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Please Need An answer ASAP! Thanks QUESTION 5 A 0.35kg object is placed at the e

ID: 586617 • Letter: P

Question

Please Need An answer ASAP! Thanks

QUESTION 5

A 0.35kg object is placed at the end of a vertical spring with a spring constant k=21.85N/m . Assuming the spring is massless, what would be the period of oscillations (in s) ?

1 points   

QUESTION 6

Assume that after completing the first activity in this lab you have this graph for the weight (force of the spring) in N as a function of change in the length in meters. What is the spring constant?

21.9 N/m

3.47 N/m

0.03 N/m

2.2 N/m

1 points   

QUESTION 7

In the previous problem, what is the theoretical value for the period of oscillations, if the hanging mass is 0.2kg and the spring is massless?

2s

1s

0.6s

1.4s

1 points   

QUESTION 8

In the previous problem what is the theoretical value for the period of the oscillations, if the mass is 0.2 kg and the mass of the spring is 0.15 kg?

2.1 s

1.2 s

0.67 s

1.5 s

1 points   

QUESTION 9

The period of oscillations for a mass spring system would ___ if the experiment was done on the surface of the moon. NOTE: The acceleration due to gravity on the surface of the moon is 1/6th of the acceleration due to gravity on Earth.

increase by a factor of 6

double

quadruple

stay the same

1 points   

QUESTION 10

The potential energy of a mass-spring system when the spring is fully compressed and the mass is at rest is 200 J. After releasing the mass, assuming there is no dissipative force, the system will oscillate. At a point during the oscillation the potential energy of the system is 50 J. What is the kinetic energy of the mass at that point, assuming the spring is massless?

50 J

250 J

150 J

200 J

21.9 N/m

3.47 N/m

0.03 N/m

2.2 N/m

Explanation / Answer

As per the Chegg rules only 4 question can post once.

Question 5

T = 2 x sqrt (m / k)

k =21.85N/m

m= 0.35kg

T = 2x3.14x sqrt ( 0.35 / 21.85)

T=6.28 x 0.127

T=0.795Sec

Question 6:

F=ma = -kx

(0.35 x 9.8 ) = F =3.43N

x= F/ -k = 3.43/ (-21.85) = -0.157m

F= 21.9; 3.47; 0.03; 2.2 N

x= F/ -k = -1.0023m ; -0.1588m; -1.37x10-3m; -0.1007 m

Question 7:

T = period = 2*pi*sqrt(L/g)

question 9:

period of oscillation is double