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3 years ago

7

Why should flexibility exercises be done in conjunction with strength-building exercises?

2 answers:

bixtya [17]3 years ago

7 0

<span>Flexibility exercises should be done in conjunction with strength-building exercises because:

</span><span>They work together to maintain your overall level of fitness.
</span>
I hope my answer has come to your help. God bless and have a nice day ahead!

agasfer [191]3 years ago

7 0

Answer:

The answer is flexibility exercise allows muscles to perform strength exercises with greater capacity.

Explanation:

The strength-building exercise should be done in conjunction with the flexibility exercises because it allows us to protect our muscles and joints from possible injuries, in addition to providing us with a greater and better range of movement since a relaxed muscle has more ease of performing a rapid contraction and therefore more possibility of developing a greater force.

There is a direct relationship between flexibility and the ability to execute movements with power since a flexible muscle has an adequate capacity to exert its full power.

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in which type of wave are vibrations at right angles to the direction in which the wave is travelling

Pani-rosa [81]

Answer:

longitudinal waves have those properties

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3 years ago

A student has a rectangular block. It is 2 cm wide, 2 cm tall, and 25 cm long. It has a mass of 600 g. First, calculate the volu

WINSTONCH [101]

Answer:

6g/cm³

Explanation:

Density is the mass per unit volume of any substance. To solve this problem:

Density = $\frac{mass}{volume}$

Since mass = 600g

Let us find the volume;

Volume = length x width x height

Volume = 25cm x 2cm x 2cm = 100cm³

Therefore;

Density = $\frac{600}{100}$ = 6g/cm³

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A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"

Mashutka [201]

The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g

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Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

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3 years ago

Read 2 more answers

Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.

amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0

3 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

3 years ago