Answer:
1000 N
Explanation:
The magnitude of the electrostatic force between two charged object is given by

where
k is the Coulomb constant
q1, q2 is the magnitude of the two charges
r is the distance between the two objects
Moreover, the force is:
- Attractive if the two forces have opposite sign
- Repulsive if the two forces have same sign
In this problem:
are the two charges
r = 3000 m is their separation
Therefore, the electric force between the charges is:

We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
Answer:
t=67.7s
Explanation:
From this question we know that:
Vo = 6m/s
a = 1.8 m/s2
D = 1500m
And we also know that:
Replacing the known values:
Solving for t we get 2 possible answers:
t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:
t = 67.7s