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galben [10]
3 years ago
13

A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *

Physics
1 answer:
frez [133]3 years ago
8 0

With velocity, you have to be careful.  With velocity, it doesn't matter how much total distance you covered while you were moving.  All that matters is the straight-line distance between the place you started from and the place where you stopped.  

If you ended in the same place where you started from, then it doesn't matter whether you drove around town all day and then came home, or ran around laps on a circular track, or zig-zagged back and forth a hundred times.  The straight distance from your start-point to your end-point is zero.  So technically, according to the defintion of velocity, it was <em>ZERO</em>.  

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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
2 years ago
When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi
ryzh [129]

Answer:

<u>The pendulum bob swing past the mean position because:</u>

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to  the restoring toque the bob starts to accelerates towards its mean postion. The  maximum acceleration of the pendulum's bob  is -w^{2} Aand the the acceleration decreases as -w^{2} x  towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0
3 years ago
The diagram below shows the stars that are nearest to our solar system.
abruzzese [7]

Answer:

no u tried of the same dam

thing

Explanation:

7 0
3 years ago
Heartburn is a form of indigestion felt as a burning sensation in the chest when stomach acid comes back up into the esophagus.
xenn [34]
The antacid is basic so it neutralizes acidity or lowers it. Then if it goes into the esophagus, it's not as strong and it doesn't hurt, and it also calms your stomach because the acidity in your stomach is also lower. Antacids are therefore taken by many people, especially as they grow older and things like heartburn become more common.
5 0
2 years ago
Read 2 more answers
Calculate how much work is required to launch a spacecraft of mass mm from the surface of the earth (mass mEmE, radius RERE) and
SIZIF [17.4K]

Answer:

Work done = (1/2)[(Gmm_e)/(R_e)]

Explanation:

I've attached the explanations below.

5 0
3 years ago
Read 2 more answers
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