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kirill [66]
3 years ago
7

12in 4in 5in a. 40in2 b. 216in c. 256in d. 176in? What is the answer

Physics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

a

Explanation:

because 20 + 20 = 40 and 40 = 8 * 5

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6 0
3 years ago
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

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3 years ago
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Answers:

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2 years ago
When devising a model, scientists can only use the information available during their lifetime. This means that the current mode
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