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NNADVOKAT [17]
3 years ago
13

When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp

ring?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

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Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

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The phase at the second point is  \phi _2  = 1.57 \  rad

Explanation:

From the question we are told that

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   \phi _2 -  \phi _1 =   \frac{2 \pi }{\lambda } (r_2 - r_1 )

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    \phi _2  =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) +   3.142

   \phi _2  = 1.57 \  rad

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