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3 years ago

A certain wave has a compressions and rarefactions.How should this wave be classified?

Physics

1 answer:

AlexFokin [52]3 years ago

7 0

A) as a longitudinal wave

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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:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

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

4 0

3 years ago

A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire

gladu [14]

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

$V_T^2=v_N^2-v_W^2$

$V_T = \sqrt{v_N^2-v_W^2}$

Travel north 2mph and west to 1mph, then,

$V_T = \sqrt{2^2-1^2}$

$V_T = \sqrt{3}$

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

$v= \frac{\Delta x}{t}$

$t = \frac{\Delta x}{v}$

$t = \frac{2*(1mile)}{\sqrt{3}mph}$

$t = 1.15hour$

Therefore the total travel time for the man is 1.15hour.

3 0

3 years ago

A particle of mass 2kg resting on a smooth table attached to a fixed point on the table by a rope 1.0m making 300revolution per

zepelin [54]

Answer is: 1973.17N aprox.
step by step in the pic below

7 0

3 years ago

The energy photon affects which property of light

Triss [41]

Answer:

photoelectric effect

Explanation:

When the energy from photons is absorbed by matter, the matter can emit electrons. This process is called the photoelectric effect. The photoelectric effect is a property of light that is not explained by the theory that light is a wave.

4 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :