At about 55 meters/sec, a falling parachuter (before the parachute opens) no longer accelerates. Air friction opposes acceleration. Although the effect of air friction begins gradually, imagine that the parachuter is free
falling until terminal speed (the constant falling speed) is reached. How long would that take?

Answers

Answer 1

Answer: Well, it takes one second to get every 10 meters (if $g=10 m/s^2$ ), so one needs just 5.5 seconds. He would fall 137.5 meters (if my calculations are right) during that time.

Answer 2

Answer:

Answer:

5.6 seconds

Explanation:

I did the worksheet and got it right

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If 600 J of work is done in 3 s, what is the power? (Power: P = W/t)

Answers

P = W/t
= 600 / 3
Power = 200 Watt

Answer: 200 W

Explanation:

In physics, power is defined as the amount of work done divided by the time taken to do that work:

$P=(W)/(t)$

where W is the work done and t is the time.

In this problem, the work done is W=600 J, while the time taken is t=3 s. Substituting into the formula, we find the power:

$P=(600 J)/(3 s)=200 W$

A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applied to the string, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s, what is the moment of inertia of the pulley

Answers

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

What effect does adding more loops to the wire coil have on the electromagnet?

Answers

The strength of the electromagnet depends on the magnitude of the current in the wire and the number of turns of wire around the core. If you add more turns to the coil and keep the magnitude of the current constant, the electromagnet becomes stronger.

a 5kg mass resting on a smooth horizontal, frictionless table is connected to a cable which passes over a pulley and then is fastened to a hanging 10kg mass. find the acceleration of the two objects and the tension of the string.

Answers

T= 5a

-T +10g = 10a

-15a =10g

a=-10g/15

a= -1/2 g
Negative a means block is accelerating downwards.

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Answers

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Answers

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