It is a set of measure designed to determine the level of physical fitness?

The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross

Oksana_A [137]

The speed and pressure at another end will be 9 m/sec and -31500 pascals.

<h3>What is gauge pressure?</h3>

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

For horizontal pipe, Z₁=Z₂

Pressure at end 1, P₁ = 4500Pa

Speed at end 1, V₁ = 3 m/sec

Speed at end 2, V₂ = ? m/sec

Area of cross-section at end 1,A₁ =A

Area of cross-section at end 1,A₂ = A/3

From the continuity equation;

A₁V₁=A₂V₂

A× 3 = (A/3)×V₂

V₂ = 9 m/sec

From Bernoulli's equation;

$\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2 \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} = \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} = \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 = - \ 31,500 \ pa$

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

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7 0

2 years ago

A car starting at rest accelerates to a speed of 6miles/hr in 5 sec what is the average acceleration

Luden [163]

Answer:

1.2 mi/hr/sec

Explanation:

<em>a = Vf - Vi / t (formula)</em>

-----------------------------------

6mi/hr - 0 mi/hr 6 mi/hr

a = ____________ = _______ = 1.2 mi/hr/sec

5 sec 5 sec

5 0

3 years ago

Read 2 more answers

What is the average power consumption in watts of an appliance that uses 5.00 KW.h of energy per day?

abruzzese [7]

(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules

Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts

(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules

3 0

3 years ago

A non-reflective coating that has a thickness of 198 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What

spin [16.1K]

Answer:

The wavelength is $\lambda_ 1 = 574.2 nm$

Explanation:

From the question we are told that

The thickness is $t = 198 nm = 198 *10^{-9 }\ m$

The refractive index of the non-reflective coating is $n_m = 1.45$

The refractive index of glass is $n_g = 1.50$

Generally the condition for destructive interference is mathematically represented as

$2 * n_m * t * cos (\theta) = n * \lambda$

Where $\thata$ $\theta$ is the angle of refraction which is 0° when the light is strongly transmitted

and n is the order maximum interference

so

$\lambda = \frac{2 * n * t * cos (\theta )}{n}$

at the point n = 1

$\lambda _1 = \frac{2 * 1.45 * 198*10^{-9} * cos (0 )}{1}$

$\lambda_1 = 574.2 *10^{-9}$

$\lambda_1 = 574.2 nm$

at n =2

$\lambda _2 = \frac{\lambda _1 }{2}$

$\lambda _2 = \frac{574.2*10^{-9} }{2}$

$\lambda _2 = 2.87 1 *10^{-9} \ m$

$\lambda _2 = 287. 1 nm$

Now we know that the wavelength range of visible light is between

$390 \ nm \to 700 \ nm$

So the wavelength of visible light that is been transmitted is

$\lambda_ 1 = 574.2 nm$

6 0

4 years ago

A researcher reports an f-ratio with df = 3, 36 for an independent-measures experiment. how many treatment conditions were compa

user100 [1]

The correct option is B.

There were 4 treatment conditions compared in the experiment.

F ratio plays an essential role in performing a particular dataset of ANOVA. F ratio is particularly a ratio which is obtained by the between-group variance or it is also called MSB and also within-group variance known as MSW. Any F-ratio which is computed is compared with the critical F-ratios from the table as it will check out if there are any variations available between the groups or not.

The researcher reports an F-ratio where the degrees of freedom = 3, 36.

F-ratio is obtained by the calculation of dividing the Mean squared errors because of the treatment by the Mean squared error which occurs due to error.

For the particular case, the researcher reports an F-ratio having degrees of freedom = 3, 36. It is indicating that the treatments are being distributed with degrees of freedom which is 3 and specified errors are distributed with degrees of freedom which is 36.

Let the treatments which were involved in the study can be denoted by k.

Let the total number of individuals involved in the study can be taken as N.

Then, the treatments will be having degrees of freedom as,

$df_{treatments}$ = k-1

3=k-1

k=3+1=4

Therefore, the required treatment condition number that was compared is 4.

Learn to know more about degrees of freedom on

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6 0

1 year ago