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3 years ago

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Melanie watched the path a baseball followed after a pitcher threw it. She noticed that the ball traveled horizontally away from

the pitcher as well as downward toward the ground. What force caused the ball to accelerate in the downward direction when it was thrown?

1 answer:

Alex17521 [72]3 years ago

7 0

Answer:gravity

Explanation:

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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.

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

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3 years ago

A stationary siren emits sound of frequency 1000 Hz and wavelength 0.343 m. An observer who is moving toward the siren will meas

Ainat [17]

Answer:

f>1000Hz and wavelength=0.343 m

Explanation:

We are given that

Frequency of stationary siren,f=1000 Hz

Wavelength of stationary sound, $\lambda=0.343 m$

When a observer is moving towards the siren then the frequency increases.

Therefore,an observer who is moving towards the siren measure a frequency >1000 Hz.

The wavelength depends upon the speed of source.

But we are given that siren is stationary.

Therefore, source is not moving and then the wavelength remains same.

f>1000Hz and wavelength=0.343 m

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3 years ago