How many coulombs pass through a wire that carries a current of 3.2 A for 5.0 h?

Why does a balloon filled with the air from your lungs sink to the ground but a balloon filled with Helium floats ?

hoa [83]

Helium is lighter than air from the lungs.

Explanation:

A balloon filled with the air from lungs will sink to the ground but a balloon filled with Helium floats because helium is lighter than air.

Air is a mixture of many gases.

The air from the lungs is predominantly made up of carbon dioxide and water vapor.

Gram per mole of carbon dioxide is 44. Water vapor is 18.
Compared to helium which is 4, we see that the gas is lighter.
Since the gas is lighter, it will float in air.
The main component of air around us is nitrogen gas with a gram per mole of 28.
A balloon filled with more of air from the lungs weigh more than that will filled air around and even far more than helium gas.
This will cause a helium filled balloon to float.

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4 0

3 years ago

After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?

faust18 [17]

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.

8 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$