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3 years ago

The rate at which energy is transferred is called

Physics

2 answers:

shusha [124]3 years ago

7 0

Answer:

your answer is power..

makvit [3.9K]3 years ago

6 0

Answer:

The rate at which energy is transferred is called power and the amount of energy that is usefully transferred is called efficiency.

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Explain how objects in motion have kinetic energy, use examples

Ann [662]

Kinetic energy is energy that a body possesses by virtue of being in motion, there for if an object is moving, it has kinetic energy.
Example; A roller coaster sitting on top of hill has potential energy. When it starts to move and is going down the hill, it has kinetic energy. :)

8 0

3 years ago

What allows the bimetallic coil to turn on or off a heating or cooling system? The two metals contract the same amount. The ther

Vedmedyk [2.9K]

Answer:

The two metals expand differently.

Explanation:

The bimetallic strip has two metal strips positioned like a bridge, these strips connect the electrical circuit to the heating system. When these strips are linear or "down" they allow the electricity to move through the circuit to the heating system to turn the heat on. When the strips are "up" the disconnect the electricity flow, thus turning the heating system off, thus the room becomes cool/cold.

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2 years ago

Which of the following are true about centripetal force? Check all that apply. A. Without centripetal force, an object cannot ac

Katyanochek1 [597]

B. Friction can be a centripetal force, such as when it keeps a car on the road going around a curve.

C. Gravity can be a centripetal force, such as when it pulls a satellite in its orbit.

Explanation:

The centripetal force is any force that keeps an object moving in circular motion, "pulling" the object towards the centre of the circular trajectory.

Several forces can act as centripetal force. Examples are:

- friction: when a car is going around the curve, is moving by circular motion. The force that keeps the car in circular motion is, in fact, the friction between the tires and the road.

- Gravity: when a satellite moves around the Earth, it is moving by circular motion. The force that keeps the satellite in circular motion is the gravitational attraction between the Earth and the satellite, that pushes the satellite towards the Earth.

The other two options are not correct because:

A) An object can also accelerate if there is no centripetal force (for example, a car speeding up on a straight road is accelerating, but there is no centripetal force since there is no circular motion

D) Centripetal force is not an outward force, since it pushes the object inwards (towards the centre of the trajectory).

3 0

3 years ago

Read 2 more answers

Which the answer It's science

earnstyle [38]

The answer is A. ive done a 5-k race, so its for sure 3 miles.

3 0

3 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$