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yaroslaw [1]
3 years ago
8

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

he moment arm of the elbow flexor muscles is 4 cm. how much force must be produced by these muscles to hold the 200 n barbell in static equilibrium?
Physics
1 answer:
CaHeK987 [17]3 years ago
4 0
Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is                  τ = F*r                     = 200 N*0.4 m                     = 80 N.mat equilibrium condition ,    Στ = force*distance - 80 N.m = 0             F'*0.4 - 80 N.m = 0             F'*0.4 = 80          force F' = 200 N
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Answer:

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Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

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Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
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D. The liver's role is to remove harmful substances from the blood.
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The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at
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Answer:

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vovangra [49]

Answer:

Net work done, W=3.08\times 10^5\ J      

Explanation:

It is given that,

Mass of the plane, m=7.5\times 10^3\ kg  

Acceleration of the plane, a=1.2\ m/s^2 (upwards)

Distance covered, d = 34.3 m

We need to find the net work done on the plane as it accelerates upward. The product of force and the distance covered is equal to work done. It is given by :

W=F\times d

W=m\times a\times d

W=7.5\times 10^3\ kg \times 1.2\ m/s^2 \times 34.3\ m

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W=3.08\times 10^5\ J

So, the net work done on the plane is 3.08\times 10^5\ J. Hence, this is the required solution.

3 0
3 years ago
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Shalnov [3]

Answer:

D. 66.4

Explanation:

So this problem uses SOHCAHTOA or the three trig functions.

Specifically this uses cosine, because it has an adjacent and a hypotenuse.

First you would determine what to do on the calculator, and since the problem is saying so, use the inverse cosine button. This will give you a angle measure from the decimal.

On a calculator, type in cos^-1(6/15). I used 6/15 because cosine is adjacent over hypotenuse. This will give you 66.4, which is D on the answers.

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3 years ago
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