Answer:

Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :

Put all the values,

So, the work done is
.
Answer: Brittle
Explanation:
took the test and I chose Soft, Soft is the wrong answer don't choose it. The CORRECT ANSWER IS BRITTLE
Answer:
Height, H = 25.04 meters
Explanation:
Initially the ball is at rest, u = 0
Time taken to fall to the ground, t = 2.261 s
Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

Here, a = g
H = 25.04 meters
So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.
Answer:
1.40 m/s^2
Explanation:
Given data
Velocity= 17.4 m/s
time= 12.4 seconds
We want to find the acceleration of the rock
We know that
acceleration = velocity/time
Substitute
acceleration= 17.4/12.4
acceleration=1.40 m/s^2
Hence the acceleration is 1.40 m/s^2
Answer:
t = 12,105.96 sec
Explanation:
Given data:
weight of spacecraft is 2000 kg
circular orbit distance to saturn = 180 km
specific impulse = 300 sec
saturn orbit around the sun R_2 = 1.43 *10^9 km
earth orbit around the sun R_1= 149.6 * 10^ 6 km
time required for the mission is given as t
![t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cmu_sun%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28R_1%20%2B%20R_2%29%5D%5E%7B3%2F2%7D)
where
is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2.![t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%201.32712%20x%2010%5E%7B20%7D%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28149.6%20%2A%2010%5E%206%20%2B1.43%20%2A10%5E9%20%29%5D%5E%7B3%2F2%7D)
t = 12,105.96 sec