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Y_Kistochka [10]
3 years ago
5

What are some sources of resistance? (Check all

Engineering
1 answer:
Zarrin [17]3 years ago
7 0

Answer:

All of them.

Explanation:

I take this class rn too and those are answers

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I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!
allsm [11]

Answer:

the answer how you analyzs the problwm

6 0
3 years ago
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1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and
belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of  sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = \frac{wet mass - dry mass }{wet mass} = \frac{0.525 - 0.49}{0.525} = 0.07 or 7%

water content =  \frac{wet mass - dry mass}{wet mass} = 7%

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3 years ago
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Mr. Blue lives in a blue house, Mrs. Pink lives in a pink house and Mr. Red lives in a red house. Who lives in the White House?
Cerrena [4.2K]

Answer:

the president and mr.white my history teacher lol

6 0
3 years ago
Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of
Orlov [11]

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

7 0
3 years ago
Select the correct answer.
Elodia [21]
I think balance




Can I get Brainlyist
3 0
3 years ago
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