A Light spiral spring is loaded with a mass of 50g and it extends by
10cm.
(is calculate the period of
small vertical oscillations.

0.63 s

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 50 g

Extention (e) = 10 cm

Period (T) =?

Next, we obtained 50 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

50 g = 50 g × 1 Kg / 1000 g

50 g = 0.05 kg

Next, we shall convert 10 cm to m. This is illustrated below:

100 cm = 1 m

Therefore,

10 cm = 10 cm × 1 m / 100 cm

10 cm = 0.1 m

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass = 0.05 Kg

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

Force (F) =?

F = mg

F = 0.05 × 9.8

F = 0.49 N

Next, we shall determine the spring constant of the spring.

Extention (e) = 0.1 m

Force (F) = 0.49 N

Spring constant (K) =?

F = Ke

0.49 = K × 0.1

Divide both side by 0.1

K = 0.49 /0.1

K = 4.9 N/m

Finally, we shall determine the period as follow:

Mass = 0.05 Kg

Spring constant (K) = 4.9 N/m

Pi (π) = 3.14

Period (T) =?

T = 2π√(m/k)

T = 2 × 3.14 × √(0.05 / 4.9)

T = 6.28 × √(0.05 / 4.9)

T = 0.63 s

Thus, the period of oscillation is 0.63 s

the rearrangement of particles in a physical change. hope this

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.