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Marat540 [252]
3 years ago
10

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from

its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position
Physics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

615 N

Explanation:

If θ is the angle from vertical and T is the rope tension

θ = arcsin (2.0 / 3.5) = 34.85°

Summing vertical forces to zero

Tcosθ - mg = 0

T = 90.0(9.81) / cos34.85 = 1,075.85 N

Summing horizontal forces to zero

F - Tsinθ = 0

F = 1075.85sin34.85 = 1075.85(2.0/3.5) = 614.772... ≈ 615 N

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Answer:

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Explanation:

We use the kinematics equation to solve this question:

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v_{o}=0    because the ball is dropped

a=-g         the acceleration is the gravity, negative because it points downwards

y_{o}=h     initial height

y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

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Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?
MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

<h3><u>Data:</u></h3>
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  • t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:

\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }

\bf{d=46 \ m}

Therefore, the speed at 10 seconds is 46 meters.

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6 0
2 years ago
You are testing a new amusement park roller coaster with an empty car with a mass 120 kg. One part of the track is a vertical lo
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Answer:

- 5436 J

Explanation:

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radius of loop, r = 12 m

velocity at the bottom (A) = Va = 25 m/s

Velocity at the top(B) = Vb = 8 m/s

Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m

by use of Work energy theorem

Work done by all the forces = change in kinetic energy of the body

Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A

- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)

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4 0
3 years ago
If the phase angle for a block–spring system in SHM is ϕ and the block's position is given by x = xm cos(ωt + ϕ), what is the ra
matrenka [14]

<h2>K.E/P.E = m/k  tan²φ x ω²</h2>

Explanation:

The given position of block x = x₀ cos(ωt + φ)

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The potential energy of spring = 1/2 k x² , where k is the spring constant

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When t = 0

K.E = 1/2 m x₀²sin²φ x ω²

P.E = 1/2 k x₀² cos²φ

Dividing these , we have

K.E/P.E = m/k  tan²φ x ω²

7 0
3 years ago
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