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2 years ago

What is the monthly cost of using a 750 W refrigerator that runs for 8 hours every day, if the cost per kWh is $0.23?

Physics

1 answer:

disa [49]2 years ago

8 0

Answer:

Assuming there are 28 days in each month,

750W = 0.75kW

Cost of electric bill = 0.75 × 8 × 28 × $0.23

= $38.64

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A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t

Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

F = mg sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

W = F d = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

W = K = 1/2 mv ^2

Solve for v :

v = √(2W/m) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0

3 years ago

Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. w

PSYCHO15rus [73]

1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

$F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N$

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

$F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m$

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s$

And the frequency of oscillation is given by:

$\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz$

3) 2.19 m/s

The velocity at time t of the block is given by:

$v=v_0 cos (\omega t)$

where

$v_0 = 4.4 m/s$ is the initial velocity of the block

$\omega=5.81 rad/s$ is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

$v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s$

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

$a_0 = \omega^2 A$

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

$K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m$

So, the maximum acceleration is

$a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2$

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

$F=mg-kx$

First, we need to find the position x at t=0.36 s, which is given by

$x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m$

And so, the net force is

$F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N$

And the negative sign means the direction of the force is upward.

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8_murik_8 [283]

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