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Grace [21]
2 years ago
5

Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type. 7.7 7.2 10.7 7.4 8.7 6.5 6.8 7.0

9.7 11.3 7.5 6.8 5.9 7.9 6.3 8.2 6.3 9.7 11.6 7.8 7.7 9.0 7.3 11.8 7.0 8.2 8.1 (a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. [Hint: Σxi = 220.1.] (Round your answer to three decimal places.)
Engineering
1 answer:
aleksandr82 [10.1K]2 years ago
8 0

Answer:

8.152.

Explanation:

The "point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion" that we are told to calculate is very direct. What we is only require of us is to find the average of the whole values of x that is;

7.7 + 7.2 + 10.7 + 7.4 + 8.7 + 6.5 + 6.8 + 7.0 + 9.7 + 11.3 + 7.5 + 6.8 + 5.9 + 7.9 + 6.3 + 8.2 + 6.3 + 9.7 + 11.6 + 7.8 + 7.7 + 9.0 + 7.3 + 11.8 + 7.0 + 8.2 + 8.1 = 220.1

So, the addition of all the x-values gives 220.1. The average = 220.1/ n.

Where n = total numbers of x entry = 27.

Therefore, 220.1/ 27 = 8.152.

Hence, "point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion" = 8.152 to three decimal places.

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A liquid with a specific gravity of 2.6 and a viscosity of 2.0 cP flows through a smooth pipe of unknown diameter, resulting in
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2 years ago
simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The
Roman55 [17]

Answer:

A. ) 591.7 v

B.) 991.7v

C.) 59.7%

D.) 47.9 Kw

E.) 247925 W

F.) 59.7 %

Explanation:

Given that a simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The transmission line is 100 km copper wire of 3 cm diameter. If the voltage at the load side is 400 V,

Let first calculate the resistance in the wire.

The resistivity (rho) of a copper wire is 1.673×10^-8 ohm metres

Resistance R =( L× rho)/A

Where Area = πr^2 = π × 0.015^2

Area = 0.00071 m^2

R = (100000 × 1.673×10^-8) / 0.00071

Resistance in wire = 2.367 ohms

Then let calculate the resistance in the load.

Also, since Power P = V^2 /R

Make R the subject of formula

R = V^2/ P

R = 400^2/100000

Resistance in load = 1.6 Ohms

Current l = V / R

I = 400/1.6 = 250 Ampere

a.) Voltage drop across the line V line will be achieved by using Ohms law.

V = I R

V = 250 × 2.367

V = 591.7 v

B.) Voltage at the source side Vsource will be

V = V line + V load

V = 400 + 591.7

V = 991.7 v

C.) Percentage of the voltage drop Vline /Vsource

591.7/991.7 × 100 = 59.7%

D.) Line losses

P = I V

P = 250 × 591.7

P = 147925 W

Power loss = 147925 - 100000

Power loss = 47,925 W

Power loss = 47.9 Kw

E.) Power delivered by the source

P = IV

P = 250 × 991.7 = 247925 W

F.) System efficiency

Efficiency = power line / power source × 100

Efficiency = 147925 / 247925 × 100

Efficiency = 59.7 %

5 0
2 years ago
The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to
skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 kN/m^{2}

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = 0.02\ m^{3}/s

(Since, 1 litre = 10^{-3} m^{3})

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, H_{loss} = 45 cm = 0.45\ m

Height, h_{roof} = 2.5\ m

Now,

The velocity in the bore is given by:

v = \frac{Q}{\pi (\frac{d}{2})^{2}}

v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s

Now, using Bernoulli's eqn:

\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k                  (1)

The velocity head is given by:

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Now, by using energy conservation on the surface of water on the roof and that in the tank :

\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}

\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45

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4 0
2 years ago
As the porosity of a refractory ceramic brick increases:
ivolga24 [154]

Answer:

A) strength decreases, chemical resistance decreases, and thermal insulation increases

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A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav
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Answer:

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We know that

1\ lbf=32.17 \ lmb.ft/s^2

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

1\ ft/s^2= 0.304\ m/s^2

5.48\ ft/s^2= 1.66\ m/s^2

a=6 g's

a=9.99\ m/s^2

So

F = 90 x 9.99 N

F= 899.59 N

3 0
2 years ago
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