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Westkost [7]
3 years ago
5

A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo

ng. Next you pull down on the weight so the spring is 25.2 cm long and you release the weight from rest. What is the period of oscillation?
Physics
1 answer:
Naddika [18.5K]3 years ago
5 0

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

\Delta L = L_f - L_i

\Delta L = 21.5 - 11.9 = 9.6 cm

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

mg = kx

0.037 (9.81) = k(0.096)

k = 3.78 N/m

Now the period of oscillation of spring is given as

T = 2\pi \sqrt{\frac{m}{k}}

Now plug in all values in it

T = 2\pi \sqrt{\frac{0.037}{3.78}}

T = 0.62 s

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Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
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Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

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c) I = 16.8 kg m²

Explanation:

a) Given data:

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a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

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2 years ago
In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar
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Time period of any moon of Jupiter is given by

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from above formula we can say that mass of Jupiter is given by

M = \frac{4 \pi^2 r^3}{GT^2}

now for part a)

r = 4.22 * 10^8 m

T = 1.77 day = 152928 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}

M = 1.9* 10^{27} kg

Part B)

r = 6.71 * 10^8 m

T = 3.55 day = 306720 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}

M = 1.9* 10^{27} kg

Part c)

r = 10.7 * 10^8 m

T = 7.16 day = 618624 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}

M = 1.89* 10^{27} kg

PART D)

r = 18.8 * 10^8 m

T = 16.7 day = 1442880 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}

M = 1.889* 10^{27} kg

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