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The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
It is Real,Virtual,The Same Size, Inverted
Answer:
- The work made by the gas is 7475.69 joules
- The heat absorbed is 7475.69 joules
Explanation:
<h3>
Work</h3>
We know that the differential work made by the gas its defined as:

We can solve this by integration:

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


This give us

As n, R and T are constants

![\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}](https://tex.z-dn.net/?f=%20%5CDelta%20W%3D%20%5C%20n%20%5C%20R%20%5C%20T%20%20%5Cleft%20%5B%20ln%20%28V%29%20%5Cright%20%5D%5E%7Bv_2%7D_%7Bv_1%7D%20)



But the volume is:



Now, lets use the value from the problem.
The temperature its:

The ideal gas constant:

So:


<h3>Heat</h3>
We know that, for an ideal gas, the energy is:

where
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

where
is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So:

