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2 years ago

A satellite with a mass of 110 kg and a kinetic energy of 3.0 ´ 109 J must be moving at a speed of

2 answers:

8 0

Answer: A satellite with a mass of 110 kg and a kinetic energy of 3.08×10^9 J must be moving at a speed of 7483 m/s.

Explanation: To find the answer we need to know about the kinetic energy of a body.

<h3>How to solve the problem the equation of kinetic energy?</h3>

We have the expression for kinetic energy of a body as,

$KE=\frac{1}{2}mv^2$

Given that,

$m=110kg\\KE=3.08*10^9J\\$

We have to find the speed of the satellite,

$v=\sqrt{\frac{2KE}{m} } =\sqrt{\frac{2*3.08*10^9}{110} } =7.483*10^3 m/s$

Thus, we can conclude that, the velocity of the satellite will be 7438m/s.

Learn more about Kinetic energy here:

brainly.com/question/28105739

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pshichka [43]2 years ago

8 0

In order to move at a speed of 7483 m/s, a satellite with a mass of 110 kg and a kinetic energy of 3.08× 10^-9 J must be travelling.

Understanding a body's kinetic energy is necessary in order to determine the solution.

<h3>How can the kinetic energy equation be solved?</h3>

The phrase "kinetic energy of a body" might be used as,

$KE=\frac{1}{2}mV^2$

Given that,

$m=110kg\\KE=3.08*10^9J$

We must determine the satellite's speed.

$V=\sqrt{\frac{2KE}{m} }=7.48km/s$

Thus, we can infer that the satellite will move at a speed of 7438 m/s.

Learn more about the kinetic energy here:

brainly.com/question/28105739

#SPJ1

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What best describes a sonar wave

lawyer [7]

Sonar<span> (originally an acronym for Sound Navigation And Ranging) is a technique that uses sound propagation (usually underwater, as in submarine navigation) to navigate, communicate with or detect objects on or under the surface of the water, such as other vessels.</span>

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3 years ago

Sort the sentences based on whether they describe radio waves, visible light waves, or both.

valentina_108 [34]

The electromagnetic spectrum includes a continuous spectrum of wavelengths that include:

Radio waves, microwaves, infrared light, visible, ultraviolet, X-rays, gamma rays

The wavelength decreases from radio waves to gamma rays, whereas the energy increases along the same direction.

In the given example, radio waves have a lower energy and higher wavelength than visible light. The latter can be perceived by the human eye, whereas radio waves are not visible to the human eye.

1) They have colors = visible light

2) They can travel in a vacuum = both

3) They have energy = both

4) They’re used to learn about dust and gas clouds = radio waves

5) They’re used to find the temperature of stars = visible light

6)They’re invisible = radio waves

6 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .