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Sergio039 [100]

2 years ago

7

How are electrical energy, thermal energy, and mechanical energy related?

1 answer:

Lera25 [3.4K]2 years ago

4 0

Answer:

Mechanical energy is the ordered movement of the molecules as a single unit, while Thermal energy is the random movement of the molecules. Mechanical energy can be 100% converted to thermal energy, but thermal energy cannot be fully converted to mechanical energy.

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Imagine that Earth stops orbiting the Sun but continues to rotate in place about its own axis at its current rate. In this case,

otez555 [7]

Answer:

The length of the solar day will get shorter.

Explanation:

The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
Since the speed of the Earth's rotation on it's own axis and around the Sun is constant we don't feel the effects of the rotation.We can only feel the motion if the earth changes it's rotation speed.
If by any means or chance the Earth stopped spinning (stopped rotation) then the atmosphere surrounding the Earth would be in motion and all the Earth's land would be scoured clean.

7 0

3 years ago

Aspirin is a compound composed of carbon, hydrogen, and oxygen atoms

vekshin1

Answer:

A homogeneous Mixture

Explanation:

The acid that contains the acetylsalicylic acid is a <u>mixture,</u> but it isnt a compound. though aspirin is. (hopefully this helps? qwq)

8 0

2 years ago

Help me with this please!

alexgriva [62]

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3 0

3 years ago

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

Citrus2011 [14]

Answer:

The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

$\tau=I\times\alpha$

Here, I = mk²

$\tau=mk^2\times\alpha$

Put the value into the formula

$\tau=3\times(0.2)^2\times0.5$

$\tau=0.06\ N-m$

$\tau=6.0\times10^{-2}\ N-m$

Hence, The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

7 0

3 years ago

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

Darya [45]

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$ , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$

8 0

3 years ago