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Vera_Pavlovna [14]
2 years ago
8

If a 110-volt appliance requires 20 amps, what is the total power consumed?

Engineering
2 answers:
juin [17]2 years ago
4 0

Answer:

  2200 W

Explanation:

Use the given relation between current, voltage, and power to find the power requirement:

  P = IV

  P = (20 A)(110 V) = 2200 W

Novay_Z [31]2 years ago
4 0
Unit for power is watts (w)
p=IV
110x20=2200W
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Explain the differences among sand, silt, and clay, both in their physical characteristics and their behavior in relation to bui
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Answer:PHYSICALLY- sand is spherical in shape with large particles up to 0.18g

Silt also contains more of spherical particles with weights around 0.0029g.

Clay contain plate like particles,which are flat or layered. It's particles are generally lower that silt and sand below 0.0029g.

BEHAVIORS Sand particles are coarse and very porous,water can easily penetrate with no particles cohesion,does not retain water, difficult to expand,it is IDEAL FOR BUILDINGS.

Silt particles are also porous with some coarseness and no particles cohesion. retains water and can expand. NOT IDEAL FOR BUILDINGS.

Clay contain particles that are not coarse,they have high cohesiveness,they are not porous as it is difficult for water to penetrate. NOT IDEAL FOR BUILDINGS.

Explanation: Sand particles are coarse,very porous,contains spherical particles with large particles sizes and IDEAL FOR BUILDINGS

Silt particles are also porous, with some coarseness and with little or no particles cohesion, NOT IDEAL FOR BUILDINGS.

Clay soils are not porous water can not easily flow thought it,it is hard when dry and soft when wet. IT IS NOT A GOOD OPTION FOR BUILDINGS.

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3 years ago
You start your car and begin to pull out of a parking space. After leaving the space, You notice that the brake light on your in
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3 years ago
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
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79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

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