3 years ago

Which of the following does NOT describe the conversion of potential to kinetic energy?

Physics

1 answer:

ivann1987 [24]3 years ago

8 0

I believe it would be A. (:

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Physics help please..............

adoni [48]

Answer:

3.68 m/s

Explanation:

Full answer in the attached picture

7 0

3 years ago

Which best explains satellites? The moon is a satellite of the Earth, and the Earth is a satellite of the Sun. Mercury and Pluto

expeople1 [14]

A satellite is a body that revolves around another larger body. Where the larger body has a stronger gravitational pull on the smaller body, keeping it in orbit. Thus the moon is the satellite of the Earth is a good example of this.

4 0

3 years ago

The force between two charges, q, and 92, is F. If the distance between the

Sholpan [36]

Answer:

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

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3 years ago

Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?

oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

6 0

3 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

hodyreva [135]

Answer : The correct option is, (c) $3.7\times 10^2J/^oC$

Explanation :

First we have to calculate the energy or heat.

Formula used :

$E=V\times I\times t$

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

$E=(3.6V)\times (2.6A)\times (350s)$

$E=3276J$

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

$C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}$

where,

C = heat capacity of the calorimeter

$T_{initial}$ = initial temperature = $20.3^oC$

$T_{final}$ = final temperature = $29.1^oC$

Now put all the given values in this formula, we get:

$C=\frac{3276J}{(29.1-20.3)^oC}$

$C=372.27J/^oC=3.7\times 10^2J/^oC$

Therefore, the heat capacity of the calorimeter is, $3.7\times 10^2J/^oC$

7 0

3 years ago