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Degger [83]
3 years ago
8

Engineering solutions use scientific laws to predict behaviors of objects in certain situations. Will used Newton's Laws of Moti

on to predict the path of an arrow he was about to shoot. His prediction was way off. Why? A) Scientific laws can only be applied to lab experiments. Eliminate B) Scientific laws are NOT valid for some common objects. C) Scientific laws can only be applied when objects are at rest. D) Scientific laws do not account for unseen variations, like wind.
Physics
1 answer:
Alinara [238K]3 years ago
6 0

D) Scientific laws do not account for unseen variations, like wind

Explanation:

Will model in predicting the path of an arrow he was about to shoot failed because scientific laws most times do not account for unseen variations like wind.

Scientific laws are the description of an observed phenomenon in nature.

  • Most scientific laws have exceptions.
  • Exceptions in scientific laws are conditions in which the law will not hold true.
  • There are exceptions to newton's law of motion which Will did not take into account.

learn more:

Newton's law brainly.com/question/11411375

#learnwithBrainly

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Explanation:

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25 POINTS FOR CORRECT ANSWER
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You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find
wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0
3 years ago
You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m
sergey [27]

Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

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<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

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N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

v=12.65\ m.s^{-1} is the maximum velocity at which the vehicle can turn without skidding.

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