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VashaNatasha [74]
3 years ago
11

Peter was holding a Human Physiology textbook that weighs about 4lbs with his non- writing arm for one minute. His elbow remaine

d exactly at a 45° isometric position. He did not complain of any muscle fatigue or muscle weakness. a) i. Is the number of activated motor units stay the same while he was holding the book for two minutes? (0.5 pt) ii. Why? (0.5 pt) Increasing the period Peter is holding the texting will only cause the motor units to fire more frequently to keep the muscles from relaxing. b) i. Are the same motor units activated while he was holding the book for two minutes? (0.5 pt) ii. Why? (0.5 pt)
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

a

When peter held the book for one minute the rate at which motor unites are fired increase steeply but as the duration increases the increese in the rate at which it is being fired becomes linear so the number of activated motors stay the same but are being activated at a more rapidly

b

The same motors are activated whilest he is hold the book for 2 minutes this is because for peter to hold the book in one fixed position one specific motor units  need to be activated

Note changing the motor unites would change the positon of the hand

Explanation:

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10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
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Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

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If two monochromatic light waves undergo destructive interference, the amplitude of the resultant wave is
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Answer: The amplitude is 0. (assuming that the amplitude ot both initial waves is the same)

Explanation:

When two monochromatic light waves of the same wavelength and same amplitude undergo destructive interference, means that the peak of one of the waves coincides with the trough of the other, so the waves "cancel" each other in that point in space.

Then if two light waves undergo destructive interference, the amplitude of the resultant wave in that particular point is 0.

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A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
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The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
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Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

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