Part C

Physics

1 answer:

7nadin3 [17]3 years ago

5 0

Answer:

yu counter the arguement by either walking out or confronting whats the problem then fixing that problem

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A perfectly elastic collision occurs between a 15.0-kg mass traveling at 3.50 m/s and a 9.00-kg mass traveling at 2.35 m/s. if t

BaLLatris [955]

Momentum is conserved in a collision. Momentum is mass*velocity, so you can find your answer by calculating initial and final momentums and setting them equal to each other.

15kg * 3.50 m/s + 9kg * 2.35 m/s = 73.65 kg m/s

73.65 = 9 * 2.8 + 15x

solve for x
x= 3.23

The final velocity is 3.23 m/s

5 0

3 years ago

Ike is at the beach watching the waves in the ocean. Ike notices that some of the waves are short. Other waves are very tall and

aniked [119]

Ike is at the beach watching the waves in the ocean. Ike notices that some of the waves are short. Other waves are very tall and come up high above the water. Two waves that are different heights because They have different amplitudes.

Answer: Option (D) is correct

Explanation:

The different heights of the waves are due to their different amplitudes. The Amplitude of a particular wave depends upon the amount of energy being carried by waves. It the waves carry more energy than their amplitude will be higher.

But if energy carried by a wave is less than the wave will have a low amplitude. The Amplitude shows the distance covered from the rest position to peak position.

3 0

3 years ago

What is neptune's greatest features?

Alex777 [14]

Neptune was named after the Roman god of the sea and it is the last known of the planets

8 0

4 years ago

The distance from the Earth to the Sun is known as an astronomical unit (AU). This distance is also equal to 1.496e+011 m. Neptu

Lerok [7]

Answer:

The distance from the Sun to Neptune is 29,41 AU.

Explanation:

We know, from the sentence, that the orbit of Neptune has an average diameter around 8.80*10⁹km.

Now, we can calculate the radius of this orbit, which is equivalent to the distance from thsi planet to the Sun. Let's recall tha the radius is the half of the diameter.

$R=\frac{8.80\cdot 10^{9}}{2}=4.4\cdot 10^{9}km=4.4\cdot 10^{12}m$