u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2
Hope it helps
a 10 kg block reaches a point with a velocity of 15 m per second and slides down a rough track my the coefficient of the kinetic energy between the two surface ab and the block iis0.52
Answer:
H = 6.93 m
Explanation:
given data
velocity v = 35 m/s
horizontal component Vx = 33 m/s
solution
we get here maximum height so first we get vertical component here that is express as
Vy = 
.........................1
put here value
Vy = 
Vy = 11.66 m/s
and
now we get height
H = 
.............................2
put here value
H = 
H = 6.93 m
weight = mg acts
downwards <span>
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is
acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left
Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ
We let the value of µs be equal to a value such that any F
will not be able to move the crate. Then, if we increase F by an amount F',
then the force pushing the crate towards the right also increases by F' cosΘ. Additionally,
the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase
by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds
by F' sin Θ.
<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>
