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Leokris [45]
3 years ago
9

PLEASE HELP!!

Physics
1 answer:
BartSMP [9]3 years ago
4 0

Answer:A) 19.8 seconds

B) 467.5 metres

Explanation: 100km/hr = 27.78m/s

So, it covers 27.78 metres in one second

Take the time it would cover 550m as x

Therefore 27.78m = 1second

550m = x

Cross multiplying

27.78x = 550m

Divide both sides by the coefficient of x

x = 550÷27.78

x = 19.8seconds

B) distance = speed × time

Speed = 85km/hr = 23.61m/s

Time = 19.8seconds

Therefore,

Distance = 23.61 × 19.8

Distance = 467.5m

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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
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The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

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To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

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<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

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<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

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Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

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3 years ago
A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

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3 years ago
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