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ehidna [41]
3 years ago
5

A young man exerted a force of 9000N on a stalled car but was unable to move it. How much work was done?

Physics
1 answer:
rodikova [14]3 years ago
7 0

Answer:

No work was done at all.

Explanation:

Work is only done when a force is exerted and it moves an object. However, when the man exerted his force on the car, the car did not move at all. Therefore, no work was done.

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evablogger [386]

Answer:

A

Explanation:

The asteroid belt is the farthest from the sun, so that means it will take the longest to orbit.

3 0
3 years ago
List five situations in which a lesser amount of friction would be beneficial. Overall, would a high or low coefficient of frict
Natalka [10]
Low coefficient of friction

1. flying a plane (friction between air and plane)
2. ice skating (friction between ice and skate blade)
3. swimming (water & skin)
4. rowing a boat (water and boat)

6 0
3 years ago
Read 2 more answers
Which of the following intermolecular forces explains why iodine (I2) is a solid at room temperature?
egoroff_w [7]
"Dispersion forces" is the one intermolecular force among the following choices given in the question that <span>explains why iodine (I2) is a solid at room temperature. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.</span>
3 0
3 years ago
A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
BartSMP [9]

Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

5 0
3 years ago
A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the
Natali [406]
There are three forces acting on the book. 
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.
4 0
3 years ago
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