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kherson [118]
2 years ago
15

If there were no external forces acting on the two pucks, their complex motion could be described as the combination of the unif

orm linear motion of the center of mass and a uniform circular motion of the pucks about the center of mass. Describe how well your results agree with this expectation, and explain any deviations that you observe from the predicted behavior
Physics
1 answer:
mote1985 [20]2 years ago
3 0

The absence of external forces will make the pucks move in the form of a uniform circular motion.

<h3>What is a circular motion?</h3>

It should be noted that a circular motion simply means the movement of an object along the circumference of the circle.

In this case, the absence of external forces will make the pucks move in the form of a uniform circular motion.

If the friction is absent, the pucks will continue to move on the same path due to the first law of Newton and the law of conversation of energy. In this case,the results will match the predictions until there's loss in energy.

Learn more about circular motion on:

brainly.com/question/106339

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If I was born in 1998 how old will I be
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3 years ago
A mechanic uses a mechanical lift to raise a car. The car weighs 11,000 N. The lift raises the car 2.5 m.
KengaRu [80]

Potential energy = (weight) x (height)

After the car has been raised 2.5 meters, it has

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5 0
3 years ago
Read 2 more answers
A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d)  Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
3 years ago
002 10.0 points note: report your answer in joules here. a proton and an antiproton, both at rest with respect to one another, m
Ulleksa [173]
The mass of the proton is:
m_p = 1.67 \cdot 10^{-27} kg
and the mass of the antiproton is exactly the same, so the total mass of the two particles is 2m_p.

In the annihilation, all the mass of the two particles is converted into energy, and the amount of this energy is given by Einstein's equivalence between mass and energy:
E=Mc^2
where M is the mass converted into energy and c is the speed of light. In this example, M=2m_p, therefore the energy released is
E=2m_p c^2 = 2 (1.67 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2=3 \cdot 10^{-10}J
3 0
3 years ago
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