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4 years ago

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

2 answers:

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]4 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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An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

Svetllana [295]

Answer:

$P_G = 14.03 \ psig$

$h_m = 0.148 \ m$

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is $P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2$

The level of mercury is $h = 950 \ mm = 0.950 \ m$

The drop in the mercury level at the visible arm is $d = 39.0 = 0.039 \ m$

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

$P_g = P_m = g * \delta h * \rho$

Here $\rho$ is the density of mercury with value $\rho = 13.6 *10^{3} kg/m^3$

and $\delta h$ is the difference in the level of gas in arm one and two

$\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }$

$\delta h = 0.802 \ m$

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

$h_m = 0.950 - 0.802$

=> $h_m = 0.148 \ m$

Generally from manometry principle we have that

$P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0$

Here $P_G$ is the pressure of the gas

$P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0$

$P_G = 9.6724 04 *10^{4} \ N/m^2$

converting to psig

$P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}$

$P_G = 14.03 \ psig$

6 0

3 years ago

If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?

poizon [28]

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel. The 2R and 2R resistors are in parallel. The 4R and 4R resistors are in parallel. Each parallel combination is in series with each other. Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

6 0

3 years ago

A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho

spayn [35]

<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A ------------------------(i)

Where;

A = πd² / 4 [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

4 0

3 years ago

The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.

qaws [65]

Density is the characteristic property of a substance. It is the measure of mass of the substance divided by its volume (density= mass/volume). Manipulate the given formula to come up with the formula for the volume. Therefore, volume is equals to mass of a substance divided by its density (Vol= mass/density). Given 12.6 g/ml as density and 7.65 g mass, volume is equals to 0.60714 ml, since 1 ml = 1cm^3, volume is equals to 0.60714 cm^3 then extract the cube root of the volume to get the length of the cube in cm which is equal to 0.84677 cm.

4 0

4 years ago

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aleksandr82 [10.1K]

The answer is 24 atoms.

7 0

3 years ago

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