Consider, please, this solution:
The final heat is:
Q=Q₁+Q₂+Q₃≈726.116 kJ
All the details are in the attachment.
Answer:
Explanation:
Given
Let us suppose police car and motorist travel in straight line and police car catches motorist after s distance
Distance travel by motorist
----1
Distance traveled by Police car


----2
from 1 & 2 we get

(a)Velocity of Police car after t sec



(b)time taken by police car is

(c)Distance travel by police car
Answer:

Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.


Initial speed of the projectile is v and final speed is 0.5 v.


g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :



So, the height of the projectile above the ground is
. Hence, this is the required solution.
Answer:
v=0.60 m/s
Explanation:
Given that
m ₁= 390 kg ,u ₁= 0.5 m/s
m₂ = 250 kg ,u₂ = 0.76 m/s
As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.
Pi = Pf
m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v
Now putting the values in the above equation
390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

v=0.60 m/s
Therefore the velocity of the system will be 0.6 m/s.
Answer:
Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.
Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.
Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.