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kondaur [170]
3 years ago
5

On a windless day, two identical balls P and Q are dropped from the top and the middle of a tower at exactly the same time as sh

own .Which will be more: the time taken by P to reach the middle of the tower ,or the time taken by Q to reach the ground?
Physics
1 answer:
Kruka [31]3 years ago
3 0

Answer: time is the same

Explanation: the distance(H) is the same in each case .

we drop the balls , no drag force using basic kimnematics

y =gt*t/2  , yo=0 , vo=0 , y=H  , so : t= sqrt(2H/g)

comment: if distance H starts to grow....we could begin to note a difference  because of gravity g is smaller as we go up

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Here I come and we wanna go home!!!
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The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa
Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

3 0
2 years ago
A cabbie is trying to stop when he notices a fare is whistling them over. The
liberstina [14]
  • K.E=18750J
  • Mass=m=2100kg
  • Velocity=v

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

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3 years ago
The exact speed of an object in a specific instant is?
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Im not so sure but it should be the
instantaneous speed 

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<span>the arrangement of the outer planets is 
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