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3 years ago

14

new york, Vermont, Massachusetts are specifically what region of the united states? and what two wind directions do you combine?

(north, south, west, east)

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

if you mean the regions during slavery, North, but if in general, East. (think East Coast)

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a heavy jar sits on top of a 3.4 m shelf with a gravitational potential energy of 180 j. What is the mass of the jar?

sp2606 [1]

Answer:

5.2941176471 kg or 5294.1 grams

Explanation:

g.p.e= mgh

g.p.e/gh=m

180j/10×3.4= m

180/34= m

5.2941 kg= m

6 0

3 years ago

Which force causes all objects with mass to attract one another?

Scrat [10]

Gravitational force

5 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

Read 2 more answers

What is the definition of the word PER?

Gnesinka [82]

Answer:

Per is a Latin preposition that means "through" or "by means of". Per (pronounced like "purr") is also an English preposition that can mean "for each", "to each", "by means of", or [lastly, annoyingly] "in accordance with"; somewhat interchangeably with the [even more annoying] "as per". In the example expressions here, it conveys "because of the rationale presented at/by".

3 0

2 years ago

Read 2 more answers

You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal

RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

$K_{total} = K_{traslation} + K_{rotation}$

let's look for each energy

linear

$K_{traslation}$ = ½ m v²

rotation

$K_{rotation}$ = ½ I w²

the moment of inertia of a solid sphere is

I = 2/5 m r²

we substitute

$K_{total}$ = ½ mv² + ½ I w²

angular and linear velocity are related

v = w r

we substitute

K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

K_{total} = m w² r² (½ + 1/5)

K_{total} = $\frac{7}{10}$ m w² r²

let's calculate

K_{total} = $\frac{7}{10}$ 6.40 16.0² 0.130²

K_{total} = 19.4 J

6 0

3 years ago