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Ludmilka [50]
3 years ago
10

An engine has a piston with a surface area of 17.31 in2 and can travel 3.44 inches. What is the potential change in volume, disp

lacement (\Delta VΔV) , of the cylinder?
Engineering
1 answer:
Katena32 [7]3 years ago
5 0

Answer:

$$\begin{align*}

P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$

Explanation:

\eqalign{

 P(Y-X=m\mid Y\gt X)

   &=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr

   &=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr

   &=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr

}

P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$

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A _____ satellite system employs many satellites that are spaced so that, from any point on the Earth at any time, at least one
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Answer:

d. low earth orbit (LEO)

Explanation:

This type of satellites form a constellation deployed as a series of “necklaces” in such a way that at any time, at least one satellite is visible by a receiver antenna, compensating the movement due to the earth rotation.

Opposite to that, a geostationary satellite is at an altitude that makes it  like a fixed point over the Earth´s equator, rotating synchronously with the Earth, so it is always visible in a given area.

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3 years ago
Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default
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Answer:

Explanation:

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4 years ago
With the help of diagrams explain the differences in the functions of electrostatic
ololo11 [35]

Answer:

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Explanation:

5 0
3 years ago
An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat
IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

b)

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature T_L = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature T_H = 99° F = ( 99 + 460 )R = 559 R

Cooling Load Q_L = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

a)

Coefficient of performance of the air conditioner;

COP_{air-condition = Cooling Load Q_L  / power P

we substitute

COP_{air-condition = 30000 Btu/h / 8396.619 Btu/h

COP_{air-condition = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

b)

- Power input required ( in hp )

Q_L / P_{required = T_L / ( T_H - T_L )

we substitute

30000 Btu/h / P_{required = 530 R / ( 559 R - 530 R )

30000 Btu/h / P_{required = 530 R / 29 R

we solve for P_{required

P_{required  = ( 30000 Btu/h × 29 R ) / 530 R

P_{required  = ( 870000 Btu/h / 530 )

P_{required  = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

P_{required  = ( 1641.5094 / 2544.43 ) hp

P_{required  = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

COP_{rev-air-condition = T_L / ( T_H - T_L )

we substitute

COP_{rev-air-condition = 530 R / ( 559 R - 530 R )

COP_{rev-air-condition = 530 R / 29 R

COP_{rev-air-condition = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0
3 years ago
Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
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