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2 years ago

What is a table saw for

Engineering

2 answers:

bezimeni [28]2 years ago

7 0

Cutting the table because a table saw is for cutting tables.......

svet-max [94.6K]2 years ago

5 0

Answer:

a table

Explanation:

because you can saw the table

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Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a

stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0

2 years ago

Convert the unit Decimeter (dm) into Micrometer (um).

oksian1 [2.3K]

Answer:

86701 Micrometers.

Explanation:

Multiply 0.86701 dm by 100,000 to get 86701 um.

7 0

2 years ago

A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

7 0

2 years ago

You installed a new 40 gallon water heater with a 54,000 BTUh burner. The underground water temperature coming into the house is

kipiarov [429]

14256000. Kanjiuijhgg

5 0

2 years ago

Read 2 more answers

Problem 1: A catchment has the following Horton’s infiltration parameters: f0=280 mm/hr, fc=25 mm/hr and k = 2.5 hr-1 . For the

harkovskaia [24]

Answer:

Ponding will occur in 40mins

Explanation:

We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.

Consider checking attachment for the step by step solution.

6 0

3 years ago