Can someone pls give me the answer to this?

A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2

IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length $l$ for the deformation, we use the following relation;

$l$ = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

$l$ = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

$l$ = 3161.025289 / 7240

$l$ = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0

3 years ago

A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok

galben [10]

Answer:

The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.

When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.

By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.

In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.

Also, a Flow control valve is provided tho control the velocity of clamping

Explanation:

find attached the figure

4 0

3 years ago

Breh/bro <br><br>what is this, finally I can learn programming

LenKa [72]

Answer:

It's an intoduction to hacking and systematic programming.

Explanation:

Yes, you might be able to grasp a few things from it, but it also may be a way hackers could hack you, by luring you to click it.

8 0

3 years ago

A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio

Mama L [17]

Answer:

a) 570 kWh of electricity will be saved

b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

$P_{incandescent}$ = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find the total power consumption by the fluorescent lamp

$P_{fluorescent}$ = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

$P_{saved}$ = $P_{incandescent}$ - $P_{fluorescent}$

$P_{saved}$ = 750 - 180

$P_{saved}$ = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved = 570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

3 0

3 years ago

Opinion: Choose a side and defend it in four to five sentences. Given the choice between a vehicle that runs on gasoline or one

vaieri [72.5K]

Answer:

my opinion would be electric

Explanation:

because when it comes down to the bare minimum the best choice for the world in the long run would be electric because it puts a big dent back for global warming and the burning of gasses so the more people who drive electric cars are the people who are trying to save the world a little at a time by preventing the burning of gasses. hope this helps :)

6 0

3 years ago

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