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Kruka [31]
3 years ago
8

When a rubberband is stretched all the way back, it is an example of which type of energy?

Physics
2 answers:
Murljashka [212]3 years ago
8 0
Elastic potential energy. because elastic is were something is pulled and when let go it pulls its self back to the original position so when you pull the rubberband it builds up energy to pull its self back together
 
avanturin [10]3 years ago
5 0
The answer is C. elastic potential energy
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A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?
ivolga24 [154]

Answer:

Explanation:

Given

mass of rock m=40\ kg

Elevation of Rock h=10\ m

Distance traveled by rock with time

h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

5=0\times t+\frac{1}{2}(9.8)(t^2)

t^2=\frac{10}{9.8}

t=1.004\approx 1\ s

velocity at this time

v=u+at

v=0+9.8\times 1.004

v=9.83\ m/s

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3 years ago
If a vibrating string is made shorter (i.e., by holding your finger on it), what effect does
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4 0
3 years ago
A rocket is fired at 100 m/s at an angle of 37, how many seconds did it take to get to the top?
Kobotan [32]

Answer:

6.14 s

Explanation:

The time the rocket takes to reach the top is only determined from its vertical motion.

The initial vertical velocity of the rocket is:

u_y = u sin \theta = (100)(sin 37^{\circ})=60.2 m/s

where

u = 100 m/s is the initial speed

\theta=37^{\circ} is the angle of launch

Now we can apply the suvat equation for an object in free-fall:

v_y = u_y +gt

where

v_y is the vertical velocity at time t

g=-9.8 m/s^2 is the acceleration of gravity

The rocket reaches the top when

v_y =0

So by substituting into the equation, we find the time t at which this happens:

t=-\frac{u_y}{g}=-\frac{60.2}{-9.8}=6.14 s

7 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
2 years ago
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3 years ago
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