The second stone hits the ground exactly one second after the first.
The distance traveled by each stone down the cliff is calculated using second kinematic equation;

where;
- <em>t is the time of motion </em>
- <em />
<em> is the initial vertical velocity of the stone = 0</em>

The time taken by the first stone to hit the ground is calculated as;

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as


Thus, we can conclude that the second stone hits the ground exactly one second after the first.
"<em>Your question is not complete, it seems be missing the following information;"</em>
A. The second stone hits the ground exactly one second after the first.
B. The second stone hits the ground less than one second after the first
C. The second stone hits the ground more than one second after the first.
D. The second stone hits the ground at the same time as the first.
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Answer:
1.07 nT
Explanation:
We know that E/B = c where E = electric field amplitude = 320 mV/m = 0.32 V/m, B = magnetic field amplitude and c = speed of light = 3 × 10⁸ m/s.
So, B = E/c
Substituting E and c into B, we have
B = E/c
= 0.32 V/m ÷ 3 × 10⁸ m/s
= 0.1067 × 10⁻⁸ T
= 1.067 × 10⁻⁹ T
= 1.067 nT
≅ 1.07 nT
Answer:
a. one line down one line to the right one live to the northwest from the object
b. t1=190 t2=310
Explanation: