Answer:
Explanation:
A 40kg child throw stone of 0.5kg
At a direction of 5m/s
Recoil can be calculated using recoil of a gun formula
m_1•v_1 + m_2•v_2
m_1•v_1 = -m_2•v_2
The negative sign show that the momentum of the boy is directed oppositely to that of the stone
m_1 Is mass of boy
v_1 is the recoil velocity of the boy
m_2 is mass of stone
v_2 is the velocity of stone
Then,
m_1•v_1 = -m_2•v_2
40•v_1 = -0.5 × 5
40•v_1 = -2.5
v_1 = -2.5 / 40
v_1 = -0.0625 m/s
The recoil velocity of the boy is 0.0625 m/s
Answer:
t = 1.05 s
Explanation:
Given,
The distance between your vehicle and car, 100 ft
The constant speed of your vehicle, u = 95 ft/s
Since, the velocity is constant, a =0
If the car stopped suddenly, time left for you to hit the brake, t = ?
Using the second equation of motion,
S = ut + ½ at²
Substituting the given values in the equation
100 = 95 x t
t = 100/95
= 1.05 s
Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s
work is done by the pulling force which is same as the tension force in the rope. the net work done is zero for the crate since crate moves at constant velocity. but there is work done by the tension force which is equal in magnitude to the work done by the frictional force.
T = tension force in the rope = 115 N
d = displacement of the crate = 7.0 m
θ = angle between the direction of tension force and displacement = 37 deg
work done on the crate is given as
W = F d Cosθ
inserting the values given above
W = (115) (7.0) Cos37
W = 643 J
A. The switch within the closed circuit is opened.
If the switch is opened current will stop flowing, as the current will meet a gap, that is not connected by a conductor.
Answer:38.675 m/s
Explanation:

t=4.2 s

Solving in x direction


ucos25=25
u=27.584 m/s
Initial velocity in vertical direction usin25
Let h be the height of Rooftop

h=37.47 m
Therefore final vertical velocity is 
=29.51 m/s
Final Resultant velocity

