The energy stored per unit volume of the inductor is called the energy density of the magnetic field. Why?

As a box is pushed 30 meters across a horizontal floor by a constant horizontal force of 25 newtons, the kinetic energy of the b

irakobra [83]

Answer:

1,050 Joules

Explanation:

Step 1: work done in moving the box 30 meters

work done = force X distance

= 25N X 30 = 750 Joules

Step 2: calculate total internal energy

Total internal energy = work done + kinetic energy

= 750 Joules + 300 Joules

= 1,050 Joules = 1.05 KJ

5 0

3 years ago

A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally

leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration a of

a = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ F = n - mg = 0

where ∑ F is the net force, n is the magnitude of the normal force, and mg is the weight of the block - it follows that n = mg ; and

• the net horizontal force is

∑ F = F - f = ma

where F is the applied force, f is kinetic friction, m is the block's mass, and a is the acceleration found earlier. F stretches the spring by x = 0.250 m, so we have

F - f = kx - µn = kx - µmg = ma

where k is the spring constant and µ is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ F = n - mg = 0

so that n = mg again; and

• the net horizontal force is

∑ F = F - f = 0

This time, F stretches the spring by y = 0.0544 m, so we have

F - f = ky - µmg = 0

Solve the equations in boldface for k and µ :

kx - µmg = ma

ky - µmg = 0

==> k (x - y) = ma

==> k = ma / (x - y)

==> k = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

ky - µmg = 0

==> µ = ky / (mg)

==> µ = (182 N/m) (0.0544 m) / ((17.6 kg) g) ≈ 0.0574

3 0

3 years ago

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt$