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Rufina [12.5K]
3 years ago
11

Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a meta

l target, producing X-rays. (a) How many electrons per second strike the target if the current is 0.500 mA? (b) What charge strikes the target in 0.750 s?
Physics
1 answer:
Paul [167]3 years ago
3 0

Answer:

(a) Number of electrons per second strike the target is 3.125 x 10¹⁵.

(b) Charge that strikes the target in 0.750 s is 3.75 x 10⁻⁴ C .

Explanation:

(a) Let n be the number of electrons per second strike the target. We know that current (I) is given by the relation :

I = n x e

Here e is charge of electron.

Substitute 0.500 x 10⁻³ A for I and 1.6 x 10⁻¹⁹ C for e in the above equation.

0.500 x 10⁻³ = n x 1.6 x 10⁻¹⁹

n = 3.125 x 10¹⁵

(b) Let q be the charge that strikes the target. We know that :

q = I x t

Here I is current and t is time.

Substitute 0.500 x 10⁻³ A for I and 0.750 s for t in the above equation,

q = 0.500 x 10⁻³ x 0.750

q = 3.75 x 10⁻⁴ C

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Answer

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where as activation energy of the product and the reactant decreases.

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The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
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Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

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The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

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3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
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A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

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At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

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The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

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Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

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