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3 years ago

Find the tension in the two ropes that are holding the 4.2 kg object in place.

Physics

1 answer:

Blababa [14]3 years ago

7 0

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

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A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

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3 years ago

How do you solve 0.004 dm + 0.12508 dm?

Effectus [21]

0.004 of something added to 0.12508 of the same thing
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3 years ago

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Anit [1.1K]

Answer:

Option A

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Explanation:

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