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faltersainse [42]
3 years ago
14

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons

tant force of 2650 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s.
Engineering
1 answer:
sammy [17]3 years ago
3 0

Answer:

  t = 0.113 s

Explanation:

In this exercise we can use Newton's second law

X Axis            F - fr = ma

Y Axis           N –W = 0

                    N = W = m g

The force of friction is

                  fr = μ N

We replace

                 F - μ m g = m a

                  a = (F- μ mg) / m

                 a = (2650 - 0.45 300 9.8) / 300

                 a = 4,423 m / s2

Now we can use the kinematics expressions, where we assume that the system starts with zero speed

                v = v₀ + a t

                t = v / a

                t = 0.5 / 4.423

                t = 0.113 s

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In consolidated AB Co 300 shares.

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3 years ago
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Explanation:

5 0
3 years ago
Problem a) – c): Use the method of joints, the method of sections, or both to solve the following trusses. Draw F.B.Ds for all y
Ghella [55]

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This is confusing sorry

Explanation:

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3 years ago
Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L
Travka [436]

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

Reactive power Q1 = P tan(\phi)

                               = 14.91\times tan(cos^{-}0.8) = 11.18 kvar


Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power P_3 = 0 due to purely capacitiveload

           Reactive power Q_3 = -20 Var

a) since all three loads are connected in parallel therefore

    The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

I = \frac{P}{\sqrt{3}V_{L}}

= \frac{34910}{\sqrt{3}\times 480} = 41.99 A

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A certain working substance receives 100 Btu reversibly as heat at a temperature of 1000℉ from an energy source at 3600°R. Refer
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Answer:

Explanation:

t1 = 1000 F = 1460 R

t0 = 80 F = 540 R

T2 = 3600 R

The working substance has an available energy in reference to the 80F source of:

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B1 = 100 * (1 - 540 / 1460) = 63 BTU

The available energy of the heat from the heat wource at 3600 R is

B2 = Q1 * (1 - T0 / T2)

B2 = 100 * (1 - 540 / 3600) = 85 BTU

The reduction of available energy between the source and the 1460 R temperature is:

B3 = B2 - B1 = 85 - 63 = 22 BTU

6 0
3 years ago
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