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adell [148]
3 years ago
7

PLS HELP

Engineering
1 answer:
RSB [31]3 years ago
4 0

Answer:

T=kg·m^2/s^2

Explanation:

T = kg (m^2/s^2) m^3 /m^3

Here I wrote down the unit for every dimension.

T=kg m^2 / s^2

m^3 is divided between m^3, this is equal to 1.

Result: T=kg·m^2/s^2

PD: I'm not sure if this is what you ask for. I hope it helps

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An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisput
evablogger [386]

Answer:

The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisputed support for Stalin both among the government and the common people. In turn, this fueled self-censorship and made political change harder. This cult of personality was achieved through propaganda and censorship, as the Communist Party had control of all mass media. This desire to make himself a "god-like" figure was also an attempt to increase acceptance of communism among the people and to boost morale.

Explanation:

7 0
3 years ago
2. The metal to be welded is called the
Sveta_85 [38]

Answer:

I believe the answer is "filler metal"

5 0
3 years ago
A horizontal 20-mm diameter tube is to carry a laminar flow of SAE 30 oil. What pressure drop. (p1-p2) is needed to produce a fl
notsponge [240]

Answer:

P_1-P_2=795.18 Pa

Explanation:

Given that diameter of tube= 20 mm

flow rate = 72 lt/hr

The properties of SAE 30 oil at 30°C

\mu =155.31 mPa

\rho =872.5 \frac{kg}{m^3}

We know that volume flow rate Q= AV

2\times 10^{-5}=\dfrac{\pi}{4}\times{0.02^2}V

V= 0.064 m/s

Now lets find Reynolds number to check the type of flow

Re=\dfrac{\rho VD}{\mu }

Re=\dfrac{872.5\times 0.064\times0.020}{0.15531 }

Re=7.18 So we can say that this is laminar flow and we know that pressure drop for laminar flow is given as

P_1-P_2=\dfrac{32\mu VL}{d^2}

Now by putting the values the pressure drop for per unit length

P_1-P_2=\dfrac{32\times 0.15531\times 0.064 }{0.02^2}

P_1-P_2=795.18 Pa

7 0
3 years ago
What are watts equal to?
Umnica [9.8K]

Answer:

joules per second

equivalent to one ampere under a pressure of one volt.

4 0
3 years ago
Read 2 more answers
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