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iren2701 [21]
3 years ago
5

The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.

Physics
1 answer:
Eva8 [605]3 years ago
8 0

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

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Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apar
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The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

We have to calculate  much work is needed to move the charges apart and double the distance between them.

<h3>What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?</h3>

The potential energy of the system of two charges separated by a distance is given by -

U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}

In order to solve this question, it is important to remember the work - energy theorem which states -

"The change in the energy of the body is equal to work done on it"

Hence, using this work -energy theorem in the question given to us we get -

U_{f} -U_{i} =W_{net}

In our case -

U_{f}  = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} =   \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

To solve more question on potential energy, visit the link below -

brainly.com/question/15014856

#SPJ4

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