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jasenka [17]
3 years ago
10

What is the length of a simple pendulum with a period of 11.5 s?

Physics
1 answer:
givi [52]3 years ago
4 0
Using the equation for period length for a pendulum, you get 32.829 meters.
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Suspend a heavy weight by two pieces of rope. Tension is greatest when the ropes
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3 years ago
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring
Dennis_Churaev [7]

Answer: 6067.5 N

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Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

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5 0
2 years ago
Momentum is usually not exactly conserved in a real world demonstration of momentum conservation. What is a possible reason for
Maksim231197 [3]

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5 0
3 years ago
Two trains A and B of length 400 m each are moving on two parallel tracks with
Vinvika [58]

<u>Answer:</u>

<em>The initial distance between the trains is 1450 m. </em>

<u>Explanation:</u>

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration  is 1m/s^2   and it accelerated for 50 seconds.

<em>a=1 m/s^2</em>

<em>t=50 s </em>

<em>initial speed u=72km/h </em>

<em>we have to convert this speed into m/s  </em>

<em>u=72 \times 5/18=20 m/s</em>

<em>Distance covered in accelerating phase  S=ut+1/2  at^2  </em>

<em>=20 \times 50+1/2 \times 1 \times 50^2</em>

<em>=1000+1250=2250 m </em>

If  a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

<em>Here length of train A+length of train B=400+400=800 m</em>

<em>Hence the initial distance between the trains = 2250-800=1450 m</em>

6 0
3 years ago
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