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2 years ago

8

True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'

s Law.

1 answer:

defon2 years ago

8 0

Answer:

true

Explanation:

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1. A drawing of a cabinet shows that its dimensions are 9cm. by 4cm. The drawing indicates 1:50 scaling. What are the actual dim

yanalaym [24]

Explanation:

This means that for every 1 cm on the drawing, there is 80 cm in reality. To put it another way, take this

1:80 means that the building is 80 times the size of the drawing

80:1 means that the drawing is 80 times the size of the building

If it were 80:1, the drawing itself would be over 100m long.

5 0

3 years ago

How can the direction of rotation of a split-phase motor be changed? *

S_A_V [24]

Nejejenshsjsjsjs:

Explanation:rjejsjjejej

8 0

3 years ago

Read 2 more answers

If the load parameters are: Vln=600kV, Il=100A (resistive), calculate the source voltage and current when the line is 50Miles (s

Archy [21]

s 0Miles (short), 150 Miles(medium), and 300 Miles (long).

Explanation:

4 0

3 years ago

Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p

oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0

3 years ago

A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is

Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = $\frac{A}{A-C}$

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = $\frac{A}{A-C}$

= $\frac{1034}{1034-675 \cdot 6}$

Apparent specific gravity = 2.88

b) Bulk specific gravity $G_{B}^{O D}=\frac{A}{B-C}$

$G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}$

= 2.76

c) Bulk specific gravity (SSD):

$G_{B}^{S S D}=\frac{B}{B-C}$

$=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}$

$G_{B}^{S S D}$ = 2.80

d) Absorption% :

$=\frac{B-A}{A} \times 100 \%$

$=\frac{1048 \cdot 9-1034}{1034} \times 100$

Absorption = 1.44 %

e) Bulk Volume :

$v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}$

$=\frac{1048 \cdot 9-675 \cdot 6}{1}$

= $373.3 cc$

5 0

3 years ago