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Leno4ka [110]
3 years ago
15

A body moves due north with velocity 40 m/s. A force is applied

Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer:

the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Explanation:

The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:

\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v

p: momentum

m: mass

\Delta v:  change in the velocity

The sign of the change in the velocity determines the direction of rate of change. Then you have:

\Delta v=v_2-v_1

v2: final velocity = 35m/s

v1: initial velocity = 40m/s

\Delta v =35m/s-40m/s=-5m/s

Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

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Answer:

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Explanation:

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         Em_f = K = ½ m v²

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to find the friction force let's use Newton's second law

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        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

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          θ = 5.74º

         sin 5.74 = Wₓ / W

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         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

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Work is friction force is

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Let's use the relationship of work with energy

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        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

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Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm
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Answer:

a

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b

The distance is  D = 22.4  \  m

Explanation:

From the question we are told that

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     The time taken is  t = 1.4 \ s

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substituting values

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Answer:

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This is how Newton's second law of motion is verified.

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Answer:

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2. True

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Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

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