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BlackZzzverrR [31]
2 years ago
12

Why is it important to assess flexibility both before and during a workout regimen? a. to assess flexibility progress b. to keep

muscles warm c. to establish flexibility baselines d. none of the above please select the best answer from the choices provided. a b c d
Physics
2 answers:
Inga [223]2 years ago
8 0

Answer:

A

Explanation:

lidiya [134]2 years ago
5 0

It important to assess flexibility both before and during a workout regimen because to assess flexibility progress.

<h3>What is a flexibility exercise?</h3>

Flexibility exercises, as the name implies, help to increase the stretch of the muscles, improving the range of motion. More flexibility is also important to help prevent possible post-workout soreness.

In this case we have that the reason to continue doing stretching before and during the exercises is about  to assess flexibility progress.

See more about flexibility at brainly.com/question/15395713

#SPJ4

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A 70 kg football player running at 3m/s north tackles an 80kg player running at 1.5 m/s south. what is the magnitude and directi
dimulka [17.4K]
It could never actually happen like this, but the question is
looking for you to 'conserve' the momentum.

Momentum of a moving object is (mass) x (velocity).
Like velocity, momentum has a direction.
Momentum is one of those things that's 'conserved'. 
That means that momentum can't appear out of nowhere, and
it doesn't disappear.  The total after the collision is the same as
the total was before the collision.  

Momentum of the skinny player:

                     (70 kg) x (3 m/s north) = 210 kg-m/s north.

Momentum of the heavy player:

                     (80 kg) x (1.5 m/s south)  =  120 kg-m/s south .

Total momentum before the collision is

                     (210 kg-m/s north) + (120 kg-m/s south)

                 =        90 kg-m/s north  .

It has to be the same after the collision.

                       (mass) x (velocity)  =  90 kg-m/s north.

The  mass after the collision is 150 kg, because they get
tangled up and stuck together, and they move together.

                                (150 kg) x (velocity)  =  90 kg-m/s north .

Divide each side
by  150 kg :                   velocity  =  (90 kg-m/s north) / (150 kg)

                                                =  (90/150)  (kg-m/s / kg  north)

                                                 =        0.6 m/s  north  .
3 0
3 years ago
A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters​ [cm]. The rover picks up the rock and lifts it
Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid SG (also called relative density) is the ratio of the density of that solid \rho_{rock} to the density of water \rho_{water}=1 kg/m^{3} (normally at 4\°C):

SG=\frac{\rho_{rock}}{\rho_{water}} (1)

On the other hand, the density of the rock is calculated by:

\rho_{rock}=\frac{m_{rock}}{V_{rock}} (2)

Where:

m_{rock} is the mass of the rock

V_{rock}=\frac{4}{3} \pi r^{3} is the volume of the rock, since is spherical

Well, we already know the value of \rho_{water}, but we need to find \rho_{rock} in order to find the rock's specific gravity; and in order to do this, we firsly have to find m_{rock} and then calculate V_{rock}:

In the case of the mass of the rock, we can calclate it by the following equation:

W_{rock}=m_{rock}g_{mars} (3)

Where:

W_{rock} is the weight if the rock in mars

g_{mars}=3.7 m/s^{2} is the acceleration due gravity in Mars

Isolating m_{rock}:

m_{rock}=\frac{W_{rock}}{g_{mars}} (4)

m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}} (5)

To find W_{rock} we can use the following equation of the potential gravitational energy U:

U=W_{rock}H (6)

Where:

U=2 J=2 Nm is the potential energy

H=20 cm \frac{1m}{100 cm}=0.2 m is the height at which the rock has the mentioned potential energy

Isolating W_{rock}:

W_{rock}=\frac{U}{H} (7)

W_{rock}=\frac{2 Nm}{0.2 m} (8)

W_{rock}=10 N (9)

Substituting (9) in (5):

m_{rock}=\frac{10 N}{3.7 m/s^{2}} (10)

m_{rock}=2.702 kg (11)

Substituting (11) in (2):

\rho_{rock}=\frac{2.702 kg}{V_{rock}} (12) At this point we only need to find the volume of the rock, knowing its diameter is d=10 cm, hence its radius is r=\frac{d}{2}=5 cm

V_{rock}=\frac{4}{3} \pi (5 cm)^{3} (13)

V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3} (14)

Substituting (14) in (12):

\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}} (15)

\rho_{rock}=5166.34 kg/m^{3} (16)

Substituting (16) in (1):

SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}} (17)

Finally we obtain the specific gravity of the​ rock:

SG=5166.347

7 0
3 years ago
Which of these statements best proves that electromagnetic forces are weaker than strong nuclear forces? Neutrons do not repel o
musickatia [10]
Answer: <span>Protons do not move out of the nucleus of atoms although they repel each other.

Reasoning:

Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.

That repulsion force tends to destabilize the atom. The only reason why the atom does not explode is because there is a there exists a force of attraction that exceeds the repulsion of the like charged protons. That is the strong nuclear force. Therefore, the strong nuclear forces are stronger than the electromagnetic force that tends to get the protons apart.
</span>
7 0
3 years ago
Puck A, of inertia mm, is attached to one end of a string of length ℓℓ , and the other end of the string is attached to a pivot
algol13

Answer:

L_{B} / L_{A} = \frac{5mvl}{mvl}= 5

Explanation:

Find the given attachment

8 0
3 years ago
Which model is defined atomic model that describes an atom has a spherical object containing a certain number of electrons trapp
victus00 [196]

Answer:

C! Raisin Bread

Hope this helps you!

4 0
3 years ago
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