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elena55 [62]
3 years ago
14

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 1783

N with an effective perpendicular lever arm of 2.80 cm , producing an angular acceleration of the forearm of 120.0 rad / s 2 . What is the moment of inertia of the boxer's forearm
Physics
1 answer:
denis23 [38]3 years ago
6 0

Answer:I=0.416kgm^2

Explanation:

Torque is the rotational equivalence of Force and mathematicallly, it is given by

T = r X F

Where

T=Torque

r= perpendicular distance between the applied force and the axis

F =force

T= 0.028 X 1783 =49.924 Nm

T=αI

Where α=angular acceleration produ ed

I=moment of inertia

I= T/α

I= 49.924/120

I= 0.416kgm^2

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2 years ago
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Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,
Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of  motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of  motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0
3 years ago
All of the following measure mass except
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A centimeter cannot measure mass because mass is measured by grams. 

Answer: Centimeter 
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3 years ago
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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
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