Asked by Damon
i need help with this one
thanks
A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t. The speed of the particle is proportional to:
a. sq. root t
b. t
c. t^2
d.1/sq. root t
e. 1/t
Work done is proportional to the integral of a constant times t. That means the work done is proportional to t^2. It will equal the final kinetic energy, which is proportional to v^2. Therefore v^2 is proportional to t^2, and v must be proportional to t.
thanks
A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t. The speed of the particle is proportional to:
a. sq. root t
b. t
c. t^2
d.1/sq. root t
e. 1/t
Work done is proportional to the integral of a constant times t. That means the work done is proportional to t^2. It will equal the final kinetic energy, which is proportional to v^2. Therefore v^2 is proportional to t^2, and v must be proportional to t.
Answers
Answered by
jonah
i choose answer d.
Answered by
ethan
i got the same problem, and the answer is A. I have no idea why
Answered by
Jacksq.
the answer is A
work is proportional to t.
work is the change in kinetic energy.
then work is of the order of V^2.
then V^2 is prop. to t.
therefore, V is prop. to sq.root t
work is proportional to t.
work is the change in kinetic energy.
then work is of the order of V^2.
then V^2 is prop. to t.
therefore, V is prop. to sq.root t
Answered by
Kiya
Draw a sketch to show the force acting on the climber
Answered by
Gary Yeager
the sum of 3 and C
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