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3 years ago

A medium is a substance or material in which a wave can travel through. True or False

Physics

2 answers:

pogonyaev3 years ago

8 0

The answer is:

True

Hope this Helps

Please mark my nswer as brainliest?

Andru [333]3 years ago

7 0

The answer is true , for your questions hope this helps

You might be interested in

Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this

LenaWriter [7]

Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

5 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to Newton's 2nd law of motion:

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

3 years ago

Astronomers estimate that a 2

tia_tia [17]

M)³ / 6 = 4.2e9 m³
so its mass is 
M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg 
and so its KE at 16 km/s = 16000 m/s is 
KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J

# of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs
give or take.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

6 0

3 years ago

To which category does galaxy #1 belong? Why does it belong in this category?

vodka [1.7K]

Answer:

This galaxy belongs to the elliptical galaxy category. This is because it does not have spiral arms.

5 0

2 years ago

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

8 0

2 years ago