The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m
Answer:
0.19m/s²
Explanation:
Initial velocity(u) = 50×1000/60×60
=13.88 m/s
Final velocity(v) = 36.5×1000/60×60
=10.13 m/s
Acceleration(a) = v-u/t
=10.13-13.88/19.5
a= -0.19m/s²
-a = 0.19m/s²
The magnitude of retar dation is 0.19m/s²
Answer:
0.72
Explanation:
= Time period of oscillation = 1.5 s
Angular frequency is given as
= Amplitude of oscillation = 40 cm = 0.40 m
= Coefficient of static friction = ?
= acceleration of the block
= mass of the block
Maximum acceleration of the block is given as
frictional force is given as
As per newton's second law
Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
