3 years ago

Which statement accurately describes impulse?

2 answers:

7 0

Answer:

Impulse is related to momentum because impulse is a term that describes object's change in momentum. The measurable quantity of momentum changing is called impulse.

netineya [11]3 years ago

4 0

Answer:

Impulse is equal to the change in momentum.

Explanation:

It was a risk,but I took it and it was the right answer

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Explain how you think an asteroid impact could affect the tilt of Earth’s axis. Explain how this effect would change Earth’s sea

a_sh-v [17]

Answer:

An asteroid impact could affect the tilt of the Earth due to the force it applies onto the planet. This would change Earth's seasons due to the fact that Earth's tilt causes seasons.

4 0

3 years ago

A cat is being chased by a dog. Both are running in a straight line at constant speeds. The cat has a head start of 3.8 m. The

Mars2501 [29]

In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s

7 0

3 years ago

Iron, nickel, and cobalt can all be made into magnetic?

ikadub [295]

Answer:

ferromagnetism

Explanation:

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2 years ago

Read 2 more answers

Winds are the main cause of ___ currents in the ocean

raketka [301]

Winds are the main cause of ocean currents in the ocean.

5 0

3 years ago

Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob

Mars2501 [29]

Answer:

$a=2.77*10^{13}m$

$R_a=5.49*10^{13}m$

Explanation:

The period of the comet is the time it takes to do a complete orbit:

T=1951-(-563)=2514 years

writen in seconds:

$2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s$

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

$T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m$

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

$R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m$

8 0

3 years ago