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alexgriva [62]
3 years ago
10

Which of the following is not an example of kinetic energy being converted to potential energy?

Physics
2 answers:
KengaRu [80]3 years ago
6 0

The list of choices you provided with your question
is utterly devoid of any such examples.

Bumek [7]3 years ago
6 0
You did not give any answers to choose from. So I'll just explain to you what the difference is.

Kinetic energy means the object is in motion, or is moving.

Potential energy means the object is not moving, but it has the potential to.

Example:
                You are holding your phone in your hand one foot above the ground and you are not moving it at all (Potential energy). You then drop your phone and it begins to fall (Kinetic energy). 

Now, you hold your phone in your hand 6 feet above the ground and you are not moving it at all (More Potential energy). You then drop your phone and it begins to fall.. very very hard... (More Kinetic Energy).

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Describe the movement of sperm in the female reproductive system. State the different structures that it must pass in order to m
saw5 [17]

Answer:

After a male ejaculates, many sperms move to the upper vagina (via contractions from the vagina) through the cervix and across the length of the uterus, reaching the fallopian tubes. Here they will meet the egg cell ready to be fertilized

5 0
3 years ago
A small car with mass of 0.800 kg travels at a constant speed
Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

7 0
2 years ago
. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?
Hoochie [10]

The acceleration of the object is 6.7 m/s^2

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
3 years ago
Which one of the following statements does not accurately describe vibrations?
Paladinen [302]
A. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.
4 0
3 years ago
PLEASE HELP and actually help plz
frutty [35]

Answer:

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Explanation:

3 0
3 years ago
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