For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:
where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:
, so the work done by the brakes to stop the bus is
And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
Answer:
35.3 N
Explanation:
U = 0, V = 0.61 m/s, s = 0.39 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0.61 × 0.61 = 0 + 2 × a × 0.39
a = 0.477 m/s^2
Force = mass × acceleration
F = 74 × 0.477 = 35.3 N
Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = 
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
= -K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
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Answer:
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Explanation:
