The work done on the bullet is force•distance. This is also equivalent to the energy it has leaving the muzzle (work energy theorem). This kinetic energy is (1/2)(0.006kg)(336m/s)^2=338.7J. Then 338.7J=F•0.83m
F=338.7/0.83=408N
The speed of the vehicle after the push is 2.7 m/s.
<h3>What is velocity?</h3>
velocity is the rate of change of displacement.
To calculate the velocity of the vehicle, we use the formula below.
Formula:
- v = √[2(W-W')/m]............... Equation 1
Where:
- W = work done by the pusher
- W' = Work done by friction
- m = mass of the car
From the question,
Given:
- W = 6500 J
- W' = 2000 J
- m = 1200 kg
Substitute these values into equation 1
- v = √[2(6500-2000)/1200]
- v = √(7.5)
- v = 2.7 m/s
Hence, the speed of the vehicle after the push is 2.7 m/s.
Learn more about velocity here:
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Yes they do. So if you need to add or subtract a pair, then you have to first change one or the other so that they both have the same power of 10.
Electricity flows from positive to negative
Answer:
-7500 N
Explanation:
v₀ = 108 km/h = 30 m/s
v = 0 m/s
t = 4 s
a = Δv / Δt
a = (0 m/s − 30 m/s) / 4 s
a = -7.5 m/s²
F = ma
F = (1000 kg) (-7.5 m/s²)
F = -7500 N