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IrinaK [193]
2 years ago
12

Explain mathematically whether the divergence of the curl of a vector field is

Engineering
1 answer:
svp [43]2 years ago
6 0

Answer:

yes

Explanation:

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Sarah needs to create an architectural drawing for a museum building with an inclined surface. Which presentation view will be t
prohojiy [21]

Answer: auxiliary

Explanation: got it right

7 0
2 years ago
Chemical manufacturers must present which information on the product's label?
Wewaii [24]

Answer:

hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement

Explanation:

this is so you know what chemicals are in it

6 0
3 years ago
Which material has the highest cp value?
Nataly [62]

Answer: B. Water

Explanation:

6 0
2 years ago
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A
uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0
3 years ago
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