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2 years ago

11

A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m

ass is released, separating it 5 cm from its equilibrium position. Find the period of the motion

1 answer:

Finger [1]2 years ago

8 0

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

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Natasha2012 [34]

Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

ΣPbefore = ΣPafter

$(m_{1}*v_{1}) - F*t = (m_{1}*v_{2})$

where:

m₁ = mass of the hammer = 0.15 [m/s]

v₁ = velocity of the hammer = 8 [m/s]

F = force [N] (units of Newtons)

t = time = 0.0015 [s]

v₂ = velocity of the hammer after the impact = 0

$(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]$

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A child swings on a playground swing. How many times does the child swing through the swing's equilibrium position during the co

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<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

Explanation:

One period means time taken to complete one revolution.

In case of swings in one period time it travels the same position through two times.

Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

For 1 period = 2 times

For 3 periods = 3 x For 1 period

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