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dusya [7]
3 years ago
14

A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin

al pressure? 1. 0.75 atm 2. 3.0 atm 3. 0.063 atm 4. 0.38 atm 5. 0.12 atm
Physics
1 answer:
ollegr [7]3 years ago
8 0
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
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HACTEHA [7]

Answer:

Explanation:

Force on the electron = q ( v x B )

q = - 1.6 x 10⁻¹⁹

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v x B = (5.9i−6.4j)×10⁴  x (−0.63i+0.65j)

= (3.835  - 4.032 ) x 10⁴ k

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2 years ago
suppose you increase your walking speed from 5m/s to 14m/s in a period of 1 . what is your acceleration?
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QUESTION 4.
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Answer:

Follows are the solution to this question:

Explanation:

Please find the correct question in the attachment file.

Let:

\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\

Calculating the value of  \overrightarrow{R} \times \overrightarrow{S}:

\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j]

Calculating the value of \overrightarrow{R}  \cdot (\overrightarrow{R} \times \overrightarrow{S}):

\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])

by solving this value it is equal to 0.

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Answer:

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