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2 years ago

5

A bar extends perpendicularly from a vertical wall. The length of the bar is 2 m, and its mass is 10 kg. The free end of the rod

is attached to a point on the wall by a light cable, which makes an angle of 30° with the bar. Find the tension in the cable.

1 answer:

Simora [160]2 years ago

6 0

Answer:

T = 98.1 N

Explanation:

Given:

- mass of bar m = 10 kg

- Length of the bar L = 2 m

- Angle between Cable and wall Q = 30 degres

Find:

- Find the tension in the cable.

Solution:

- Take moments about intersection of bar and wall, to be zero (static equilibrium)

(M)_a = 0

T*sin( 30 )*L - m*g*L/2 = 0

T*sin(30) - m*g / 2 = 0

T = m*g / 2*sin(30)

T = 10*9.81 / 2*sin(30)

T = 98.1 N

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Answer:

$90.2^{\circ}C$

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Upon re-arranging

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$T=90.2^{\circ}C$

Therefore, the temperatures at which the metals are joined is $90.2^{\circ}C$

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Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

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From the question we are told that

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$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

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Here $T_1$ is the period of Deimos

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So

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muminat

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