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EastWind [94]
3 years ago
10

At its natural resting length, a muscle is close to its optimallength for producing force. As the muscle contracts, the maximumf

orce it can deliver decreases. When a muscle is at approximately80\% of its natural length, it cannotcontract much more and the force it can produce drops drastically.For a muscle stretched beyond its natural length, the same is true.At about 120\% of its natural length, the forcethat a muscle can exert again drops drastically.
This muscle length to force relationship can be demonstrated bydoing a chin-up. As you hang from the bar, your biceps muscles arestretched and can produce only a relatively small force. As you getclose to the bar, your biceps muscles contract substantially, andyou again experience difficulty. The easiest part of the chin-upoccurs somewhere in between, when your muscles are close to theirnatural length.



Imagine hanging from a chin-up bar and beginning a chin-up.Which of the following velocity versus time graphs best representsthe first part of your motion (from being at rest to beingapproximately halfway to the bar)? Assume you are trying to do thechin-up as quickly as possible.
Physics
1 answer:
lesantik [10]3 years ago
5 0

Answer:

Figure E is the correct representation of the first part of the motion. When in a hanging position from the chin-up bar, the bicep muscles are stretched beyond their normal length already. So at this point they are at the peak of their capacity and you are at rest (this corresponds to the velocity v = 0 at t = 0). On contracting the bicep muscles and pulling your whole body up, you begin to gain speed and v increases. This increase in velocity is exponential. Soon the bicep muscles contract up to 80% their normal length reducing the force they can produce to keep you rising up to zero. The velocity change happens because the body is accelerating and the muscles can still supply a net force to lift you up. The acceleration is present because of this net force. The moment this force reduces to zero, the acceleration too reduces to zero. (From Newton's second law of motion). This reduction in acceleration is responsible for the reduction of the curvature of the v curve in figure E above. The point where the velocity becomes horizontal corresponds to the point where the muscles reach their maximum contraction unit and can supply no more net force and as a result no acceleration. This further results inba constant velocity which is the flat nature of the curve seen in diagram E.

Thank you for reading.

Explanation:

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Using a complete sentence state what would most likely happen to the production of oxygen by duckweed plans if the intensity and
jeka94

Answer:

This question will be answered based on general photosynthetic understanding. The answer is:

The production of oxygen would increase

Explanation:

The characteristics of most plant forms is their ability to photosynthesize i.e. use solar energy (from sunlight) to make food (chemical energy). The product of this photosynthetic process is OXYGEN gas, which is released as a waste product via the stomata on their leaves. Note that, photosynthesis cannot occur without LIGHT as it provides the energy needed for the process.

Hence, in the duckweed plant like every other photosynthetic plant, the increase in the intensity and duration of exposure to light means the rate at which photosynthesis occurs will be increased. An increased photosynthetic rate means the synthesis of the products will also be increased i.e. glucose and OXYGEN.

6 0
3 years ago
How do valence electrons relate to the chemical reactions of an element?
11111nata11111 [884]
Valence electrons are required for certain bonding processes. Without them you cannot bond with certain elements. For example, carbon bonds really well with carbon because it has the same amount of valence electrons. However carbon would not bond well with uranium due to the massive differences in valence electrons. Hope this helps!
6 0
3 years ago
Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.
Mars2501 [29]

Answer:

Rms speed of the particle will be  38.68\times 10^8m/sec

Explanation:

We have given mass of the air particle m=5\times 10^{-16}kg

Gas constant R = 8.314 J/mol-K

Temperature is given T = 27^{\circ}C=273+27=300K

We have to find the root mean square speed of the particle

Which is given by v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec

So rms speed of the particle will be 38.68\times 10^8m/sec

5 0
3 years ago
An airplane is flying at a speed of 45 m/s when it drops a 40 kg food package to the Polar exploration team. If the plane drops
Alina [70]

To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,

PE = KE

mgh = \frac{1}{2}mv^2

Here,

m = Mass

g = Gravity

h = Height

v = Velocity

Rearranging to find the velocity

v = \sqrt{2gh}

Replacing,

v = \sqrt{2(9.8)(120)}

v = 48.49m/s

Using the vector properties the magnitude of the velocity vector would be given by,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{45^2+48.42^2}

|V| = 66.2m/s

Therefore the package is moving to 66.2m/s

3 0
3 years ago
ASAP please!! <br>The force vectors on an aircraft are as shown. Find the net (resultant force).
ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

W_D=3600 N\cdot\sin 27^\circ=1634N

acting in the direction of the Drag, and the plane-perpendicular component is:

W_L=-3600N\cdot\cos 27^\circ=-3208N

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N

The magnitude of the  net force is:

|F|=\sqrt{5366^2+829^2}N = 5430N


6 0
3 years ago
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