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cestrela7 [59]
3 years ago
10

The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. Suppose a 180

,000 kg locomotive is rolling at 25 m/s on level rails. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?
Physics
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

t = 0.354 hours

Explanation:

given,

coefficient of rolling friction μr=0.002

mass of locomotive = 180,000 Kg

rolling speed = 25 m/s

The force of friction = μ mg

                                 = (.002) x (180000) x (9.8)

                                 = 3528 N

F = m  a

now,

m a =  3528 N

180000 x a = 3528

a = 0.0196 m/s²

Then apply

v = u + at  

0 = 25 - 0.0196 x t

t = 1275.51 sec

t = 1275.61/3600 hours

t = 0.354 hours

time taken by the locomotive to stop = t = 0.354 hours

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When two volleyball players strike the ball on opposite sides of the net with the same amount of force, what happens?
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3 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
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Answer:

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Explanation:

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             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

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4 0
3 years ago
3–101 It is estimated that 90 percent of an iceberg’s volume is below the surface, while only 10 percent is visible above the su
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Answer:

The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

Explanation:

Given that,

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We need to calculate the density of the iceberg

Using equilibrium condition

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Put the value into the formula

\rho_{I}\times V_{I}\times g=\rho_{w}\times0.9V_{I}\times g

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Using equilibrium condition

W_{I}=F_{B}

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Put the value into the formula

\rho_{I}\times V_{I}\times g=\rho_{w}\times0.1V_{I}\times g

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5 0
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