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Fantom [35]
3 years ago
10

Metal sphere having an excess of +5 elementary charges has a net electric charge of

Physics
1 answer:
Natasha2012 [34]3 years ago
5 0

Answer:

q=+8.01\cdot 10^{-19}\ coulombs

Explanation:

<u>Elementary charge</u>

The elementary charge, denoted by the symbol e is the electric charge carried by a proton or, equivalently, the magnitude of a negative electric charge carried by an electron, which has charge −e.

The value of the elementary charge is a fundamental constant in physics:

\mathbf{e}=1.60217662 \cdot 10^{-19}\ coulombs

If a metal sphere has an excess of +5 elementary charge, then it has a net charge of:

q=5*\mathbf{e}=+5*1.60217662 \cdot 10^{-19}\ coulombs

\boxed{q=+8.01\cdot 10^{-19}\ coulombs}

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What is (a) the wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz?
yulyashka [42]

The wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz will be  31.25 * 10^{2} m^{-1} and 0.032 * 10^{2} m respectively

Forms of electromagnetic radiation like radio waves, light waves or infrared (heat) waves make characteristic patterns as they travel through space. Each wave has a certain shape and length. The distance between peaks (high points) is called wavelength.

Wavenumber, also called wave number, a unit of frequency, often used in atomic, molecular, and nuclear spectroscopy, equal to the true frequency divided by the speed of the wave and thus equal to the number of waves in a unit distance.

wavelength = ?

frequency = 92 m Hz = 92 * 10^{6} Hz

speed of light = 3 * 10^{8} m/s

speed of light = frequency * wavelength

wavelength = speed of light  / frequency

                     = 3 * 10^{8}  / 92 * 10^{6}

                     = 0.032 * 10^{2} m

wavenumber = 1 / wavelength

                      = 1 / 0.032 * 10^{2} m

                      = 31.25 * 10^{2} m^{-1}

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5 0
1 year ago
Which statement about dwarf planets is true?
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"Pluto was the first dwarf planet to be discovered" is the one statement among the following choices given in the question that is true <span>about dwarf planets. The correct option among all the options that are given in the question is the first option or option "a". Pluto was classified as a planet at first but in the year 1930 it was classified as a dwarf planet.</span>
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Brainliest if correct Question 10 of 10
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D thermal energy moves from the marshmallow to your fingers as you touch the marshmallow

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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

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For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.

The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:

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For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.

The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.

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