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3 years ago

Define, in words the following i relative humidity ii) dew point temperature

Engineering

1 answer:

defon3 years ago

7 0

Explanation:

i) When we divide water vapor's partial pressure by the water's equilibrium vapor pressure at a given temperature we get relative humidity.

It is denoted by ∅

$\phi ={\frac {p_ {H_{2}O} }{p_{H_{2}O} ^{*}}}$

ii) When air is cooled so that it becomes saturated with water vapor the temperature where this occurs is called the dew point.

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If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the

Vitek1552 [10]

Answer:

0.25 J/K

Explanation:

Given data in given question

heat (Q) = 100 J

temperature (T) = 400 K

to find out

the change in entropy of the given system

Solution

we use the entropy change equation here i.e

ΔS = ΔQ / T ...................a

Now we put the value of heat (Q) and Temperature (T) in equation a

ΔS is the entropy change, Q is heat and T is the temperature,

so that

ΔS = 100/400 J/K

ΔS = 0.25 J/K

4 0

2 years ago

You are considering purchasing a compact washing machine, and you have the following information: The Energy Guide claims an est

RSB [31]

Answer: $15.34

Explanation: see image below

8 0

2 years ago

A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have $volume of box = 25ft*10ft*12ft= 3000ft^{3}$

$Volume= 84.95m^{3}$

$Since 1ft^{3} =0.028m^{3}$

Now the weight of water displaced = $Weight =\rho \times Volumewhererho$ is density of water = 1000kg/ $m^{3}$

Thus weight of liquid displaced = $\frac{84.95X1000}{1000}tonnes=84.95 tonnes$ ..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0

2 years ago

Find: factor of safety (n)for point A and B by using both MSS and DE (you can neglect shear stress due to shear force and also n

gladu [14]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : Factor of safety for point A :

i) using MSS

(Fos)MSS = 3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS = 3.04

ii) using DE

(Fos)DE = 3.604

Explanation:

Factor of safety for point A :

i) using MSS

(Fos)MSS = 3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS = 3.04

ii) using DE

(Fos)DE = 3.604

Attached below is the detailed solution

8 0

3 years ago

A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring

ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0

3 years ago