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3 years ago

A 17 H inductor carries a current of 1.8 A. At what rate must the current be changed to produce a 70 V emf in the inductor?

Physics

1 answer:

pogonyaev3 years ago

5 0

Answer:

at rate the current change is 6.75 A / sec

Explanation:

given data

inductance L = 17 H

current I = 1.8 A

emf e = 70 V

to find out

At what rate must the current be changed

solution

we will apply here emf formula that is

emf = inductance (di/dt)

so here (di/dt) will be

di/dt = emf / inductance .......................1

put value of emf and inductance in equation 1 we get rate

di/dt = 81 / 12

di/dt = 6.75 A / sec

so at rate the current change is 6.75 A / sec

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Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars inc

Leno4ka [110]

Let m be the mass of the second car, so the first car's mass is 2m.

Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.

Let u and v be the speeds of the first car and the second car, respectively. At the start,

• the first car has kinetic energy

K/2 = 1/2 (2m) u ² = mu ² ==> K = 2mu ²

• the second car starts with kinetic energy

K = 1/2 mv ²

It follows that

2mu ² = 1/2 mv ²

==> 4u ² = v ²

When their speeds are both increased by 2.76 m/s,

• the first car now has kinetic energy

1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²

• the second car now has kinetic energy

1/2 m (v + 2.76 m/s)²

These two kinetic energies are equal, so

m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²

==> 2 (u + 2.76 m/s)² = (v + 2.76 m/s)²

Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.

7 0

3 years ago

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The c

Dominik [7]

Answer:

2.55 m/s

Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

The work done by friction is given as:

$W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J$

The work done by gravity is:

$W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s$

4 0

3 years ago

A bullet is fired straight up from a gun with a

melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

$V=V_{o}+gt$ (1)