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gtnhenbr [62]
4 years ago
6

25) A 10 kg box is sitting on a shelf 15 feet in the air. How could you increase the potential energy of the box? (check all tha

t apply)  Raise the box to 20 feet.  Lower the box to 10 feet  Increase the mass of the box to 15 kg  Decrease the bass of the box to 5 kg  Drop the box off of the shelf.
Physics
1 answer:
irakobra [83]4 years ago
7 0
Potential energy is the energy possessed by a body or an object while at rest while kinetic energy is the energy while the body is at motion.
Potential energy is given by Mgh where m is the mass of the object, g is the gravitational acceleration (a constant) and h is the height.
Therefore, potential energy varies directly proportional to the mass of the object and the height.
Thus in this case, an increase in the mass of the box by 15 kg and raising the height of the box by 20 feet, will result to an increase in potential energy of the box.
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3 years ago
4. What is the efficiency of a pulley that does 571 J of work if the input work is 1,694 J?
kvasek [131]

Explanation:

Efficiency = work out / work in

e = 571 J / 1694 J

e = 0.337

8 0
4 years ago
A rock falls of a cliff and hits the ground after 3 seconds. What is it’s velocity right before it hits the ground? A.0.31 m/s B
LekaFEV [45]
Hopefully this will help you.

8 0
3 years ago
The pressure at the bottom of a full barrel of water is Poriginal . Determine what happens to the pressure when the radius or he
Sholpan [36]

Answer:

a)   P' = P_original, b)  P ’= P_original + ρ  g Δh

Explanation:

The expression for nanometric pressure is

          P = ρ g h

where ρ  is the density of the liquid and h is the height

a) we change the radius of the barrel, but keeping the same height

as the pressure does not depend on the radius it remains the same

        P' = P_original

b) We change the barrel height

         h ’≠ h

we substitute in the equation

      P ’= ρ  g h’

      h ’= h + Δh

      P ’= ρ  g (h + Δh)

      P ’= (ρ  g h) + ρ  g Δh

      P ’= P_original + ΔP

In this case, the pressure changes due to the new height,

*if it is higher than the initial one, the pressure increases

*if the height is less than the initial one, the pressure is less

3 0
3 years ago
023 (part 1 of 2) 10.0 points
Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

\Delta KE_{rotational} = \dfrac{1}{2}  \cdot I \cdot \omega ^2 = \dfrac{1}{2}  \cdot m \cdot v ^2

I  = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

\Delta KE_{rotational} = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0
3 years ago
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