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
Answer:
Currently in the united states using parallel system
Explanation:
because you can walk with the twomodes with internal combustion engine or running on electric power.
M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
Answer:
ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:
Where:
Now:
Where 
is the mass of the car, 
is the initial velocity of the car, 
is the mass of train, 
is the final velocity of the car and 
is the final velocity of the train.
Replacing data:
Solving for :
Changed to cm/s, we get:
b) The kinetic energy K is calculated as:
K = 
where M is the mass and V is the velocity.
So, the initial K is:
And the final K is:
Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J
Explanation:
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