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xz_007 [3.2K]
3 years ago
6

Which could describe the motion of any object

Physics
1 answer:
Alex17521 [72]3 years ago
7 0

AnsA body is said to be at motion if it changes it's position with respect to it's environment .

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A ball is thrown straight up from point A, reaches a maximum height at point B, and then falls back to point C. Which of the fol
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A father racing his son has 1/4 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.2 m
sergejj [24]

Answer:

Explanation:

KE_s: Kinetic Energy Son

KE_f: Kinetic Energy Father.

Relationship

KE_f: =  (1/4) KE_s

m_s: = (1/3) m_f

v_f: = velocity of father

v_s: = velocity of the son

Relationship

1/2 mf (v_f + 1.2)^2 = 1/2 m_s (v_s)^2      Multiply both sides by 2.

mf (v_f + 1.2)^2 = m_s * (v_s)^2               Substitute for the mass of the m_s

mf (v_f + 1.2)^2 = (m_f/3) * (v_s)^2         Divide both sides by father's mass

(v_f + 1.2)^2 = 1/3 * (v_s)^2                      multiply both sides by 3

3*(v_f + 1.2)^2 = (v_s) ^2                         Take the square root both sides

√3 * (v_f + 1.2) = v_s

Note

  • You should work your way through all the cancellations to find the last equation shown about
  • We have another step to go. We have to use the first relationship to get the final answer.

KE_f = (1/4) KE_s                                                  Multiply by 4

4* KE_f = KE_s                                                     Substitute (again)

4*(1/2) m_f (v_f + 1.2)^2 = 1/2* (1/3)m_f *v_s^2   Divide by m_f

2* (v_f + 1.2)^2 = 1/6 * (v_s)^2                              multiply by 6

12*(vf + 1.2)^2 = (v_s)^2                                        Take the square root

2*√(3* (v_f + 1.2)^2) = √(v_s^2)

2*√3 * (vf + 1.2) = v_s

Use the second relationship to substitute for v_s so you can solve for v_f

2*√3 * ( v_f + 1.2) = √3 * (v_f + 1.2)                     Divide by sqrt(3)

2(v_f + 1.2) = vf + 1.2

Edit

2vf + 2.4 = vf + 1.2

2vf - vf + 2.4 = 1.2

vf = 1.2 - 2.4

vf = - 1.2

This answer is not possible, but 2 of us are getting the same answer. The other person is someone whose math I would never question. She rarely makes an error. And I do mean rarely. Could you check to see that you have copied this correctly?

5 0
4 years ago
A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar
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Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

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Answer:

?

Explanation:

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