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DedPeter [7]
3 years ago
13

A 92.0-kg skydiver falls straight downward with an open parachute through a vertical height of 325 m. The skydiver's velocity re

mains constant. What is the work done by the nonconservative force of air resistance, which is the only nonconservative force acting? (Assume that up is the positive direction.) a. +2.93 times 10^5 J b. 0J c. Answer is not obtainable, because insufficient information about the skydiver's speed is given. d. -2.93 times 10^5 J
Physics
1 answer:
natka813 [3]3 years ago
5 0

Answer: Workdone293.02KJ

Explanation: The equation to use to calculate Workdone = Change in KE + Change in PE

Assuming velocity is constant,KE becomes 0

Workdone= Change in PE=mg

W=92×9.8×325=293.02KJ

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THE ELECTROMAGNET SPECTRUM ONLY HAS WHICH SHAPE OF WAVE?
brilliants [131]

Answer: Transverse

Explanation: Transverse waves possess a vertical wave motion and a horizontal particle motion.

4 0
2 years ago
A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly
Ratling [72]

Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

v = 8.854 \ m/s

The tension in the string is:

T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N

Thus, the   tension in the string an instant before it broke = 34 N

6 0
2 years ago
What a neurology professor does = _______ brains
Natasha_Volkova [10]
That’s really easy ask your teacher and also peace happy
3 0
2 years ago
One problem when using rigid metal conduit in a residence is that the installation of the conduit may a weaken the structure b r
worty [1.4K]
The correct answer is A. Installation of rigid metal conduit requires grounding and the grounding equipment used may weaken the structure.
8 0
2 years ago
Read 2 more answers
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
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