Given:
m(mass of the box)=10 Kg
t(time of impact)=4 sec
u(initial velocity)=0.(as the body is initially at rest).
v(final velocity)=25m/s
Now we know that
v=u+at
Where v is the final velocity
u is the initial velocity
a is the acceleration acting on the body
t is the time of impact
Substituting these values we get
25=0+a x 4
4a=25
a=6.25m/s^2
Now we also know that
F=mxa
F=10 x6.25
F=62.5N
Average speed = distance traveled / time
average speed = (126.5 m * 3.5 laps) / (4.17 min)
= 106.2 m/min
Answer:
Which of the following would increase the elastic force acting on that object? Moving a spring to an unstretched position. Compressing a spring twice as much as its starting position.
Explanation:
Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
Answer:
The atmospheric pressure is 
.
Explanation:
Given that,
Atmospheric pressure 
drop height h'= 27.1 mm
Density of mercury 
We need to calculate the height
Using formula of pressure
Put the value into the formula
We need to calculate the new height
We need to calculate the atmospheric pressure
Using formula of atmospheric pressure
Put the value into the formula
Hence, The atmospheric pressure is .
