffraction (the smallest distinguishable detail) can be given by R = 1.22λ/D where λ is the wavelength of light and D is the diameter of the telescope's main mirror.
Here the wavelength given is 400 nm or 400 x 10^-9 m, and we need to calculate the mirror diameter required for a resolution of 1.0-km on Mars from Earth. Let's convert the resolution limit from kilometers to meters (which gives us 1000m), and then to radians (using the Earth-Mars distance), which results in an angle of 1.25x10^-11 radians approximately.
Substituting these values in the formula, we can solve for D and find that the telescope mirror size needed would be approximately 0.006 m or 6 mm for this resolution. Do note, that this is a theoretical value, in reality, the size might need to be larger due to factors like the diffraction limit, non-uniformities in mirrors, or aberrations in lenses.
Learn more about Telescope Resolution here:https://brainly.com/question/25821830
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Gasoline flows in a long, underground pipeline at a constant temperature of 15o C (rho = 680 kg/m3 ; ν = 4.6 × 10-7 m2 /s). Two pumping stations are located at the same elevation 13 km apart. The pressure drop between the stations is 1.4 MPa. The diameter of the pipe is 0.6 m and has a roughness of ε = 0.12 mm. Determine the volume flow rate.
Answer:
1.0416 m∧3/sec
Explanation:
check the pictures below for the solution
A particular gun (10kg) is able to fire 20 gram bullets at a speed of 350 m/s. From this information, calculate roughly how much energy contained in the powder in the bullet (Hint: you will need to calculate the recoil speed of the gun.) A. 2455 J B. 620 J C. Cannot be determined. D. 1230 J
Answer:
The answer is
D. 1230j
Explanation:
When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force
given data
Mass of gun M=10kg
Mass of bullet m=20g----kg=20/1000 =0.02kg
Propulsive speed of bullet = 350m/s
Hence the moment of the bullet will be equal and opposite to that of the gun
mv=MV
where V is the recoil velocity which we are solving for
V=mv/M
V=0.02*350/10
V=7/10
V=0.7m/s
The energy contained in the bullet can be gotten using
KE=1/2m(v-V)²
KE=1/2*0.02(350-0.7)²
KE=1/2*0.02(349.3)²
KE=1/2*0.02*122010.49
KE=1/2*2440.20
KE=1220.1J
roughly the energy is 1230J
Suppose a car approaches a hill and has an initial speed of
106
k
m
/
h
at the bottom of the hill. The driver takes her foot off the gas pedal and allows the car to coast up the hill.
A. If in actuality a
780
k
g
car with an initial speed of
106
k
m
/
h
is observed to coast up a hill and stops at a height
21.5
m
above its starting point, how much thermal energy was generated by friction in
J
?
B. What is the magnitude of the average force in newtons of friction if the hill has a slope of
27
∘
above the horizontal.
Answer:
a) 1.73*10^5 J
b) 3645 N
Explanation:
106 km/h = 106 * 1000/3600 = 29.4 m/s
If KE = PE, then
mgh = 1/2mv²
gh = 1/2v²
h = v²/2g
h = 29.4² / 2 * 9.81
h = 864.36 / 19.62
h = 44.06 m
Loss of energy = mgΔh
E = 780 * 9.81 * (44.06 - 21.5)
E = 7651.8 * 22.56
E = 172624.6 J
Thus, the amount if energy lost is 1.73*10^5 J
Work done = Force * distance
Force = work done / distance
Force = 172624.6 / (21.5/sin27°)
Force = 172624.6 / 47.36
Force = 3645 N
n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pendulum swing upward in an arc. Determine the maximum vertical displacement of the pendulum (height).(g = 9.8 m/s2)
Answer:
68cm
Explanation:
You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get
[tex]p_f=p_i\\mv_1+Mv_2=(m+M)v[/tex]
m: mass of the bullet
M: mass of the pendulum
v1: velocity of the bullet = 410m/s
v2: velocity of the pendulum =0m/s
v: velocity of both bullet ad pendulum joint
By replacing you can find v:
[tex](0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}[/tex]
this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:
[tex]E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2[/tex]
g: 9.8/s^2
h: height
By doing h the subject of the equation and replacing you obtain:
[tex](0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m[/tex]
hence, the heigth is 68cm
Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the collision is half its speed before the collision. Block B was at rest before the collision. The mass of block A is 7 kg and the mass of block B is 2 kg. What is the minimum initial speed (in m/s) that block A must have for block B to swing through a complete vertical circle?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The minimum velocity of A is [tex]v_A= 4m/s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]L = 1m[/tex]
The initial speed of block A is [tex]u_A[/tex]
The final speed of block A is [tex]v_A = \frac{1}{2}u_A[/tex]
The initial speed of block B is [tex]u_B = 0[/tex]
The mass of block A is [tex]m_A = 7kg[/tex] gh
The mass of block B is [tex]m_B = 2 kg[/tex]
According to the principle of conservation of momentum
[tex]m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}[/tex]
Since block B at initial is at rest
[tex]m_A u_A = m_Bv_B + m_A \frac{u_A}{2}[/tex]
[tex]m_A u_A - m_A \frac{u_A}{2} = m_Bv_B[/tex]
[tex]m_A \frac{u_A}{2} = m_Bv_B[/tex]
making [tex]v_B[/tex] the subject of the formula
