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Zolol [24]
3 years ago
15

Ordan has _ 5 8 can of green paint and _ 3 6 can of blue paint. If the cans are the same size, does Jordan have more green paint

or blue paint? Explain.
Engineering
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

Jordan has more green paints

Explanation:

Given

Green = \frac{5}{8}

Blue = \frac{3}{6}

Required

Which paint does he have more?

For better understanding, it's better to convert both measurements to decimal.

For the green paint:

Green = \frac{5}{8}

Green = 0.625

For the blue paint:

Blue = \frac{3}{6}

Blue = 0.5

By comparison:

0.625 > 0.5

<em>This means that Jordan has more green paints</em>

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A cartridge electrical heater is shaped as a cylinder of length L=200mm and outer diameter D=20 mm. Under normal operating condi
lara31 [8.8K]

Answer:

T(water)=50.32℃

T(air)=3052.6℃

Explanation:

Hello!

To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.

The equation is as follows!

Q=ha(Ts-T\alpha )

Q = heat

h = heat transfer coefficient

Ts = surface temperature

T = fluid temperature

a = heat transfer area

The surface area of ​​a cylinder is calculated as follows

a=\pi D(\frac{D}{2} +L)

Where

D=diameter=20mm=0.02m

L=leght=200mm)0.2m

solving

a=\pi (0.02)(\frac{0.02}{2} +0.2)=0.01319m^2

For water

Q=2Kw=2000W

h=5000W/m2K

a=0.01319m^2

Tα=20C

Q=ha(Ts-T\alpha )

solving for ts

Ts=T\alpha +\frac{Q}{ha}

Ts=20+\frac{2000}{(0.01319)(5000)} =50.32C

for air

Q=2Kw=2000W

h=50W/m2K

a=0.01319m^2

Tα=20C

Ts=20+\frac{2000}{(0.01319)(50)}=3052.6C

3 0
3 years ago
One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p
kupik [55]

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)  

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

3 0
3 years ago
It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less
QveST [7]

Answer:

BCDE

Explanation:

just look at the link, it tells you.

7 0
3 years ago
Read 2 more answers
You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen
VashaNatasha [74]

Answer:

Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

6 0
3 years ago
The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome
TEA [102]

Answer:

(a) Relative Humidity = 48%,

Specific humidity = 0.0095

(b) Enthalpy = 65 KJ/Kg of dry sir

Specific volume = 0.86 m^3/Kg of dry air

(c/d) 12.78 degree C

(e) Specific volume = 0.86 m^3/Kg of dry air

8 0
3 years ago
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