Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
As the distance between two charged objects increases, the strength of the electrical force between the objects <em>decreases</em>.
Percent error is calculated as follows:
% = ( |15-15.6| / 15.6 ) * 100%
% = (0.6/15.6) * 100%
% = 0.0385 * 100%
% = 3.85%
Hope this helped!
Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.
Red shift of distant galaxies