[tex]v_B =m_A \frac{u_A}{2 m_B}[/tex]
Substituting values
[tex]v_B =\frac{7 u_A}{4}[/tex]
This [tex]v__B[/tex] is the velocity at bottom of the vertical circle just at the collision with mass A
Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of [tex]v__B'[/tex] at the top of the vertical circle
The angular centripetal acceleration would be mathematically represented
[tex]a= \frac{v^2_{B}'}{L}[/tex]
Note that this acceleration would be toward the center of the circle
Now the forces acting at the top of the circle can be represented mathematically as
[tex]T + mg = m \frac{v^2_{B}'}{L}[/tex]
Where T is the tension on the string
According to the law of energy conservation
The energy at bottom of the vertical circle = The energy at the top of
the vertical circle
This can be mathematically represented as
[tex]\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L[/tex]
From above
[tex](T + mg) L = m v^2_{B}'[/tex]
Substitute this into above equation
[tex]\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = T + 5mgL[/tex]
The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero
This is mathematically represented as
[tex]\frac{49 mv_A^2}{16} = 5mgL[/tex]
making [tex]v_A[/tex] the subject
[tex]v_A = \sqrt{\frac{80mgL}{49m} }[/tex]
substituting values
[tex]v_A = \sqrt{\frac{80* 9.8 *1}{49} }[/tex]
[tex]v_A= 4m/s[/tex]
you and your roommates are studying hard for you physics exam. you study late into the night and then fall into your bed for some sleep. you all wake early before the exam and scramble groggily around making breakfast. you can’t agree on what to have, so one of you cooks waffles on a 990 w waffle iron while another toasts bread in a 900 w toaster. you want to make a cup of coffee with a 650 w coffeemaker, and you plug it into the same power strip into which the waffle iron and toaster are plugged. will the 20 a circuit breaker trip? explain.
Answer:
The circuit breaker will trip because the current drawn is in excess of its capacity
Explanation:
The 20 A circuit breaker is designed to not carry current in excess of its capacity.
Total power demand on power strip = 900 + 990 + 650 = 2540 w
Let us assume a standard American voltage outlet of 120 V
Recall that electric power is
P = I x V
Where I is current, and
V = voltage.
For the 2540 W power drawn, the current I is
I = P/V = 2540/120 = 21.16 A
This is 1.16 A in excess of its capacity.
1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What torque is transmitted by gear B to its shaft? Give your answer in kN-m.
Answer:
2.5kN.m
Explanation:
Torque is directly proportional to pitch diameter
= Ta/Tb= Da/Db
=120/Tb= 0.25/0.5
Tb= 2.469kN.m approx 2.5kN.m
Answer:
Explanation:
answer explain below
You are using a hand-held sound level meter to measure the intensity level of the roars produced by a lion prowling in the high grass. To decrease the measured sound intensity level by 36.2 dB requires the lion move away from you until its distance from you has increased by what factor? g
Answer:
64.57
Explanation:
We are given that
For decreasing measured sound intensity level=36.2 dB
We have to find the factor by which the distance increase.
Let initial distance=x
Final distance=x'
According to question
[tex]36.2=10log(\frac{x'^2}{x^2})[/tex]
[tex]36.2=10log(\frac{x'}{x})^2[/tex]
[tex]36.2=10\times 2log\frac{x'}{x}[/tex]
[tex]log\frac{x'}{x}=\frac{36.2}{20}=1.81[/tex]
[tex]\frac{x'}{x}=10^{1.81}[/tex]
[tex]x'=10^{1.81}x=64.57x[/tex]
Hence, the distance is increases by factor of 64.57
STATE THE LAWS OF REFLECTION?
Answer:
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A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.77 m from the center). 1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? kg-m2 2)What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? kg-m2 3)What is the final angular velocity of the disk? rad/s 4)What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) J 5)What is the centripetal acceleration of the person when she is at R/3? m/s2 6)If the person now walks back to the rim of the disk, what is the final angular speed of the disk? rad/s
Answer:
1) 883 kgm2
2) 532 kgm2
3) 2.99 rad/s
4) 944 J
5) 6.87 m/s2
6) 1.8 rad/s
Explanation:
1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:
[tex]I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2[/tex]
If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:
[tex]I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2[/tex]
2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):
[tex]I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2[/tex]
3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:
[tex]I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}[/tex]
[tex]\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}[/tex]
[tex]\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s[/tex]
4)Kinetic energy before:
[tex]E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J[/tex]
Kinetic energy after:
[tex]E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J[/tex]
So the change in kinetic energy is: 2374 - 1430 = 944 J
5) [tex]a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2[/tex]
6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.
