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3 years ago

If a wave has a wavelength of 10 m and a speed of 2.5×108 m/s (this is 250000000 m/s) what is the frequency of the wave?

Physics

1 answer:

lakkis [162]3 years ago

5 0

Answer:

The wavelength result is 3 m. Most common velocities: Light in vacuum (air) = 300,000 km/s. Light in water = 225,000 km/s. Sound in air = 340 m/s. Sound

Explanation:

the answer is 3m

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Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d

xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

$F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F = 0.0111 \ N$

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

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3 years ago

And 8 kg bowling ball is rolling along the frictionless alley

VLD [36.1K]

It will stop eventually

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3 years ago

Choose ALL of the answers that would support the theory of Continental Drift?

saveliy_v [14]

The answers are B, D, E!

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3 years ago

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On a distant planet, the GPE of a 65 kg astronaut is 11,115 j when they are on a 46 m tall cliff. What is the acceleration due t

Ainat [17]

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Explanation:

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3 years ago

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

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3 years ago