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igomit [66]
3 years ago
13

How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

60 mph?
Physics
1 answer:
skelet666 [1.2K]3 years ago
8 0
There is more kinetic energy to stop in a 60 mph car and therefore you would have a longer stop distance.
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The continental crust has the average composition of _____.
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An airplane flies 1760 km in 2 hours 45 minutes. What is its average speed in kilometers per hour?
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If the atomic weight- # of protons = # of neutrons, then how many neutrons does neon have?
chubhunter [2.5K]

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atomic weight 20.180

protons 10

neutrons 2,8

Explanation:

3 0
3 years ago
Use the formula from Newtons second law, namely Fnet=∆t=∆P that the Net force exerted on the object is equal to the product of t
Aleksandr-060686 [28]

Answer:

Explanation:

Fnet = Δt = ΔP

Fnet = ΔP/Δt

= mv - mu / t

= m(v - u) / t

But we know v = u + at

=> a = v - u / t

Therefore,

Fnet = ma

6 0
2 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

8 0
3 years ago
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