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SVEN [57.7K]
3 years ago
6

Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg

object is at the origin, and a 3.99-kg object is at -0.496 m.
Where is the center of mass of these objects?
Physics
1 answer:
Anna [14]3 years ago
3 0

Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28  kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4

11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806

y = \frac{10.806}{11.28} = 0.958 m

So the center of mass is 0.958 m from the origin

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What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
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The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

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F = m(\frac{v_{2}  - v_{1} }{t})

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where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, v_{1} is its initialvelocity and v_{2} is the final velocity of the object.

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A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute
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it can be said that  the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

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<h3> the speed of the east wind</h3>

Generally the equation for the distance  is mathematically given as

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Therefore

the speed of the east wind

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v=0.3608

For more information on this visit

brainly.com/question/22568180

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