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Alexeev081 [22]

4 years ago

10

A house painter uses the chair-and-pulley arrangement of (Figure 1) to lift himself up the side of a house. The painter's mass i

s 55 kg and the chair's mass is 10 kg .
If the painter is at rest, what is the tension in the rope?

1 answer:

Annette [7]4 years ago

6 0

Answer:

Explanation:

Draw a free body diagram for the painter. There are three forces acting on him. The tension in the rope pulling him up, the normal force of the chair pushing him up, and gravity pulling him down.

Apply Newton's second law:

∑F = ma

T + N - W = ma

Since the painter is at rest, a = 0.

T + N - W = 0

T = W - N

T = Mg - N

Now draw a free body diagram of the chair. The chair has three forces acting on it also. The tension in the rope pulling it up, the normal force pushing down, and gravity pulling down.

Newton's second law for the chair:

∑F = ma

T - W - N = ma

Since the chair is also not accelerating, a = 0:

T - W - N = 0

T = W + N

T = mg + N

Now we have two equations and two variables. If we add the equations together, we can eliminate N:

2T = Mg + mg

2T = (M + m) g

T = (M + m) g / 2

Given that M = 55 kg, m = 10 kg, and g = 9.8 m/s²:

T = (55 + 10) (9.8) / 2

T = 318.5 N

If we round to two sig-figs, the tension in the rope is 320 N.

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Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

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The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

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R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

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$=39.31mol/m^3$

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$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

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The initial velocity with which the body was thrown up = u = 39.2 m/s.

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