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dsp73
3 years ago
7

What is the velocity of an object that has a momentum of 4000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.

Physics
1 answer:
insens350 [35]3 years ago
5 0
The answer is C. You divide 4000 kg/s by 115 kg.
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Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds
solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0
2 years ago
Newton’s 3rd law of motion states that “every action has an equal and opposite reaction.” A golf ball was hit with a force of 20
DedPeter [7]

Answer:

<u>Example of Newton's III law</u>

  • In the, golf the ball was hit by a club with certain force. As the club hits the ball it's the action. When the ball flies away its the reaction.
  • When a person swings a golf club at the ball, when it hits the ball, it causes the ball to roll up the face of the club and into the air towards the target.
8 0
3 years ago
Does A= 6.4 B=12 C=12.2 Form a RIGHT TRIANLGE?
Lina20 [59]
To answer this question, you must remember the equation:

a²+b²= c²

(6.4)² + (12)²=   (12.2)²

<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84

This cannot be a triangle

hope this helps
3 0
3 years ago
Help !!!! Estimate the number of breaths taken by a person during 44 years?
Lisa [10]
44 x 12. I got the 12 from the total of 12 months in a year.


44 > 40

x
12 > 10
----------
The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.

40 x 10 = 400.

Therefore, your estimate is 400.

The real answer is 520 breaths.
6 0
3 years ago
A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
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