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Naily [24]
2 years ago
6

When a rubber band is pulled back on your finger but not yet let go how is that potential energy?

Physics
1 answer:
loris [4]2 years ago
4 0
It is potential energy because the band is not in movement, th band has the potential to move.
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Which of the following statements are true?Check all that apply.
MaRussiya [10]

Answer:

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

Explanation:

As per Kepler's law of planet motion we know that all planets revolve around the sun in elliptical path in such a way that position of Sun must be at one of the focii of the path

So all planets are in elliptical path always

Position of sun is always at one of the focus

so correct answer will be

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

4 0
2 years ago
A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

brainly.com/question/25959744

brainly.com/question/14028892

#SPJ1

5 0
1 year ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.
Gekata [30.6K]
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4 0
1 year ago
Read 2 more answers
A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes
Sholpan [36]

Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

mass of the softball, m = 720 g = 0.72 kg

velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\

Therefore, the time it takes the ball to stop is 0.021 s.

5 0
3 years ago
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