The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B=0.500 T and E=1.2x105 V/m. The strength of the magnetic field that seperates the ions is Bo=0.750 T. A stream of single charged Li ions is found to bend in a circular arc of radius 2.32 cm. What is the mass of the Li ions?
Answer:
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Explanation:
Given,
Magnetic field, B = 0.5 T
Electric field, E = 1.2 x 10⁵ V/m
strength of the magnetic field that separates the ions, Bo=0.750 T
Radius, r = 2.32 cm
Relation of charge to mass ratio is given by
[tex]\dfrac{q}{m}=\dfrac{E}{BB_0R}[/tex]
[tex]m=\dfrac{qBB_0R}{E}[/tex]
Substituting all the values
[tex]m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}[/tex]
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Mass of Li ions is equal to [tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
The mass of the Li ions is approximately [tex]\( 1.64 \times 10^{-26} \)[/tex] kg.
To find the mass of the Li ions, we can use the formula for the radius of the circular path of a charged particle moving in a magnetic field:
[tex]\[ r = \frac{m \cdot v}{q \cdot B} \][/tex]
Where:
- r is the radius of the circular path
- m is the mass of the ion
- v is the velocity of the ion
- q is the charge of the ion
- B is the magnetic field strength
We can also use the formula for the force experienced by a charged particle moving in both electric and magnetic fields:
[tex]\[ F = q \cdot (E + v \times B) \][/tex]
Since the particle moves in a circular path, the electric force F must be equal to the magnetic force [tex]\( q \cdot v \cdot B \)[/tex], where v is the speed of the particle.
We can rearrange the formulas to solve for m:
[tex]\[ m = \frac{q \cdot r \cdot B}{v} \][/tex]
Now, we need to find \( v \). Since the particle passes through both the electric and magnetic fields without deviation, the forces acting on it must be balanced. Therefore:
[tex]\[ F_{electric} = F_{magnetic} \]\[ q \cdot E = q \cdot v \cdot B \][/tex]
Solving for v:
[tex]\[ v = \frac{E}{B} \][/tex]
Now, substituting v into the equation for m :
[tex]\[ m = \frac{q \cdot r \cdot B}{\frac{E}{B}} \]\[ m = \frac{q \cdot r \cdot B^2}{E} \][/tex]
Now we can calculate \( m \) using the given values:
[tex]\[ m = \frac{(1.6 \times 10^{-19} C) \times (0.0232 m) \times (0.75 T)^2}{1.2 \times 10^5 V/m} \]\[ m= 1.64 \times 10^{-26} kg \][/tex]
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.2 mm from the center of the central bright fringe. What is the separation of the two slits?
Answer:
Explanation:
wave length of light λ = 502 nm
screen distance D = 1.2 m
width of one fringe = 10.2 mm / 20
= .51 mm
fringe width = λ D / a , a is separation of slits
Puting the values given
.51 x 10⁻³ = 502 x 10⁻⁹ x 1.2 / a
a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³
= 1181.17 x 10⁻⁶ m
1.18 x 10⁻³ m
= 1.18 mm .
Albert Einstein won the Nobel Prize in Physics for his explanation, in 1905, of the photoelectric effect. Einstein showed that the results of the experiment can only be explained in terms of a particle model of light - light must be acting as if it is made up of particles (known as photons) rather than waves. This was especially fascinating since many other experiments can only be explained in terms of light acting as a wave. (a) What are the predictions of the particle theory regarding this experiment
The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10-2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?
Answer:
Explanation:
Given that, .
Frequency
f = 60Hz
Number of turns
N = 125turns
Surface area of coil
A = 3 × 10^-2 m²
Magnetic field
B = 0.12T
Voltage peak to peak? I.e the EMF
EMF is given as
ε = —dΦ/dt
Where Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Where N is number of turns
B is magnetic field
A is the cross sectional area
And θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, your magnetic flux becomes
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
I.e Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
The peak to peak voltage is 169.65 V
The required value of peak output voltage is 169.65 Volts.
Given data:
The frequency of ac generator is, f = 60 Hz.
The number of turns of coil is, n = 125 turns.
The area of each coil is, [tex]A =3.0 \times 10^{-2} \;\rm m^{2}[/tex].
The strength of magnetic field is, B = 0.12 T.
To start with this problem, we need to find the peak emf first. The peak output voltage is nothing but the value of this peak emf only. The expression for the peak EMF is given as,
ε = —dΦ/dt
here,
Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Here,
θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, the expression for the magnetic flux becomes,
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
Thus, we can conclude that the required value of peak output voltage is 169.65 Volts.
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