(1.6 x 10⁻¹⁹ C) x (300 N/C) = <em>4.8 x 10⁻¹⁷ Newton</em>
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Answer:
UP TO four inches but usually a little slower than that
Explanation:
Answer:
d' = d /2
Explanation:
Given that
Distance = d
Voltage =V
We know that energy in capacitor given as



If energy become double U' = 2 U then d'



2 d ' = d
d' = d /2
So the distance between plates will be half on initial distance.
Answer:
the one with v = 25 m/s
Explanation:
Momentum = m * v
if they both have the same mass (15000 kg) , then the one with the higher v has more momentum...I think A= 25 m/s
Define
g = 9.8 m/s², acceleration due to gravity, positive downward.
Assume that wind resistance may be neglected.
Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.
Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s
Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.