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antiseptic1488 [7]
3 years ago
9

The loaded cab of an elevator has a mass of 3.0 x 10 3 kg and moves 200 m up the shaft in 20 s at constant speed. At what averag

e rate does the cable do work on the cab
Physics
1 answer:
Alexeev081 [22]3 years ago
4 0

The average rate at which the cable does work is 294,000 J/s.

The given parameters:

  • <em>mass, m = 3000 kg</em>
  • <em>height, h = 200 m</em>
  • <em>time of motion, t = 20 s</em>

The average rate at which the cable does work is calculated as follows;

P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{3000 \times 9.8 \times 200}{20} \\\\P = 294,000 \ J/s

Thus, the average rate at which the cable does work is 294,000 J/s.

Learn more about energy and power here: brainly.com/question/13387946

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
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Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber
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Answer:

The correct answers are

(a) It decreases to 1/3 L

(ii) is (c) It is constant

Explanation:

to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem

The given variables are as follows

Initial volume V1 = 1L

V2 = Unknown

Initial Temperature T1 = 300K

let us assume that the balloon is perfectly elastic

At 300K the balloon is filled and it stretches to maintain 1 atmosphere

at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

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8 0
3 years ago
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