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yanalaym [24]
3 years ago
12

4. As Juan is going to take a shower, the soap falls out of the soap dish on to the

Physics
1 answer:
LiRa [457]3 years ago
5 0

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

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A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,
Svetlanka [38]

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d) 289.31 m

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Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

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4 0
3 years ago
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Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let \alpha be the angular acceleration of the end of the rod

Taking torque about the link, we have:

\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

\tau = I_{rod}\ \alpha......(ii)

From equations (i) and (ii) we have:

mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }

The acceleration of the end of the rod farthest from the link is given by:

a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}

7 0
2 years ago
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