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3 years ago

What happens to the pressure in a tire if air is slowly leaking out of the tire? explain your answer?

1 answer:

3 0

It's pretty simple. When air is leaking out of a tire, like a tiny hole or something, the pressure in the tire decreases, because without air in the tire, there is no pressure.

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3 years ago

Read 2 more answers

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

2 years ago