PHYSICS COLLEGE

A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.20 ?

Answers

Answer 1

Answer:

Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

Time, t = 1 s

resistance of the coil, R = 1.2 ohm

The induced emf is given by

$e=(d\phi)/(dt)=A(B' - B)/(t)$

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

i = e/R

i = 2.184/1.2

i = 1.82 mA

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A 24.1 N solid sphere with a radius of 0.151 m is released from rest and rolls, without slipping, 1.7 m down a ramp that is inclined at 34o above the horizon. What is the total kinetic energy of the sphere at the bottom of the ramp?What is the angular speed of the sphere at the bottom of the ramp? How many radians did the sphere rotate through as it rolled down the ramp What was the angular acceleration of the sphere as it rolled down the ramp

Answers

Answer

given,

weight of solid sphere = 24.1 N

m = 24.1/g = 24.1/10 = 2.41 Kg

radius = R = 0.151 m

height of the ramp = 1.7 m

angle with horizontal = 34°

acceleration due to gravity = 10 m/s²

using energy conservation

$(1)/(2)I\omega^2 + (1)/(2)mv^2 = mgh$

I for sphere

$I = (2)/(5)mr^2$ v = r ω

$(1)/(2)\ (2)/(5)mr^2* (v^2)/(r^2) + (1)/(2)mv^2 = mgh$

$(7)/(10)mv^2 = mgh$

$h = (0.7 v^2)/(g)$

$v = \sqrt{(h * g)/(0.7)}$

$v = \sqrt{(1.7 * 10)/(0.7)}$

v = 4.93 m/s

b) rotational kinetic energy

$KE=(1)/(2)I\omega^2$

$KE=(1)/(2)\ (2)/(5)mr^2* (v^2)/(r^2)$

$KE=(1)/(5)mv^2$

$KE=(1)/(5)* 2.41 * 4.93^2$

KE = 11.71 J

c) Translation kinetic energy

$KE=(1)/(2)mv^2$

$KE=(1)/(2)* 2.41 \time 4.93^2$

$KE=29.28\ J$

A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upwards a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block

Answers

Answer:

423m/s

Explanation:

Suppose after the impact, the bullet-block system swings upward a vertical distance of 0.4 m. That's means their kinetic energy is converted to potential energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the total mass and h is the vertical distance traveled, v is the velocity right after the impact at, which we can solve by divide both sides my m

Let g = 9.81 m/s2

$gh = v^2/2$

$v^2 = 2gh = 2 * 9.81* 0.4 = 7.848$

$v = √(7.848) = 2.8m/s$

According the law of momentum conservation, momentum before and after the impact must be the same

$m_uv_u + m_ov_o = (m_u + m_o)v$

where $m_u = 0.01, v_u$ are the mass and velocity of the bullet before the impact, respectively. $m_ov_o$ are the mass and velocity of the block before the impact, respectively, which is 0 because the block was stationary before the impact

$0.01v_u + 0 = (0.01 + 1.5)*2.8$

$0.01v_u = 4.23$

$v_u = 4.23 / 0.01 = 423 m/s$

Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3neutrons. The mass of the uranium-235 is greater than the total mass of the
products. Which statement explains this difference in mass?
A. Some of the mass was transformed into neutrons during the
process.
O B. Mass was destroyed and disappeared during the process.
C. Some of the mass was transformed into gases during the
process.
D. Mass was transformed into energy during the process.

Answers

Answer:

D. Mass was transformed into energy during the process.

Answer:

Explanation:

Some of the mass

A radio station broadcasts its electromagnetic (radio)waves at a frequency of 9.05 x 107 Hz.
These radio waves travel at a speed of 3.00 x 108 m/s.
What is the wavelength of these radio waves?

Answers

Wavelength = speed/frequency
Wavelength = 3.00x108/9.05x107=
3.3x10 risen to the power of -1

Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.

Answers

Answer:

Both cars travel at < 10 , 4 > m/s

Explanation:

Conservation of Linear Momentum

The total momentum of a system of particles of masses m1 and m2 traveling at velocities v1 and v2 (vectors) is given by

$\vec p_1=m_1\vec v_1+m_2\vec v_2$

When the particles collide, their velocities change to v1' and v2' while their masses remain unaltered. The total momentum in the final condition is

$\vec p_2=m_1\vec v'_1+m_2\vec v'_2$

We know the collision is perfectly inelastic, which means both cars stick together at a common final velocity v'. Thus

$\vec p_2=m_1\vec v'+m_2\vec v'=(m_1+m_2)\vec v'$

Both total momentums are equal:

$m_1\vec v_1+m_2\vec v_2=(m_1+m_2)\vec v'$

Solving for v'

$\displaystyle v'=(m_1\vec v_1+m_2\vec v_2)/(m_1+m_2)$

The data obtained from the question is

$m_1=1200\ kg$

$m_2=3000\ kg$

The first car travels north which means its velocity has only y-component

$\vec v_1=<0,5>$

The second car travels east, only x-component of the velocity is present

$\vec v_2=<5,0>$

Plugging in the values

$\displaystyle v'=(1200<0,5>+3000<5,0>)/(1200+3000)$

$\displaystyle v'=(<0,6000>+<15000,0>)/(1500)$

$\displaystyle v'=(<15000,6000>)/(1500)=<10,4>\ m/s$

The magnitude of the velocity is

$|v'|=√(10^2+4^2)=10.77\ m/s$

And the angle

$\displaystyle tan\alpha=(4)/(10)=0.4$

$\alpha=21.8^o$

Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

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