Answer:The answer must be The weight of the man and the vertical distance moved.
Explanation: you calculate it by the force you applied times the distance you moved
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Explanation:
It is given that,
Area of square coil, 
Side of the square, L = 0.02 m
Number of turns, N = 10000
Uniform magnetic field, B = 1.5 T
Speed, v = 100 m/s
An emf is induced in the coil which is given by :


E = 30000 V
Breakdown voltage of air, 
Let d is the gap between the two wires connected to the ends of the coil and still get a spark. So,
Electric field, 

d = 0.075 m
Hence, this is the required solution.
Answer:
Collision force will be same in both the cases.
Explanation:
A perfectly inelastic collision is said to take place when a system loses the amount of its Kinetic Energy at its maximum. In a perfectly inelastic collision, the colliding particles stick to each other. In such a collision, kinetic energy is lost by combining the two bodies with each other.
In situation 1:
Speed of Car A, 
Speed of Car B, 
Relative speed of car A and car B, 
Now, in the situation 2:
Speed of car A, 
Speed of car B, 
Relative speed of car A and car B, 
Therefore, Car A and Car B both have the same relative speed, v = 10 m/s
Answer:
they meet from point o at distance 50.46 m and time taken is 11.6 seconds
Explanation:
given data
acceleration = 0.75 m/s²
speed B = 6 m/s
time B = 20 s
to find out
when and where the vehicles passed each other
solution
we consider here distance = x , when they meet after o point
and time = t for meet point z
we find first Bus B distance for 20 s ec
distance B = velocity × time
distance B = 6 × 20
distance B = 120 m
so
B take time to meet is calculate by distance formula
distance = velocity × time
120 - x = 6 × t
x = 120 - 6t .................1
and
distance of A when they meet by distance formula
distance = ut + 1/2 × at²
here u is initial speed = 0 and t is time
x = 0 + 1/2 × 0.75 × t²
x = 0.375 × t² .............2
so from equation 1 and 2
0.375 × t² = 120 - 6t
t = 11.6
so time is 11.6 second
and
distance from point o from equation 2
x = 0.375 (11.6)²
x = 50.46
so distance from point o is 50.46 m