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

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

pls mark as brainliest.

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.