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3 years ago

Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them

Physics

1 answer:

denpristay [2]3 years ago

3 0

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

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What is the frequency of a photon of emr with a wavelength of 2.55x10-3m?

olga nikolaevna [1]

Answer:

f = 1.18 x 10¹¹ Hz

Explanation:

The equation used to find frequency is:

f = c / w

In this form, "f" represents the frequency (Hz), "c" represents the speed of light (3.0 x 10⁸ m/s), and "w" represents the wavelength (m).

Since you have been given the value of the constant (c) and wavelength, you can substitute these values into the equation to find frequency.

f = c / w <---- Formula

f = (3.0 x 10⁸ m/s) / w <---- Plug 3.0 x 10⁸ in "c"

f = (3.0 x 10⁸ m/s) / (2.55 x 10⁻³ m) <---- Plug 2.55 x 10⁻³ in "w"

f = 1.18 x 10¹¹ Hz <---- Divide

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1 year ago

Please help!!!

Zolol [24]

Answer:

Explanation:

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8 0

3 years ago

Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert

damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

$F = k \frac{q\cdot q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

$q+\frac{q}{2}=\frac{3}{2}q$

while the other charge will be

$q-\frac{q}{2}=\frac{q}{2}$

So, the new force will be

$F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F$

So, the force will decrease to 3/4 of its original value.

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3 years ago

Hanging from a horizontal beam are nine simple pendulums of the following lengths:

LenaWriter [7]

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

$L =\frac{g}{w^{2} }$

where g=9.8m/s

ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

$L= \frac{9.8}{2^{2} }$

L = 2.45m

when ω= 4rad/sec

$L=\frac{9.8}{4^{2} }$

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

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yawa3891 [41]

A - i think
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