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2 years ago

What does overloading your muscles accomplish?

Physics

2 answers:

Vadim26 [7]2 years ago

7 0

Answer:

Explanation:

You are putting more strain on your muscles which is ripping the fibers which will grow back bigger.

Serga [27]2 years ago

4 0

Answer:

The correct option is A

Explanation:

Overloading of muscle is caused by subjecting the muscle to more stress than they are accustomed to. This is the technique used by body builders to build the muscle mass. When the muscle is continuously subjected to this added stress, it eventually adapts to this stress by producing new myofibrils (which makes up the muscle fibre). This added stress causes the muscle fibre to grow bigger and stronger in order to handle this additional stress subsequently.

You might be interested in

A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t

nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

$\Phi = BA Cos \theta$

Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

$\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )$

At time $t = 5.71s$ , Induced emf is,

$\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))$

$\epsilon = 10.9V$

Therefore the magnitude of the induced emf is 10.9V

4 0

2 years ago

Read 2 more answers

What are the two factors that affect the friction force between two surfaces

Masja [62]

Answer:

coefficient of static friction of the surface and the normal force

Explanation:

The coefficient of static friction of the surface and the normal force exerted on the surface given by equation F = μR

5 0

2 years ago

A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because

amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0

3 years ago

what evidence can you cite that the interstellar medium contains both gas and dust? (select all that apply.)

Basile [38]

(a) The gas of interstellar medium can be detected from the radiations of photons of wavelength 21 cm.

(b) The gas of interstellar medium can be detected from the absorption lines present in the light from distant stars, which must be caused by a medium of density and temperature other than that of the stars emitting the lights.

<h3>What is interstellar medium?</h3>

Interstellar medium is the matter and radiation that exist in the space between the star systems in a galaxy.

<h3>Evidence that interstellar medium contains both gas and dust</h3>

The gas of interstellar medium can be detected from the radiations of photons of wavelength 21 cm.
The gas of interstellar medium can be detected from the absorption lines present in the light from distant stars, which must be caused by a medium of density and temperature other than that of the stars emitting the lights.

Learn more about interstellar medium here: brainly.com/question/4173326

#SPJ11

4 0

1 year ago

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup

Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$