I think it is option (C).
If the answer is helpful then mark me as brainly.
Answer:
1) The charge on the outer shell is +4·Q
2) The charge on the inner shell is +Q
Explanation:
1) The given parameters of the spherical shell are;
The net charge on the spherical shell = 3·Q
The point charge surrounded by the spherical shell = -Q
Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;
The net charge, 3·Q = -q + x
∴ x = 3·Q + Q = 4·Q
The charge on the outer shell, x = 4·Q
2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;
-Q + y = 0
∴ y = +Q
The charge on the inner shell, y = +Q
Answer:
A. 12,240.
Explanation:
1,530 times 8.0 = 12,240.
hope this helped :)
B) All nonzero digits are significant.
Answer:
When the ball hits the ground, the velocity will be -34 m/s.
Explanation:
The height and velocity of the ball at any time can be calculated using the following equations:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the ball at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity. (-9.8 m/s² considering the upward direction as positive).
v = velocity at time "t".
If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:
y = y0 + v0 · t + 1/2 · g · t²
0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 = 60 m - 4.9 m/s² · t²
-60 m / -4.9 m/s² = t²
t = 3.5 s
Now, with this time, we can calculate the velocity of the ball when it reaches the ground:
v = v0 + g · t
v = 0 m/s - 9.8 m/s² · 3.5 s
v = -34 m/s
When the ball hits the ground, the velocity will be -34 m/s.
