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2 years ago

Which of the following is the best reason to rest muscle groups between workouts?

Physics

2 answers:

Pie2 years ago

5 0

The best reason to rest muscle group between workouts is to allow the muscles time to heal.

Further Explanation:

With each strength or resistance exercise the muscle tissue get tired because of the utilization more energy obtain from anaerobic respiration which release lactic acid in the end of the reaction. During exercise the muscle tissue stretch and break down which needs some time to heal and regenerate. If the individual don’t take enough rest in between each reap; it will cause damage to the muscle tissue.

The most common are,

1. Strain: Anterior cruciate ligament is lies in the knee which is a stabilizing ligament. The ACL strain is caused by slow down movements and trying to cut or change directions and posture. In this injury the innermost ligament get tear which is present in the knee.

2. Groin Pull: The groin muscles fibers are lies in the thigh which runs from the above-inner surface to the inner surface in the foremost portion of the knee. The injury occurs by quick side-to-side movements and due to absence of flexibility.

3. Hamstring strain: The hamstring muscle is present on the rare side of the thigh. The cause of hamstring tear is poor and absence of stretching. The person with hamstring tear will acquire bruising in the back side of the thigh and knee.

In between each session the body releases hormones that help in performing two type of reactions Anabolic and Catabolic reaction. Anabolic reaction helps in building new tissues or muscles whereas; catabolic reaction helps in burning fat to produce more energy and in breaking tissue down. They help in healing process and reduce the risk of muscle damage.

Learn More:

1. Learn more about meiosis brainly.com/question/1600165

2. Learn more about the process of molecular diffusion in a cell brainly.com/question/1600165

3. Learn more about human sperm and egg cell brainly.com/question/1626319

Answer Details:

Grade: High School

Subjects: Health

Topic: Physical Fitness

Keywords:

Muscle, heal, strength, utilization, anaerobic, respiration, regenerate, strain, groin, hamstring, anabolic, catabolic, energy, damage.

vovangra [49]2 years ago

4 0

Answer;

It allows the muscles time to heal.

Explanation;

-It is very important to rest muscles between workouts. In fact, resistance training breaks down muscles causing microscopic tears. It is only during rest when the muscle-building process is stimulated and the regeneration of new tissue occurs.

-But when muscles are not allowed adequate rest and recover time, the regeneration process cannot occur. This can have several side effects such as increased soreness, decreased strength and performance, and may lead to injury.

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a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

2 years ago

(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance

Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity F=W= mg [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

=132 kg ×9.8 m/s^2

=1293.6 kg m/s^2

=1293.6 N [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

$=\frac{1}{4} *1293.6N$

$=323.4 N$

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

$F_{net} =1293.6N -323.4N$

$=970.2 N$

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e

$F_{net} =ma$ [Here a is the acceleration]

$a =\frac{F_{net} }{m}$

$= \frac{970.2}{132} m/s^2$

$=7.35 m/s^2$

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

we know that F= ma

=m×0

=0 N

Hence they will not get any force when they will descend with a uniform speed.

4 0

2 years ago

Need help solving this question.

MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

$-15+T-W=0\\T-W=15$

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

$(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]$