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MissTica
4 years ago
5

An interplanetary space probe is flying by Jupiter on its way to the Oort Cloud. The gravity of the sun is pulling it downward w

ith a force of 631 N. Jupiter is simultaneously pulling the probe sideways with a force of 255 N. What is the net force acting on the space probe?
Physics
1 answer:
miskamm [114]4 years ago
7 0

Answer:

680.6 N

Explanation:

The net force here is the resultant force. Taking that the two given forces act perpendicular to each other then the resultant is the hypotenuse. Therefore, R=\sqrt {d^{2}+s^{2}} where d is the magnitude of downward force and s is the magnitude of the sideways force. Substituting 631 N for downward force and 255 N for sideways force then R=\sqrt {631^{2}+255^{2}}=680.5776958\approx 680.6 N

Therefore, the magnitude of net force is equivalent to 680. 6 N

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Which of the following are features of the nucleus of an atom
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We r made of atom so v can’t touch anything hehe I just joking
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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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