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mariarad [96]
2 years ago
7

The body weighing 8kg moves in a straight line uniformly accelerated with an acceleration of 3m / s? on a horizontal surface, un

der the action of force 60N.Find a) What is the friction force acting on the body) The coefficient of friction between the body and the surface where it moves
Physics
1 answer:
Usimov [2.4K]2 years ago
3 0

Answer:

pls explain this to me

Explanation:

I will not answer it

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Explanation:

F = Gm1m2/r^2

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when do you use cos and sin in situations like these? is horizontal always cos and vertical always sin?
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Answer:

yes

Explanation:

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the horizontal line is adjacent

the vertical line is opposite

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The US Environmental Protection Agency issues a daily report for pollution levels called the __________.
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You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
Monica [59]

Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

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