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MAVERICK [17]
3 years ago
5

Which is the best example of potential energy?

Physics
2 answers:
svetoff [14.1K]3 years ago
8 0
First one, holding a basketball in the air. Potential energy is the energy it has mostly from gravity. The further you go from the center of mass, the more energy.
BaLLatris [955]3 years ago
5 0

Before coming into conclusion first we have to understand potential energy.The potential  energy of a body is defined as the energy possessed by a body due  to its position and configuration.

There are  generally two types of potential energy.One is gravitational potential energy and the other one is elastic potential energy.The former one is due to the position of the body at a certain height from the surface of earth.The latter one is due to the change in  configuration of the body.

There is also electric potential energy which is stored  inside the electric filed of charge particles at rest.

The first example is holding a basket ball in air.As the body is at certain height from earth surface ,hence it has gravitational potential energy which is a best example of potential energy.

The second one is a rolling ball across a flat table.It has both rotational kinetic energy due to its rolling motion as well as potential energy due to its position.

The third one  can't be considered as it is electric in nature.

The fourth one is turning on a flashlight in a dark room.The light is due to the heating effect of electric current. so it is not gravitational in nature.

Hence option  one is the correct answer.

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Answer:

a

Explanation:

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3 years ago
What does strong language mean ?
Art [367]

Answer:

noun

the style of a piece of writing or speech.

"he explained the procedure in simple, everyday language"

coarse or offensive language.

noun: strong language

"strong language"

Explanation:

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3 0
3 years ago
Read 2 more answers
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0
3 years ago
Help!!! I need it today <br> Thank you in advance
svlad2 [7]

Answer:

 F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

          F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

         W = mg

          m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

5 0
3 years ago
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