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Juli2301 [7.4K]
4 years ago
7

The force generated by a long muscle varies as it contracts through its range of movement. At which point is the greatest force

generated?
Physics
1 answer:
Sedaia [141]4 years ago
8 0

Answer:

When the muscle is completely contract.

Explanation:

Remember the that maximum force of a muscle is when is completely contract. A characteristic of a muscle is that can contract and can relax in the opposite direction. In this way, when all the microfibers of the muscle are join together (they are contract) is when the maximum tissue force is applied.

With exercise the fibers of the muscles can grow or reproduce to strength the muscle.

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Which describes newton’s law of universal gravitation?
Zielflug [23.3K]

(B) All objects attract other objects


To be specific, the following formula defines this theory very clearly:

F = G * (m1 * m2) / r^2

The Force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

8 0
3 years ago
How does the sun's energy travel to Earth?
Elena-2011 [213]

Answer: The sun’s radiation consists of small, massless packets of energy called photons. They travel seamlessly through space; whenever they strike any object, the object absorbs photons and its energy is increased, which then heats it up.

Explanation:

5 0
3 years ago
Read 2 more answers
If the answer to your calculation has units of kg.m^2/s^2. What type of quantity could it be?
Juliette [100K]

The quantity can be work or kinetic energy or potential energy. Work is same as the energy.

Work done can be described as energy transferred from and object when a force is applied along a displacement. That is work done is the product of force in the direction of the displacement and the magnitude of the displacement.

Mathematically it can be represented as,

W = F x d

where,

W ⇒ work done

F ⇒ force acting along the displacement

d ⇒ magnitude of displacement

We know,

force = mass x acceleration

∴ the unit of force is kgms⁻²

The unit of work done will be kgms⁻² x m

That is kgm²s⁻²

where m is the unit of displacement that is meter (m)

Hence kgm²s⁻² is the unit for work or energy

To more about SI unit : brainly.com/question/788548

#SPJ4

6 0
2 years ago
The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto
igor_vitrenko [27]

a) The angular velocity of the electron is 4.12\cdot 10^{16} rad/s

b) The number of revolutions per second is 6.54\cdot 10^{15} rev/s

c) The centripetal acceleration of the electron is 8.98\cdot 10^{22} m/s^2

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

\omega = \frac{v}{r}

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the angular velocity is

\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s

b)

The period of revolution of the electron is given by

T=\frac{2\pi}{\omega}

where

\omega = 4.12\cdot 10^{16}rad/s is the angular velocity

Substituting,

T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15} rev/s

c)

The centripetal acceleration of an object in circular motion is

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circle

For the electron, we have

v=2.18 \cdot 10^6 m/s

r=5.29\cdot 10^{-11} m

Therefore, the centripetal acceleration is

a=\frac{(2.18\cdot 10^6)^2}{5.29\cdot 10^{-11}}=8.98\cdot 10^{22} m/s^2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0
4 years ago
A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N.
Likurg_2 [28]
The answers to the problem are as follows:

  <span>a) use f=ma. Mass of student is 836/9.8=85.3kg, force is 935-836=99N. Put it into the formula you will get a=99/85.3 

b)Use same formula. Force is 782-836=-54 so put it into formula a=-54/85.3 

c)Stopping would take longer as the acceleration is smaller 


I hope my answer has come to your help. God bless and have a nice day ahead!</span>
5 0
4 years ago
Read 2 more answers
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