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3 years ago

Waves transport

Physics

1 answer:

Brilliant_brown [7]3 years ago

5 0

Answer:

d)energy

Explanation:

Waves can transfer energy over distance without moving matter the entire distance. For example, an ocean wave can travel many kilometers without the water itself moving many kilometers. The water moves up and down—a motion known as a disturbance. It is the disturbance that travels in a wave, transferring energy.

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17 points

const2013 [10]

The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>

MV = mv................. Equation 1

<h3>Where:</h3>

M = mass of the cannon
m = mass of the cannonball
V = velocity of the cannon
v = velocity of the cannonball

Make v the subject of the equation.

v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>

M = 500 kg
V = 3 m/s
m = 10 kg.

Substitute these values into equation 2.

v = (500×3)/10
v = 150 m/s.

Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0

2 years ago

How does the entropy of steam compare to the entropy of ice?

Dmitriy789 [7]

The answer is; The entropy of steam is larger than because it is more disordered than ice

Entropy is synonymous to the degree of disorder or randomness of molecules in a system. The molecules of steam are far apart from each other and move randomly in the system colliding with each other. Those of ice has less kinetic energy, vibrate more or less in a fixed position in the structiure, and are arranged in a orderd fashion

3 0

3 years ago

Read 2 more answers

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is: