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3 years ago

Calculate the velocity of a 1650 kilogram satellite.

1 answer:

8 0

There's not enough information given to do that calculation.

-- The question doesn't specify whether the satellite is on the
shelf in the Vehicle Assembly Building before being installed
onto the booster, or inside the nose-cone as the rocket is slowly
being rolled to the launch-pad, or on its ascent to orbit after launch,
or in orbit. Its velocity in each of these situations is different.

-- The question reveals only the satellite's mass, but the answer
doesn't depend on that number. The satellite's velocity depends
on the speed of the truck or the rocket carrying it, or the size of
the orbit it's in. The question doesn't give any of these.

==> In particular, the size of a satellite's orbit, or its speed in that
orbit, DO NOT depend on its mass.

For example:
There are hundreds of TV satellites ... the ones that match the
Earth's rotation and appear motionless in the sky. They have
many different sizes, shapes, and masses, but they're all in the
same geostationary orbit, 22,000 miles above the equator, and
they all have the same average orbital velocity, zero displacement
per (23 hours 56 minutes 4 seconds).

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A flywheel turns through 26 rev as it slows from an angular speed of 2.9 rad/s to a stop. (a) Assuming a constant angular accele

Misha Larkins [42]

a) The wheel takes 112.7 s to comes to rest

b) The angular acceleration of the wheel is $-0.0257 rad/s^2$

c) The wheel takes 33.0 s to complete 13 revolutions

Explanation:

The rotational motion of the wheel is at constant angular acceleration, so we can use the equivalent of the suvat equations for rotational motion. In particular, to find the time the wheel takes to come to a stop, we use the following:

$\theta = (\frac{\omega_f+\omega_i}{2})t$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular speed

$\omega_f$ is the final angular speed

t is the time

For the wheel in this problem:

$\omega_i = 2.9 rad/s$

$\omega_f = 0$

$\theta = 26 rev \cdot (2\pi)= 163.4 rad$

And solving for t, we find the time:

$t=\frac{2\theta}{\omega_i + \omega_f}=\frac{2(163.4)}{0+2.9}=112.7 s$

We can find the angular acceleration by using the equation:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where here we have:

$\omega_f = 0$ is the final angular speed

$\omega_i = 2.9 rad/s$ is the initial angular speed

t = 112.7 s is the time

And substituting,

$\alpha = \frac{0-2.9}{112.7}=-0.0257 rad/s^2$

And the negative sign is because the wheel is decelerating.

In order to find the time taken to complete 13 revolutions, we use the following suvat equation:

$\theta=\omega_i t+ \frac{1}{2}\alpha t^2$

where:

$\omega_i = 2.9 rad/s$ is the initial angular speed

$\theta = 13 rev \cdot (2\pi)=81.7 rad$ is the angular displacement

$\alpha=-0.0257 rad/s^2$ is the angular acceleration

Substituting into the equation:

$81.7 = 2.9 t +\frac{1}{2}(-0.0257)t^2\\0.0129t^2-2.9t+81.7 = 0$

which has two solutions:

t = 33.0 s

t = 191.8 s

And here we have to take the first solution (33.0 s), because it corresponds to the first time at which the angular displacement of the wheel is equal to 13 revolutions (the second solution is the time at which the angular displacement of the wheel returns to 13 revolutions, after the wheel has came to rest and it started to rotate in the opposite direction).

Learn more about rotational motions:

brainly.com/question/9575487

brainly.com/question/9329700

#LearnwithBrainly

5 0

4 years ago

A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away

MrMuchimi

Answer: A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.

Explanation: To find the answer, we need to know more about the Electric charges and its properties.

<h3>What is meant by electric charge?</h3>

It can be defined as the basic fundamental property of matter causes to experience a force, when it is placed in an external electric field.
The charge of the particle can be positive, negative, and zero.

<h3>What are the fundamental properties of electric charge?</h3>

Additivity of charges
Conservation of charges
Quantization of charges.
Like charges repels and unlike charges attract.
It's a scalar quantity.

Thus, from the above, we can conclude that, when a positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.

Learn more about the electric charge and its fundamental properties here: brainly.com/question/28020194

#SPJ4

3 0

2 years ago

Which happens to the magnetic field of a wire when you change the direction of the current in the wire? It becomes stronger. It

scZoUnD [109]

Answer C. The magnetic field changes direction too.

7 0

3 years ago

The photometer is a device that converts light to voltage which is read out by the digital multimeter (DMM). This is due to the

dimaraw [331]

Answer:

It converts light into electric current.

Explanation:

The first line of the description says that the photometer is a device that converts light to voltage. if the voltage can be measured then current flowing through it can be calculated by Ohms law V=IR.

<em>Hence the photometer converts light to electric current.</em>

The photometer is used to measure the strength of electromagnetic radiation falling on it and operate within the range of ultraviolet ,visible and infrared spectrum.

It converts light to electric current using photodiode (pn junction),photoresistor and photmultiplier.

8 0

3 years ago

Why sound are more audible at night then in a day ? (please hope answer will not be too long)

mojhsa [17]

<em><u>Answer:</u></em>

You can only hear it at night. The reason for this upward diversion of sound in the daytime versus the downward diversion at night is the strong dependency of the speed of sound in the atmosphere on temperature. The atmosphere acts like a lens that focuses sound energy upwards during the day, but keeps it at ground level during the night.

<em><u>Or</u></em>

<em><u>During the day, the sound bends away from the ground; during the night, it bends towards the ground. Hence at night you have additional "sound" reaching you, making it louder.</u></em><em><u> </u></em>

This is the Answer for your question :3

I hope you are having a great day ❤️❤️❤️❤️❤️

5 0

3 years ago

Read 2 more answers