Answer:
F'= 4F/9
Explanation:
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :
...(1)
Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :
.....(2)
Dividing equation (1) and (2), we get :

Hence, the correct option is (d) i.e. " 4F/9"
Answer:
6 m/s
Explanation:
12m / 2s = 6 m/s
Hope that's the answer you seek.
Speed with which initially car is moving is 21 m/s
Reaction time = 0.50 s
distance traveled in the reaction time d = v t
d = 21 * 0.50 = 10.5 m
deceleration after this time = -10 m/s^2
now the distance traveled by the car after applying bakes



so total distance moved before it stop
d = 22.05 + 10.5 = 32.55 m
so the distance from deer is 35 - 32.55 = 2.45 m
now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop
so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m
again by kinematics



so maximum speed would be 22.1 m/s
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Given parameters:
Initial velocity of Coin = 0m/s
Time taken before coin hits ground = 5.7s
Unknown:
Final velocity of the coin = ?
Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.
The fitting one of them here is shown below;
V = U + gt
where;
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
t is the time taken
Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.
Now input the parameters and solve;
V = 0 + 9.81 x 5.7
V = 55.917m/s
Therefore, the final velocity is 55.917m/s.