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kramer
3 years ago
5

A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N

/m. The object is displaced 5.56 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the magnitude of the force acting on the object 3.96 s after it is released
Physics
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s

Force on object is

F(t)=-mAw^{2}Cos(wt)

Substitute given values

So

F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N

So

Magnitude F(t)=26.6 N

Direction: -x

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A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?  

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3 years ago
Two people carry a heavy electric motor by placing it on a light board 2.45 m long. One person lifts at one end with a force of
maxonik [38]

Answer:

W=1055N

Explanation:

In order to solve this problem, we must first do a drawing of the situation so we can visualize theh problem better. (See attached picture)

In this problem, we will ignore the board's weight. As we can see in the free body diagram of the board, there are only three forces acting on the system and we can say the system is in vertical equilibrium, so from this we can say that:

\sum F=0

so we can do the sum now:

F_{1}+F_{2}-W=0

when solving for the Weight W, we get:

W=F_{1}+F_{2}

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allsm [11]

Answer:

No.

Explanation:

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λ is 600 nm and D is 3.5 mm . Put these values in the given formula

Resolving Power = \frac{1.22\times 600\times 10^{-9} }{3.5\times 10^{-3}}\\

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So minimum distance that can be resolved is 1.254 m.

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