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2 years ago

The data points you have taken on your lab graphs roughly form a straight line. How do you interpret the slope of this line?

Physics

2 answers:

nekit [7.7K]2 years ago

3 0

Answer: Steeper slopes mean lower speeds.

Explanation:

Ann [662]2 years ago

3 0

If it’s a flat slope then it’s c
But if it’s a wavy slope then it’s a

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10. Ahmad drives 60 km from his house to the petrol station in a car moving at a constant speed of

djyliett [7]

Answer:

Explanation:

Average speed is distance traveled over time taken to do so

home to station

d = 60 km

t = 60 km/60 km/hr = 1 hr

station to shop

d = 60 km

t = 60 km / 120 km/hr = ½ hr

speed

s = (60 + 60) / (1 + ½) = 120/1.5 = 80 km/h

7 0

2 years ago

What is the difference between speed and volocity?

Sliva [168]

The average speed is the distance per time ratio. velocity is the rate at which the position changes

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2 years ago

Find the work done lifting a 290-pound weight 28 feet in the air.

Nataly [62]

The answer is d.8,120 foot-pounds

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3 years ago

Read 2 more answers

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

2 years ago

Read 2 more answers

In The United States how much more sugar is the average person consume each year that in 1970?

zmey [24]

In the 1970, the average American ate only 2 pounds of sugar a year. In 1970, we ate 123 pounds of sugar per year. Today, the average American consumes almost 152 pounds of sugar in one year. This is equal to 3 pounds (or 6 cups) of sugar consumed in one week!

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3 years ago