A sled of mass 28 kg slides across the ground with a friction force of 200.3 N. What is k between the sled and the ground?
1 answer:
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Answer:
T=575.16K
Explanation:
To solve the problem we proceed to use the 1 law of diffusion of flow,
Here,
is the rate in concentration
is the rate in thickness
D is the diffusion coefficient, where,
Replacing D in the first law,
clearing T,
Replacing our values
Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s
a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t
a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio =
ratio = 2.8