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alexgriva [62]
3 years ago
10

Suggest one reason why the bricklayer needs a higher energy diet than the computer operator

Physics
1 answer:
sashaice [31]3 years ago
7 0

Answer:

he needs more because he is doing more exercise this means he is working his body more and he needs more energy than someone he is sitting in an office. The computer operator would not need as much energy as the brick layer.

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PLEASE HELP ASAP! Any absurd answers will be reported. NO LINKS, IDC IF ITS THE ONLY WAY TO GET AN ANSWER.
svetoff [14.1K]

Answer:

There is a thing called a continental drift. It started about 200 million years ago. At first the continents were all attached, this super continent was called pangaea. Continental drift occurs because of the shift of the tectonic plates within the earth's outer shell. The heat from within the earth triggers movement to occur. This a very slow process though. It took 200 million years for the continents to get where the are now and would probably take another 200 to collide.

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3 years ago
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A 30-g car rolls from a hill 12 cm high and is traveling at 154 cm/s as it travels along a 275 cm horizontal track. What is the
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Momentum of the car is 46.2 kgm/s
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A student is creating a table with properties of electromagnetic and mechanical waves.
katen-ka-za [31]

Answer:

The property of the wave marked X is related to the source of the wave

Explanation:

The source of of origin of waves

Electromagnetic wave are waves that consists of varying electric and magnetic field that vibrate perpendicular to each other and to the direction of propagation of the wave and they are therefore transverse waves and transfer energy

Electromagnetic waves originate from the vibration of charged particles that gives off varying electric and magnetic fields

Mechanical waves are defined as waves that require a material medium such as air, water, metal, plastic, stretched leather, or wood to propagate

Mechanical waves originate from vibration of the particles of a medium

Sound waves which is a form of longitudinal mechanical waves that propagates by the vibration of the particles of a given medium about a point parallel to the direction of propagation of the wave.

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A motorcyclist changes the velocity of his bike from 20.0 meters/second to 35.0 meters/second under a constant acceleration of 4
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You can use the equation V=Vo+at since the acceleration is constant. Plugging in the values you know, you will get an answer of 3.75 seconds
8 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
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