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nadya68 [22]
3 years ago
15

man with a mass of 70 kg balances on one foot while standing on a curb. The only significant forces on the foot (taken as a whol

e structure) are the ground pushing up at a point 15.0 cm in front of the ankle joint, the calf muscle pulling upward at the heel6.0 cm behind the ankle joint, and the tibia (shin bone) exerting a force at the ankle joint. Find the magnitudesof the forces exerted on the foot by the muscle and the tibia.
Physics
1 answer:
dybincka [34]3 years ago
6 0

Answer:

a. 1715 N b. 2401 N

Explanation:

Let F = force due to calf muscle, F' = force due to tibia and N = force due to ground = weight of man = mg where m = mass of man = 70 kg and g = acceleration due to gravity = 9.8 m/s².

a. Magnitude of the forces exerted on the foot by the muscle

Since the force due to the calf muscle is 6.0 cm behind the ankle joint and the normal force due to the ground is 15.0 cm in front of the ankle joint and the force due to the tibia is at the ankle joint, taking moments about the ankle joint,

F × 6 cm + F' × 0 cm = N × 15 cm

6F = 15N = 15mg

F = 15mg/6

= 15 × 70 kg × 9.8 m/s²/6

= 1715 N  

b. Magnitude of the forces exerted on the foot by the tibia

Taking moments about the calf muscle force, we have

F × 0 cm + F' × 6 cm = N × (15 cm + 6 cm)

6F' = 21N = 21mg

F' = 21mg/6

= 21 × 70 kg × 9.8 m/s²/6

= 2401 N  

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Answer:

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3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
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Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

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