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Sergio [31]
3 years ago
10

A force of 100.0N accelerates a 5.0 kg box at 15.0 m/s^2 along a level surface. Find the coefficient of sliding friction for the

box. After that, how far will the box have gone after 5.0 seconds of pulling it?
Physics
1 answer:
Phantasy [73]3 years ago
3 0

By balancing the force, we get :

100 - Frictional Force = ma

100 - μmg = ma

100 - μ(5)(10) = (5)(15)

50μ = 25

μ = 0.5  

Distance travelled by box in 5 seconds is :

d = ut + \dfrac{at^2}{2}\\\\d = 0 + \dfrac{15\times 5^2}{2}\\\\d = 187.5\ m

Hence, this is the required solution.

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In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a
Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

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Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

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We see these 'shifts' when we look at the spectra of stars.  "Red shift" is the change in the spectrum of a star when it's moving away from us, and "Blue shift" is the change when it's moving toward us.  These measurements are the only way we have of measuring the radial motion of stars, and their speeds toward or away from us.

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4 years ago
Read 2 more answers
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