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3 years ago

The velocity of a wave is primarily determined by the

Physics

1 answer:

tamaranim1 [39]3 years ago

8 0

<h2>Answer: medium </h2><h2 />

The propagation velocity of a wave depends on the medium in which it propagates.

In addition, it depends on the type of wave, for example a mechanical or electromagnetic wave.

In any case, the velocity of a wave is generally related to the value of the wavelength and the period, and can be expressed mathematically through the following formula:

$v=\frac{\lambda}{T}$

where $v$ is the velocity of the wave, $\lambda$ is the wavelength and $T$ is the period.

Therefore:

<h2>The propagation velocity of a wave depends on the <u>medium</u></h2>

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3 years ago

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2 years ago

In an arcade game, a 0.126 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and relea

bearhunter [10]

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4.156 m/s

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See attachment

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2 years ago

What is a materials ability to be dissolved in a solvent

weeeeeb [17]

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3 years ago

slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu

andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

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2 years ago