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scoundrel [369]
3 years ago
6

You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) how much work do you do on the can of pa

int? (b) you hold the can stationary for half a minute, waiting for a friend on a ladder to take it. how much work do you do during this time? (c) your friend decides not to use the paint, so you lower it back to the ground. how much work do you do on the can as you lower it?
Physics
1 answer:
MariettaO [177]3 years ago
8 0

(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

W=mg\Delta h

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and \Delta h=1.8 m is the variation of height. Substituting the numbers into the formula, we find

W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J


(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

W=Fd

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.


(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J

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Which situation would deplete freshwater?
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If freshwater consumption was greater than freshwater renewal.

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A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed
galina1969 [7]

Answer:

False

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ac = v^2/r

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3 0
2 years ago
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c
e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = e^{kt}

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = e^{-0.01t}

V = V₀e^{-0.01t}

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀e^{-0.01t}

0.1V₀ = V₀e^{-0.01t}

0.1V₀/V₀ = e^{-0.01t}

0.1 = e^{-0.01t}

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0
3 years ago
Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the
sdas [7]

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

         F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

6 0
3 years ago
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