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2 years ago

The total mechanical energy of a basketball is 400 J. If the kinetic energy is 286 J, what must the potential energy be?

Physics

1 answer:

ch4aika [34]2 years ago

7 0

Answer:

the potential energy is 114 J.

Explanation:

Given;

total mechanical energy, E = 400 J

kinetic energy, K.E = 286 J

The potential energy is calculated as follows;

E = K.E + P.E

where;

P.E is the potential energy

P.E = E - K.E

P.E = 400 J - 286 J

P.E = 114 J

Therefore, the potential energy is 114 J.

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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 = ?

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

speed of light = frequency * wavelength

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1 year ago

1. A ball initially rolling at 10m/s comes to a stop in 25 seconds. Assuming the ball has

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Answer:

B) 125 m

Explanation:

$s = \frac{u + v}{2} t \\ s \: = \frac{10 + 0}{2} (25) \\ s \: = 125m$

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3 years ago

An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

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3 years ago