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Katena32 [7]
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​
Physics
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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Answer:

a)  1321.45 N

b)  1321.45 N

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Explanation:

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First of all, we need to remember the gravitational law:

F = G \frac{m_1 m_2}{r^2}

Were

   G = 6.67428*10^-11 N(m/kg)^2

   m1 and m2 are the masses of the objects

   r is the distance between the objects.

In the present case

m1 = earth's mass =  5.9742*10^24 kg

m2 = 497 kg

r = 1.92 earth radii = 1.92 * (6378140 m) = 1.2246*10^7 m

Replacing all these values on the gravitational law, we get:

F = 1321.45 N

a)  and  b)

Both bodies will feel a force with the same magnitude 1321.45 N but directed in opposite directions.

The acceleration can be calculated dividing the force by the mass of the object

c)

a_satellite = F/m_satellite = ( 1321.45 N)/(497 kg)

a_satellite = 2.66 m/s^2

d)

a_earth = F/earth's mass = (1321.45 N)/( 5.9742*10^24 kg)

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yo this is due tomorrow help ya girl out : if a 2kg mass is accelerating at a rate of 2m/s^2 on a frictionless surface. how much
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<u>Now free fall acceleration on planet X:</u>

W=m.g'

20=0.5\times g'

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