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4 years ago

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

Physics

1 answer:

Nesterboy [21]4 years ago

8 0

Answer:

A.) 1372 N

B.) 1316 N

C.) 1428 N

Explanation:

Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:

a. The load moves downward at a constant velocity

At constant velocity, acceleration = 0

T - mg = ma

T - mg = 0

T = mg

T = 140 × 9.8

T = 1372N

b. The load accelerates downward at a rate 0.4 m/s??

Mg - T = ma

140 × 9.8 - T = 140 × 0.4

1372 - T = 56

-T = 56 - 1372

- T = - 1316

T = 1316N

C. The load accelerates upward at a rate 0.4 m/s??

T - mg = ma

T - 140 × 9.8 = 140 × 0.4

T - 1372 = 56

T = 56 + 1372

T = 1428N

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Greeley [361]

Answer:

The equivalent resistance of the combination is R/100

Explanation:

Electric Resistance

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

$\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

$\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}$

This sum is repeated 10 times. Operating each term:

$\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}$

All the terms have the same denominator, thus:

$\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}$

Taking the reciprocals:

$R_e=R/100$

The equivalent resistance of the combination is R/100

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