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viva [34]
3 years ago
15

Based on following data, Angular acceleration of the tibia 450 rad/sec 2 The mass moment of inertia for the tibia is 0.40 Nm sec

2 The perpendicular distance between patellar tendon and the center of rotation is 0.045m Determine the muscle force acting on patellar tendon. Show all works
Engineering
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:

4000 N

Explanation:

We have given angular acceleration \alpha =450rad/sec^2

Moment of inertia = 0.40 Nm

The distance d from the center of rotation is d=0.045 m

We have to find the muscle force F

From torque equation we know that Fd=I\alpha

F\times 0.045=0.40\times 450

F=4000\ N

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Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}

where,  \gamma_m = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

\gamma_w = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation,\gamma_m = 16.62 kN/m3

puttng all value to get Gs value;

16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}

Gs - 1.055*e = 1.805 .........(2)

solving both  equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

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3 years ago
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Answer:

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Explanation:

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Answer:

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Answer:

The correct option is;

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Explanation:

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4vir4ik [10]

Answer:

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