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Levart [38]
3 years ago
11

As dancers twirl faster and faster around their partners, they are demonstrating what type of energy?

Physics
1 answer:
solniwko [45]3 years ago
7 0
Your answer would be Kinetic energy
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Rest energy is the energy a body has due to its position at rest.<br> True<br> False
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Which has more PE, the same bowling ball held 5 meters in the air or the ping pong ball held at 5 meters? Explain.
anastassius [24]

Answer:

The bowling ball

Explanation:

The PE (gravitational potential energy) of an object is given by:

PE=mgh

where

m is the mass of the object

g is the gravitational acceleration

h is the height of the object above the ground

In this case, we are comparing the PE of a bowling ball and of a ping pong ball. The two balls are located at same height (h=5 m) and the gravitational acceleration is the same (g=9.8 m/s^2), so their PE depends only on their mass: since the bowling ball has a greater mass than the ping pong ball, the bowling ball will have a greater PE.

4 0
4 years ago
5. If Joe spend 250 W power to lift a box weighing 25.5 kg in 3.0 s, how high did he lift the box?
Serjik [45]

Answer:

3m

Explanation:

p=250w

m=25.5

t=3.0s

D=x

F=mxa   F= 25.5kg x 9.8m/s^2 =250N

D= 250wx3.0s/250w

3m

5 0
3 years ago
What is the electric field strength just outside the flat surface of the conductor?
inna [77]
We can find the answer step-by-step:

1) The electric charges on a conductor must lie entirely on its surface. This is because the charges have same sign, so the force acting between each other is repulsive therefore the charges must be as far apart as possible, i.e. on the surface of the conductor.

2) We consider a cylinder perpendicular to the surface of the conductor, that crosses the surface with its section. We then apply Gauss law, which states that the flux of the electric field through this cylinder is equal to the total charge inside it divided the electrical permittivity:
\Phi =  \frac{Q}{\epsilon_0}

3) The electric field outside the surface is perpendicular to the surface itself (otherwise there would be a component of the electric force parallel to the surface, which would move the charge, violating the condition of equilibrium). The electric field inside the conductor is instead zero, because otherwise charges would move violating again equilibrium condition. Therefore, the only flux is the one crossing the section A of the cylinder outside the surface: 
\Phi = E A

4) The total charge contained in the cylinder is the product between the section, A, and the charge density \sigma on the surface of the conductor:
Q=\sigma A

5) Substituting the flux and the charge density inside Gauss law, we can find the electric field just outside the surface of the conductor:
EA= \frac{\sigma A}{\epsilon_0}
therefore
E= \frac{\sigma}{\epsilon_0}
4 0
4 years ago
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