What do we call seismic waves that are transmitted along the outside of earth?

If a substance cannot be physically separated into component parts, it is likely to be

LiRa [457]

A pure substance because it cannot be separated into parts

8 0

4 years ago

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

$\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

4 0

3 years ago

Two very small spheres are initially neutral and separated by a distance of 0.66 m. Suppose that 5.7 × 1013 electrons are remove

andrezito [222]

Answer:

1.718 N , attractive

Explanation:

r = 0.66 m, n = 5.7 x 10^13

q1 = 5.7 x 10^13 x 1.6 x 10^-19 = 9.12 x 10^-6 C

q2 = - 5.7 x 10^13 x 1.6 x 10^-19 = - 9.12 x 10^-6 C

F = K q1 q2 / r^2

F = 9 x 10^9 x 9.12 x 10^-6 x 9.12 x 10^-6 / (0.66)^2

F = 1.718 N

As both the charges are opposite in nature, so the force between them is attractive.

3 0

3 years ago

In a certain state, electricity is generated by using Earth's heat. Wells are drilled and natural steam is taken out through pip

MArishka [77]

Answer:

Geothermal energy is used. Dry system plants are set up for harnessing this energy.

Explanation:

3 0

3 years ago

If the force is applied to a car, then its acceleration will ___________ because of Newton's _____________.

dolphi86 [110]

Increase; newton’s second law of motion

3 0

3 years ago