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Lady_Fox [76]
3 years ago
9

A 60.0-kg ball of clay is tossed vertically in the air with an initial speed of 4.60 m/s. Ignoring air resistance, what is the c

hange in its potential energy when it reaches its highest point?
Physics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

<em>the change in potential energy when it reaches the highest height = 634.8 J</em>

Explanation:

<em>Potential Energy</em>: This is the energy a body posses due to position.

From the law of conservation of energy,

At the highest point and lowest points, potential energy is converted to kinetic energy

I.e

Ek = Ek

Where Ek = potential energy, Ep = potential energy

Ep₁ = 1/2mu²  (potential energy at the lowest point)................ Equation 1

Ep₂ = 1/2mv² (potential energy at the highest point)............. Equation 2

ΔEp = Ep₂ - Ep₁ = 1/2mu² - 1/2mv²........................ Equation 3

Where ΔEp = change in potential energy, m = mass of the ball of clay, v = final velocity of the ball of clay, u = initial velocity of the ball of clay

Given: <em>m= 60 kg, u = 4.6 m/s v = 0 ( velocity at the maximum height)</em>

<em>Substituting these values into equation 3</em>

ΔEp = 1/2×60×4.6 - 1/2×60×0²

ΔEp = 30×21.16 - 0

<em>ΔEp = 634.8 J.</em>

<em>Therefore the change in potential energy when it reaches the highest height = 634.8 J</em>

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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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valina [46]
I believe its a mixture

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zaharov [31]
C. 90 m

30m per second... and it takes 3 seconds

3x30= 90
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A car with two people traveling down the road has a mass of 100 kg and a velocity of 5 m/s. The car pulls over and picks up two
Andrej [43]

Answer:

New_momentum = 750 [kg.m/s]

Explanation:

Momentum is defined as

M  = mass* velocity

Original momentum

M_old = 100 Kg * 5 m/s = 500 [kg.m/s]

If the car pulls over and picks up two people. The mass of the car is now 150 kg, but the velocity stays the same.

New momentum

M_new = 150 Kg * 5 m/s = 750 [kg.m/s]

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