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2 years ago

Structure and Composition of the Atmosphere

Physics

1 answer:

12345 [234]2 years ago

3 0

The correct answer is B. Nitrogen

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A scientist claimed that fabric A is better able to resist fire than fabric B. Which option describes an experiment that will pr

UNO [17]

Answer:

B. Hold each type of fabric over a candle flame and time how long it takes for the fabric to start to burn.

Explanation:

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3 years ago

If you have to apply 40 N of force to a crowbar to lift a rock that weighs 400 N, what is the actual mechanical advantage of the

Sauron [17]

The AMA is calculated as:
AMA = force obtained / force applied
AMA = 400 / 40
AMA = 10

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3 years ago

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A 5 kg pineapple is hanging completely still in mid air on a string and suddenly explodes

11111nata11111 [884]

Answer:

Explanation:

Conservation of momentum

Initial momentum is zero

3(15) + 2(v) = 0

v = - 22.5 m/s

v = 22.5 m/s downward

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3 years ago

A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The work done is?

JulijaS [17]

Work done is the distance a force acts over.

So, the work done here is 9.0N * 3.0m = 27 J

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3 years ago

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago