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erastovalidia [21]
3 years ago
8

The force pressing two stacked objects together is equal to the weight of the object on top. The friction between a piece of woo

d on a stone surface is 22.5 newtons. The same piece of wood is placed on another surface, and the friction between the wood and the surface is now 46.5 newtons.
What could be the second surface?

A) metal
B) concrete
C) brick
D) skin
Physics
1 answer:
34kurt3 years ago
3 0

Answer: B. Concrete

Explanation:

Let N = reacting force pressing the bodies in context together (units in Newtons),

The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.

We need to start by finding the weight of the piece of wood.

friction is given by

f = μN

The value of f is 22.5,

and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.

22.5 = 75. 0.30

Putting the values into the equation: 22.5 = 0.30N.

Divide both sides by 0.30 to find the value of N:

N= 22.5/0.3 = 75

Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.

To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:

46.5 = µ(75).

Divide both sides by 75 to isolate μ.

The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).

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kodGreya [7K]

Answer:

\Delta H=687.4 J

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

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Both process have an associated enthalpy as shown below:

\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J

\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J

Therefore, the required heat is:

\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

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2 years ago
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Explanation:

It is given that,

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I=0.437+0.15=0.587\ kgm^2

So, the moment of inertia of the wagon wheel for rotation about its axis is 0.587\ kgm^2. Hence, this is the required solution.

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