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NISA [10]
2 years ago
14

Why is it important for a muscle to be attached to a fixed origin at one end and a moving insertion at the other? Discuss how th

is promotes movement, using one of the muscles you observed as an example.
Physics
1 answer:
vlabodo [156]2 years ago
6 0
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
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Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u>  m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u>  m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is  <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is  <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is  <u>1.86</u> m.

Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!

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Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I
Andrews [41]
Momentum question. This is an inelastic collision, so 

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
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5 0
2 years ago
An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

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An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco
sdas [7]
A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
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