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Brut [27]
3 years ago
10

Radon is a radioactive gas found below ground. If a sample of radon occupies 29 mL at 25 oC, what volume will it occupy at 500 K

? Assume constant pressure.
Physics
1 answer:
Valentin [98]3 years ago
8 0

Answer:

48.63 mL

Explanation:

Using Charles's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 29 mL

V₂ = ?

T₁ = 25 °C

T₂ = 500 K

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using above equation as:

\frac{29\ mL}{298.15\ K
}=\frac{V_2}{500\ K}

V_2=\frac{29\cdot \:500}{298.15}\ mL

<u>New volume = 48.63 mL</u>

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uniform solid sphere of radius R rotates about a diameter with an angular speed 536 radians/second. The sphere then collapses un
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Answer:

2144 rad/s

Explanation:

R1 = R

ω1 = 536 rad/s

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ω2 = ?

Mass is M

By use of angular momentum remains constant if no external force is acting on the body.

I1 ω1 = I2 ω2

The moment of inertia of solid sphere is 12/5 MR^2

So, 2/5 x M R^2 x 536 = 2/5 x M (R/2)^2 x ω2

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

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R2 = L

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F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

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the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

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