Which refers to the distance between two crest or to drafts on a transverse wave

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

$\frac{F_E}{F_G}=\frac{qE}{mg}$ (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0

3 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

A 1.0 kg object is attached to a 0.50 m string. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s.

7nadin3 [17]

Given that,

Mass of the object, m = 1 kg

It moves in a circle of radius 0.5 m with a speed of 5 m/s

To find,

The direction of the acceleration.

Solution,

Whenever an object moves in a circular path, the only force that acts on its is centripetal force which is given by the formula as follows :

$F=\dfrac{mv^2}{r}$

The centripetal acceleration acts in the direction of force. It acts along the radius of the circular path. In this figure, the direction of the acceleration is shown at point d.

4 0

3 years ago

An object moves with constant acceleration 3.45 m/s2 and over a time interval reaches a final velocity of 14.0 m/s.

Sav [38]

Assuming the object was originally at rest, it must have been traveling for
14.0/3.45 = 4.06 seconds

4 0

3 years ago

Occurs when fuel rods produce too much heat

gizmo_the_mogwai [7]

it is known as the
nuclear meltdown

7 0

4 years ago