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vredina [299]
2 years ago
12

How much energy is needed just to boil 5kg of water at 100◦ C?

Physics
1 answer:
ddd [48]2 years ago
8 0
11285

Mark brainless pls
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When you buy U.S government saving bond you're doing so under the power given to congress too
lilavasa [31]
Your to borrow money
4 0
4 years ago
The clothes washer in your house consumes 470 kWh of energy per year. Price of the washer is $360 and the lifetime of the washer
VashaNatasha [74]

Answer:

$893

Explanation: the complete question should be

The clothes washer in your house consumes 470 kWh of energy per year. Price of the washer is $360 and the lifetime of the washer is 10 yrs. Energy price in your city is 9 cents per kWh. What is the lifecycle cost of the clothes washer? (assume a maintenance cost of $11 per year)

SOLUTION

Given:

The clothes washe power consumption (PC) is 470 kWh

Price of the washer (P) is $360

lifetime of the washer (L) is 10 yrs

Energy price in the city (E) is 9 cents per kWh (Covert to $ by dividing 100)

maintenance cost (M) is $11 per year

Lifecycle cost = P + (PC × L × E) +M + L

Lifecycle cost = $360 + (470kWh × 10years × 9cents/100) + ($11 × 10years)

=$893

7 0
3 years ago
if a cyclist is travelling a road due east at 12km/h and a wind is blowing from south-west at 5km/h. find the velocity of the wi
slamgirl [31]

Answer:

The velocity of wind with respect to cyclist is -15.5 \widehat{i} - 3.5 \widehat{j}.

Explanation:

speed of cyclist = 12 km/h east

speed of wind = 5 km/h south west

Write the speeds in the vector form

\overrightarrow{vc} = 12 \widehat{i}\\\\overrightarrow{vw} = - 5 (cos 45 \widehat{i} + sin 45 \widehat{j})\\\\\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}

The velocity of wind with respect to cyclist is

\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}\overrightarrow{v_(w/c)} = \overrightarrow{vw}-\overrightarrow{vc}\\\\\overrightarrow{v_(w/c)} = - 3.5 \widehat{i} - 3.5 \widehat{j}  - 12 \widehat{i}\\\\\overrightarrow{v_(w/c)} =-15.5 \widehat{i} - 3.5 \widehat{j}

7 0
3 years ago
PHYSICS CIRCUIT QUESTION PLEASE HELP!! 20 Points!
dimulka [17.4K]
This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.

If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.

The energy dissipated by the circuit is the energy delivered by
the battery.  We'd know what that is if we knew  I₁ .  Everything that
flows in this circuit has to go through  R₁ , so let's find  I₁  first.

-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
--  I₁ is  10volts/10Ω  =  1 Ampere.

-- R1:  1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts. 
    7 of the 10 appear across R₁ .
   So the other 3 volts appear across all the business at the bottom.

-- R₂:  3 volts across it = V₂. 
           Current through it is  I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4:  6Ω in the series combination
                     3 volts across it
                     Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere

--  Remember that the current is the same at every point in
a series circuit.  I₃  and  I₄  must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.

-- R₃:  1/2 Ampere through it = I₃ .
           1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄:  1/2 Ampere through it = I₄
           1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that  I₂  is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches  I₁  coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.

Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up.  But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.

The power supplied by the battery  = (voltage) · (current)

                                                         =  (10 volts) · (1 Amp) = 10 watts .

"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.

             (30 minutes) · (60 sec/minute)  =  1,800 seconds

             (10 joules/second) · (1,800 seconds)  =  18,000 joules  in 30 min

The power (joules per second) dissipated by each individual resistor is

                       P  =  V² / R
             or
                       P  =  I² · R ,

whichever one you prefer.  They're both true.

If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total. 

They're not asking for that.  But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.
7 0
3 years ago
Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz
amid [387]

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

7 0
4 years ago
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