ENGINEERING
COLLEGE

Answer:

Answer:

h=32.1 km

Explanation:

solution:

using newton law of gravitational attraction and newton second law:

r= distance between two masses

at sea level

a=g

.............................(1)

.........................(2)

by substituting (2) and (1) acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

COLLEGE

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the inlet. The volumetric flow rate is 0.2 m3/s, and the inlet and exit areas are 0.06 m2 and 0.03 m3, respectively. (a) Assuming the temperature to remain constant and neglecting any heat transfer, determine the power input to the pump in kW. (b) What-if Scenario: What would the necessary power input be if the change in KE were neglected in the analysis??

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

Work done by pump

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

Work done by pump

∴ Power input to the pump 20.7088 kW

MIDDLE SCHOOL

Module 42 Review and Assessment Summing Up the Key Ideas: Match the following terms to the correc

A. Inflation

D. Consumer price index G. Producer price

B. Inflation rate

I. Shoe leather costs

E. Market basket

index

J. Menu costs

C. Deflation F. Base period

H. Hyperinflation

1. A decrease in the price level.

2. A measure of the overall price level faced by a typical consumer.

3. The costs of updating prices due to inflation.

4. A bundle of goods that an average consumer might buy.

5. The annual percentage increase in the price level.

6. A period of very rapid inflation

7. A rise in the price level.

8. A period in the past used as a benchmark for comparisons.

9. A measure of the cost of a basket of goods and services bought by producers.

10. The costs of time and effort involved in frequent trips to the bank or ATM to avoid

holding much cash during periods of high inflation.

Analyze and Explain: Determine if the person described in each of the following situations

would be helped by inflation or hurt by inflation and then explain your answer.

HELPED OR HURT?

EXPLANATION

1. a retired worker with a fixed pension

during a time of inflation

2. a lending bank that expocted the

inflation rate to be 7 percent when the

actual rate of indiation for the duration

of the lon turned oun to be 4 percent

I need that

This is a lot of information but I just wanted to say I hope you have a great day!

COLLEGE

A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.90 and 0.10, respectively, estimate the upper limit for the specific stiffness of this composite given the data that follow. Specific Gravity Modulus of Elasticity (GPa) Copper 8.9 110

Tungsten 19.3 407

Answer:

Specific stiffness=37.95 GPa

Explanation:

Given that

Volume fraction of tungsten V₁ = 0.9

Volume fraction of copper V₂ = 0.1

Specific gravity for tungsten S₁ = 19.3

Specific gravity for copper S₂= 8.9

Modulus of Elasticity for tungsten E₁= 407 GPa

Modulus of Elasticity for copper E₂ = 110 GPa

Modulus of Elasticity for mixture

E= E₁V₁+E₂V₂

E = 110 x 0.1 + 407 x 0.9 GPa

E=377.3 GPa

Specific gravity for mixture

S= S₁V₁+S₂V₂

S = 19.3 x 0.1 + 8.9 x 0.9

S=9.94

Specific stiffness K

K=37.95 GPa

Specific stiffness=37.95 GPa

COLLEGE

A 208 V delta-connected source is energizing two parallel loads. One of the loads is connected in delta and the other in wye. The delta load has impedance ΖΔ-10 L-25°Ω. The wye load has impedance ZY-5 L40°Ω. Compute the following line current active and reactive power.

Given Information:

Vab = 208 V

ZY = Z1 = 10 < -25° Ω

Zdelta = Z2 = 5 < 40° Ω

Required Information:

Line current = Ia = ?

Active Power = P = ?

Reactive Power = Q = ?

Answer:

° A

Explanation:

First, convert the wye connected load into delta

° Ω

° Ω

Now both of the loads and also the source are delta connected, now we can proceed further.

Find the equivalent load impedance, since the loads are connected in parallel;

Ω

° Ω

The phase current can be found by

A

° A

The relation between phase current and line current is

° A

The Active Power is

The Reactive Power is

COLLEGE

A catch block may not have the ability to handle certain type of exceptions or may decide that the exception should be handled by the calling block or environment. In such cases the programmer has the ability to _____. rethrow the exception restart the program enter error-handling instructions during runtime remove the exception

Answer:

rethrow the exception.

Explanation:

If it is unable to tackle the specific exception that catch block is captured then we should use rethrow the exception.

The expression of rethrow allows the subject to thrown again. In Java rethrow exception enables to indicate more particular types of exceptions of declaration of method in throws clause.

