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jeka94
3 years ago
6

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

ground. There is a 0.5 kg spherical ball in the hand. The person rotates her arm upward producing an initial angular acceleration of 10 rad/s2 . The elbow and wrist remain rigid. What are the muscle moments about the shoulder and elbow joints
Physics
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

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a. The car will not hit the tractor.

b. The car would travel a distance of 52.07 m before stopping

c. At the moment when the car stopped, the tractor is 11.5 m in front of the car.

Linear motion

From the question, we are to determine of you will hit the tractor before you stop

First, we will determine the time it will take the car to stop

From the given information,

Initial velocity, u = 27.0 m/s

a = -7 m/s² (Negative sign indicates deceleration)

v = 0 m/s (Since the car will come to stop)

From one of the equations of linear motion,

v = u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration

and t is the time taken

Putting the parameters into the equation, we get

0 = 27 + (-7)t

7t = 27

t = 27/7

t = 3.857 secs

This is the time it will take the car to stop

Now, we will determine the distance the car would travel after applying the brakes

Using the formula

S = (u + v) / t

Where S is the distance traveled

S = [(27 + 0) / 2] * 3.857

S = 13.5 * 3.857

S = 52.0695 m

S ≅ 52.07 m

This means the car would travel 52.07 m after applying the brakes

Now, we will determine the distance the tractor would have traveled when the car came to a stop

Speed of the tractor = 10.0 m/s

Time taken for the car to stop = 3.857 secs

Using the formula,

Distance = Speed × Time

Distance = 10.0 × 3.857

Distance = 38.57 m

Now, we will determine the distance between the car and the tractor when the car finally stopped

Distance between the car and the tractor = Distance ahead + Distance traveled by the tractor after the car stopped - Distance traveled by car after applying the brakes

Distance between the car and the tractor = 25 m + 38.57 m - 52.07 m

Distance between the car and the tractor = 11.5 m

Therefore, the distance the tractor was ahead of the car + the distance the tractor traveled after the car stopped is more than the distance the car traveled after applying the brakes (25 m + 38.57 m > 52.07), the car will not hit the tractor.

To know more about distance, refer: brainly.com/question/10903482

#SPJ4

<u><em>[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tractor 25.0 m ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s. you immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s2 .</em></u>

<u><em>a.will you hit the tractor before you stop?</em></u>

<u><em>b.how far will you travel before you stop or collide with the tractor?</em></u>

<u><em>c.if you stop, how far is the tractor in front of you when you finally stop?]</em></u>

7 0
1 year ago
Plz help<br>I need it fast<br>​
Rufina [12.5K]

Answer:

perpendicular to

Explanation:

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7 0
3 years ago
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What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m
pychu [463]

Complete question:

A taut rope has a mass of 0.123 kg and a length of 3.54 m. What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m if the waves are to travel at 28.0 m/s ?

Answer:

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

Explanation:

Velocity = Frequency  X wavelength

V = Fλ ⇒ F = V/λ

F = 28/0.6 = 46.67 Hz

Angular frequency (ω) = 2πF = 2π (46.67) = 93.34π rad/s

Average power supplied to the rope will be calculated as follows

P_{avg} =\frac{1}{2} \mu \omega^2 A^2 V

where;

ω is the angular frequency

A is the amplitude

V is the velocity

μ is mass per unit length = 0.123/3.54 = 0.0348 kg/m

P_{avg} =\frac{1}{2} ( 0.0348)(93.34 \pi )^2 (0.2)^2 (28) = 1676.159 watts

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

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