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olga_2 [115]
3 years ago
13

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2

. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

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An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump
ruslelena [56]

Answer:

80%

Explanation:

Efficiency = Power output / Power input × 100 %

To calculate efficiency we need to find power output of electric pump.

We can use,

Work done = Energy change

Work done per second = Energy change per second

Work done per second = Power

Therefore, Power = Energy change per second

                              = Change in potential energy of water per second

                              =mgh / t

                              = 200× 10×6 / 10

                              = 1200 W = 1.2 kW

Now use the first equation to find efficiency,

Efficiency = \frac{1.2}{1.5} × 100%

                = 80 %

8 0
3 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

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Answer:

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