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tatyana61 [14]
3 years ago
7

In a shipping yard, a crane operator attaches a cable to a 1240-kg shipping container and then uses the crane to lift the contai

ner vertically at a constant velocity for a distance of 37 m. Determine the amount of work done by each of the following.
(a) the tension in the cable
(b) the force of gravity
Physics
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

a) W_T=449624\ J

b) W_g=-449624\ J

Explanation:

Given:

mass of the container, m=1240\ kg

displacement of the container, h=37\ m

Since the load is lifted using cable there will be weight acing downwards and the tension in the cable will act upward which will be equal in magnitude to the weight.

So,

T=m.g (upwards)

T=1240\times 9.8

T=12152\ N

Now the work done by the tension:

W_T=T\times h

W_T=12152\times 37

W_T=449624\ J

Now the work done by the gravity:

W_g=m.g.(-h) since displacement is in opposite direction to the gravity

W_g=1240\times 9.8\times (-37)

W_g=-449624\ J

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a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

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P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

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So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

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when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

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