All categories

3 years ago

What steps are involved in converting potential energy to kinetic energy, or kinetic energy to potential energy?

Physics

1 answer:

Ray Of Light [21]3 years ago

7 0

Explanation:

Potential energy is the energy obtained by an object due to the position of an object. Whereas kinetic energy is the energy obtained by an object due to its motion.

For example, an ball is placed over a building of height h. Then steps followed to covert its potential energy into kinetic energy are as follows.

Step 1: When ball is at placed over a building of height h, then it has only potential energy and no kinetic energy.

Step 2: A man pushes the ball and it moves in the downward direction.

Step 3: When ball starts to move or fall it gains kinetic energy, that is potential energy now becomes equal to 0.

Similarly, on reaching the ground and after bouncing a few times when it stop moving then kinetic energy converts into potential energy.

You might be interested in

#26: Why is spear fishing much more successful for scuba divers than someone spear fishing from above the water? Explain your re

frozen [14]

The water has an irregular surface so if the light rays fall on top then it will reflect oppositely or in some cases may refract, so the fish will seem a bit smaller or further or closer in anyway, example if u put a straw in water it looks bent but is it ? nope. So the main reason is that it refracts which makes the accuracy lesser which make it difficult.

6 0

3 years ago

An incline plane has Ф = 40.0° and μ k = 0.15. Starting from rest, how long will it take a 4.0 kg block to reach a speed of 12 m

lesya [120]

The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:

$P_{t}-Fat=ma \\ mgsen\O-mgucos\O=ma \\ a=g(sen\O-ucos\O)$

Using the Velocity Hourly Equation, we get:

$V=V_{o}+a\Delta t \\ \Delta t= \frac{V}{a}$

Uniting the equations:

$\Delta t= \frac{V}{g(sen\O-ucos\O)}$

Entering the unknowns:

$\Delta t= \frac{V}{g(sen\O-ucos\O)} \\ \Delta t= \frac{12}{10(sen40^o-0.15cos^o)} \\ \Delta t= \frac{12}{10(0,64-0.15x0.77)} \\ \Delta t= \frac{12}{5.27} \\ \boxed {\Delta t=2.28s}$

Obs: Approximate results

If you notice any mistake in my english, please let me know, because i am not native.

5 0

3 years ago

List five objects in your daily life that produce sounds with a high pitch and five that produce sounds with a low pitch.

drek231 [11]

Answer:

Explanation:

High pitch:

-birds chirping

-nails on a chalkboard

-flies buzzing

-children squealing

-whistle

Low pitch:

-thunder

-dogs barking

- earthquakes

- bass drum,

-cannon shot,

8 0

3 years ago

How can a force change the motion of an object that is already moving?

lara [203]

Answer:They may cause motion to change

Explanation:They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity.

7 0

3 years ago

A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The ba

tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

E = Vo / D

E = 600 / 0.3

E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

Fe = E*q

Fe = (2,000 V / m) * ( 3*10^-5 C)

Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

Fe = m*a

a = Fe / m

a = 0.06 / 0.0004

a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

s = C - A

s = ( 0.25 , 12 ) - ( 0.05 , 12 )

s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

vf^2 = vi^2 + 2*a*s

vf^2 = 0 + 2*0.2*150

vf = √60

vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

vf - vi = a*t

t = ( 7.746 - 0 ) / 150

t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

$Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)]$

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

$E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\$

- Again, apply the Newton's second law of motion and determine the acceleration (a):

Fe = m*a

a = Fe / m

$a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}$

- Where the acceleration is rate of change of velocity "dv/dt":

$\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\$

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

$c = \frac{A*eo}{d}$

Where, eo = 8.854 * 10^-12 .... permittivity of free space.

$K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\$

- The differential equation turns out ot be:

$\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\$

- Separate the variables the integrate over the interval :

t : ( 0 , t )

v : ( 0 , vf )

Therefore,

$\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )$

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

$vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s$

- The final velocity of particle while charging the capacitor would be:

vf = 7.264 m/s ... slightly less for the fully charged capacitor

7 0

3 years ago