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Andreas93
2 months ago
10

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

is mu_k. During time interval t, the sled moves a distance s, starting from rest.Find the average velocity vavg of the sled during that time interval.Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities.
Physics
1 answer:
kicyunya [3.2K]2 months ago
4 0
The average velocity of the sled can be expressed with the formula vavg = s/t. Hello! The calculation for average velocity involves determining the distance traveled over time. Thus, vavg = Δx/Δt, where vavg represents average velocity, Δx is the distance traversed, and Δt is the duration of time taken. We know both the distance (s) and the time (t) required for the sled to cover that distance, which allows us to compute the average velocity using the formula vavg = s/t. Wishing you a great day!
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2 months ago
A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0
kicyunya [3294]

Answer:

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

U_1=k\dfrac{q_1q_2}{x_1}

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

W=-\Delta U

W=-(U_2-U_1)

W=(U_1-U_2)

W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Consequently, the work done by the electrostatic force on the moving charge is kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}}). Therefore, this concludes the solution.

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2 months ago
Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
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Answer:

Jari

Explanation:

To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.

For Jari's path, starting point is (0, 0) and (6, 7) is another point.

The gradient is the difference in y divided by the difference in x:

Change in y=7-0=7

Change in x=6-0=6

Thus, the slope equals 7/6.

For Jade, her first point is (0, 10) and another is (6, 16).

Change in y=16-10=6

Change in x=6-0=6

Thus, the slope equals 6/6=1.

It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.

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The entire and thorough solution is included.

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