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djverab
19 days ago
10

A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur

e of the capacitor. The charge on the capacitor is held fixed. How is the electric field between the plates of the capacitor affected? A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structure of the capacitor. The charge on the capacitor is held fixed. How is the electric field between the plates of the capacitor affected? The electric field becomes infinite because of the insertion of the Teflon®. The electric field becomes zero after the insertion of the Teflon®. The electric field decreases because of the insertion of the Teflon®. The electric field increases because of the insertion of the Teflon®. The electric field is not altered, because the structure remains unchanged. SubmitRequest Answer Provide Feedback Next
Engineering
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A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, t
mote1985 [299]

Answer:

The calculated result is 11.7 ft

Explanation:

You can apply the combined gas law, which incorporates Boyle's law, Charles's law, and Gay-Lussac's Law, because hydrogen demonstrates ideal gas behavior under these specific conditions.

\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}

where the subscripts indicate "p" for pressure, "V" for volume, and "T" for temperature (in Kelvin) at varying moments. Let's denote t_1 as the balloon at 150,000 ft so

p_1 = 0.14 \ lb/in^2

V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3

and T_1 = -67^\circ F = 218.15\ K.

Then t_2 represents the point at which the balloon is on the ground.

p_2 = 14.7 \ lb/in^2 and T_2 = 68^\circ F = 293.15\ K.

Based on the first equation

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}, we find

V_2 = 6701.07 ft^3 and consequently the radius turns out to be

R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft.

5 0
3 months ago
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