COLLEGE

Each of these geometric shapes has a different number of sides. Arrange the shapes in order from the shape with the greatest number of sides to the shape with the fewest number of sides. Rank these shapes from greatest to fewest number of sides. To rank items as equivalent, overlap them.triangle , square, rectangle, octagon, hexagon, pentagon

Based on the given information, we can rank these shapes from greatest to fewest number of sides as ;

- Octagon

- hexagon

- Pentagon
- Rectangle
- square (overlapped)
- triangle.

In geometry, shape is very important, and they have different number of sizes. For instance, Octagon has 8 sizes, Hexagon posses 6 sizes, up to triangle which have 3 sizes.

Therefore, Octagon have the highest number of size here.

Learn more about Shapes at;

brainly.com/question/14934754

Answer:

The rank from shapes with the greatest number of sides to the shape with the fewest number of sides is:

Octagon, hexagon, Pentagon, Rectangle, and square (overlapped) and triangle.

Explanation:

We want to order these shapes for the greatest number of sides:

The Octogon has 8 sides.

Hexagon has 6 sides.

Pentagon has 5 sides.

The Rectangle and the square both have 4 sides.

The Triangle has 3 sides.

COLLEGE

A mechanical system comprises three subsystems in series with reliabilities of 98%, 92%, and 87%. What is the overall reliability of the system?

Answer:

The overall reliability of the system is found to be 0.784 or 78.4%.

Explanation:

Since, the system has three subsystems in series with following reliabilities:

Reliability of 1st system = R1 = 98% = 0.98

Reliability of 2nd system = R2 = 92% = 0.92

Reliability of 3rd system = R3 = 87% = 0.87

Thus, the overall reliability for series configuration is given by the following formula:

Overall Reliability = (Reliability of 1st system) (Reliability of 2nd System) (Reliability of 3rd System)

Overall Reliability = (0.98)(0.92)(0.87)

Overall Reliability = 0.784

Overall Reliability = 78.4%

COLLEGE

A 30-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 12. Calculate the pitch diameter, circular pitch, addendum, dedendum, tooth thickness, and clearance.

To develop the problem it is necessary to apply the concepts related to Pitch diameter, circular pitch, addendum, dedendum, tooth thickness, and clearance.

The pitch diameter of the gear is defined as

Where,

N = Number of teeth

Diametral pitch

Replacing we have

Therefore the pitch diameter of the gear is 2.500

The circular pitch of the gear is given as

Therefore the circular pitch of the gear is 0.2618

The addendum of the gear teeth is given by

Therefore the addendum of the gear teeth is 0.0833

The dedendum of the gear teeth is gives as

The thickness of the tooth is given by

Therefore the thickness is 0.1309

Finally the clearance is given by

Therefore the clearance of the gear teeth is 0.028

As you can see most of the definitions are made from a theoretical aspect of defined equations for each of the parameters.

COLLEGE

When a certain motor is started, it is noticed that its supporting frame begins to resonate when the motor speed passes through 900 rpm. At the operating speed of 1750 rpm, the support oscillates with an amplitude of 8 mm. Determine the amplitude that would result at 1750 rpm if the support were replaced with one having one-half the stiffness.

Answer:

X₂ = 6.74 mm

Explanation:

As we know that:

ω = 900 (2π/60) rad/ sec = natural frequency

ω = 1750(2π/60) rad/ sec = maximumm frequency

Also,

r ₁ = ω / ω = [1750(2π/60)]/[900(2π/60)] = 1.94

Now we know that general formula for stiffness is

ω = ---------- equation (1)

Similarly

ω = ---------- equation (2)

put k = k/2 in equation 2, we get

ω =

Now, by equating equation 1 and 2, we get

ω = ω

or

(1/)ω = ω

By putting ω = 900 (2π/60) we get

(1/)900 (2π/60) = ω

By solving it we get

ω = 636.4(2π/60)

Also we know that

r ₁ = ω / ω

By putting values in it, we get

r₂ = [1750(2π/60)] / [636.4 (2π/60)]

r₂ = 2.75

For τ = 0, we have rotating unbalanced equation as

X =[m∈/m][r²/(1-r²)]

For X=8, r = r₁ ; 8 =[m∈/m][r₁²/(1-r₁²)] ---------------equation (a)

For X=X₂, r = r₂ ; 8 =[m∈/m][r₂²/(1-r₂²)] ---------------equation (b)

Dividing equation a by b , we get

X₂/8 = (r₂/r₁)²[(1-r₁²)/(1-r₂²)]

Now, put r₁ = 1.94 ; r₂ = 2.75

we get

X₂/8 = (2.75/1.94)²[(1-1.94²)/(1-2.75²)]

= 0.842

X₂ = 6.74 mm

The amplitude of one half thickness is calculated to be 7.74mm