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vredina [299]
3 years ago
7

Using the state mileage guide table located on the top of each map page you can

Engineering
1 answer:
Ivenika [448]3 years ago
5 0
C. Route and roadways defined as class I highways
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Staying hydrated at all times
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Many BLANK apply trial and error to develop a product. Please Help! I have one hour to finish. 30 points
Julli [10]

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Entrepreneurs?

Explanation:

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3 years ago
I don’t understand this
blondinia [14]

Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of 160cm^{2} by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

7 0
3 years ago
A cubical picnic chest of length 0.5 m, constructed of sheet styrofoam of thickness 0.025 m, contains ice at 0\[Degree]C. The th
maksim [4K]

Answer:

Rate of heat transfer is 0.56592 kg/hour

Explanation:

Q = kA(T2 - T1)/t

Q is rate of heat transfer in Watts or Joules per second

k is thermal conductivity of the styrofoam = 0.035 W/(mK)

A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2

T1 is initial temperature of ice = 0 °C = 0+273 = 273 K

T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K

t is thickness of styrofoam = 0.025 m

Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s

Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr

6 0
3 years ago
Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be
hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0
3 years ago
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