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Katena32 [7]
3 years ago
14

What kind of small objects composes much of the universe?

Physics
1 answer:
Inga [223]3 years ago
7 0
It’s atoms thank me later (give me brailiest)
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Bill Nye help.. I didn’t get link to vid, hoping y’all have seen it. 20 POINTS!
Lyrx [107]

Answer:

Potential, Kinetic and Chemical energy.

Explanation:

btw, congratulations for turning into an expert.

6 0
3 years ago
Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0
3 years ago
A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit
Kipish [7]

Answer:

F = 47.6 N

Explanation:

  • Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

       F = \frac{\Delta p}{\Delta t}

  • So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and  divide it by the time interval , as follows:

       F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N

       ⇒ Fsk = 47.6 N (normal to the wall)

3 0
3 years ago
A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara
otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

r=2.3*sin70=2.161m (ball rotates in a circle)

If the system is in equilibrium, the tension is:

Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}

Replacing:

\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}

Replacing:

v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s

7 0
3 years ago
Two wires of the same metal having the same cross-
USPshnik [31]
The answer is true

Step by step explanation:
3 0
2 years ago
Read 2 more answers
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