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gtnhenbr [62]
3 years ago
12

A sample of helium behaves as an ideal gas as energy is

Physics
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

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Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T
ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

                   \vec{A}=x_A\hat{i}

Vector B , which has a magnitude of 5.5 m

                   \vec{B}=x_B\hat{i}+y_B\hat{j}

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The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A                    \vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}

Comparing we will get

                  x_A=-x_B\\\\y_B=6x_A

Substituting in x_{B}^{2}+y_{B}^{2}=30.25

                  \left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904

So we have

    \vec{A}=0.904\hat{i}

Magnitude of vector A = 0.904

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