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Nat2105 [25]
3 years ago
9

The 25-lb slender rod has a length of 6 ft. using a collar of negligible mass, its end a is confined to move along the smooth ci

rcular bar of radius 3√2 ft. end b rests on the floor, for which the coefficient of kinetic friction is b = 0.4. if the bar is released from rest when  = 30, determine the angular acceleration of the bar at this instant.
Physics
1 answer:
sp2606 [1]3 years ago
3 0
Which of the following is NOT a factor in efficiency?<span><span>A.metabolism</span><span>B.type of movement</span><span>C.muscle efficiency</span><span>D.digestion</span></span>
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Explanation:

it's a unit used to measure charge (C)

1C=1000millicoulombs

1millicoulomb=1000microcoulumbs

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A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory (ignorin
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Answer:

b) Gravity

Explanation:

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What is the importance of protein to the health of a human being
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A cup has a mass of 0.0650 kg and a
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3 years ago
(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi
nydimaria [60]

Answer:

ks= 133.2 N/m

Explanation:

  • Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
  • This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
  • When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
  • Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
  • If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
  • So, we can write the following equation:

       \Delta U + \Delta K = 0 (1)

  • where ΔU = -1/2*k*(Δx)²  (2)
  • and ΔK = 1/2*m*v² (3)
  • Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

       k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)

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