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1

Poiseuille's law will tell you that a pipe of 0.1 m diameter will achieve a flow velocity of around $v=0.1\mathrm{~m/s}$, which is actually more than I expected. For pipes a bit wider than this, the flow will be turbulent (Re>2200), for which you can't apply Poiseuille. For turbulent flow, you can look into the Darcy-Weisbach equation. The interesting part ...


1

Here are the equations for your calculation (starting from energy but not force). Wind energy (mainly kinetic energy) $$E_k=\frac12 m v^2= \frac12 (\rho V) v^2$$ Then wind power $$P=\frac {dE_k}{dt}=\frac12 (\rho \frac {dV}{dt}) v^2 =\frac12 (\rho A v) v^2=\frac12 \rho A v^3$$, which is the equation Han-Kwang showed above. Note here the velocity is the ...


0

It looks like a dubious exercise question, unless the pages preceding the exercise set up a framework of what kind of things you're allowed to assume here. Both the entrance and exit (speaker/ear) represent discontinuities in the acoustic impedance of the tube. The impededance of the speaker (will it absorb reflected acoustic energy or not?) is not described ...


0

You are seeking to much behind the question. Betz's law states that the theoretical limit for a wind turbine is 59% of the kinetic energy ($P=A\rho v^3/2$) of the air flowing through the swept area of the blades if the windmill hadn't been there. In your case, you have 50% rather than 59%, which is a realistic value for a modern wind turbine.


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You would have to know the cross-sectional area of each of the blades, not just the area swept, along with pitch angle and other information. Additionally, one thing that is not provided is the direction of the wind. Is the wind blowing directly at the turbine, or at an angle? If the wind blows at an angle, material bending stresses must also be taken into ...


0

Let say the power produced by a transmission company is 50 Giga_Watt(we can take any value really). What happens is, they convert this to very high voltage.When you use a 1 kilo_ohm resistor, the current draw should be 22 amps, but the power generated(50Kw) must be kept in mind! You cannot have a power source of 10 W and expect it to provide 10 Volts at 2 ...


0

To generate more power, an engine needs to be able to burn more fuel; laws of physics, chemistry and thermodynamics dictate this requires a larger displacement (bigger volume in which you burn the fuel). Larger volume = larger area over which you generate friction, more more importantly, more air being pulled through per stroke. Some engines with a lot of ...


0

You probably managed to get only one hotspot of maximum wave amplitude hitting your coffee while two is more usual. Putting your cup off-centre on the rotating dish would probably get the more average heating you expected.


0

'valerio92' is correct and for: $$\frac{d}{dt} u(t) + A \ u(t)^{3/2} + B = 0$$ ... there probably are no analytical solutions (wolfram alpha's DSolve for instance yielded nothing). But $F_d=C_d A \frac{\rho}{2}v^2$ is not the only model for drag forces. At low speeds: $$F_d=kv$$ ... can be used, so linear dependence of the drag force on speed. The ...


2

You will obtain a first-order nonlinear differential equation: $$P=P-P_d = P-D v^3 = \frac{m}{2} \frac{d}{dt}(v^2) $$ $$\to \frac{d}{dt} (v^2) + \frac{2D}{m} v^3 - \frac{2P}{m} = 0$$ If we define $u(t)=v(t)^2$ , $2D/m = A$ and $-2P/m = B$, we obtain $$\frac{d}{dt} u(t) + A \ u(t)^{3/2} + B = 0$$ When $A$ is small this reduces to the result you obtained ...


0

Peak level might mean "the amount of power that caused transformer failure." Rms (root mean square) might mean "typical continuous running load power".


-1

The formulas are of course all true if used and interpreted correctly. But human error is the wild card. For practical reasons, I-squared-R is the most reliable formula because it's almost impossible to apply it incorrectly.


0

After solving problems on circuits and power dissipation in them, I observed that $V^2/R$ is used when the voltage is constant across the elements in the circuit and $I^2R$ is used when current is constant through the elements in the circuit. They yield the same result when a purely resistive load is used. Even the formula $P=VI$ will give the answer. This ...


1

If the water bucket is not accelerating there should not be any resultant force on it. The bucket is moving at constant speed, so all forces acting on it cancel: it has gravity acting with $mg$ on it and tension equal and opposite to gravity. But, F in the equation for power is not the net force, it is the driving force: $$P = F_{driving} v$$, i.e. the force ...


2

From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ...


1

Along the lines of what @CuriousOne said you can use Kirchhoff's laws to understand the current and voltage drop at each circuit element with any complexity you desire. Kirchhoff's laws deal with the fact of the total sum of currents and voltages in any circuit junction and loop, respectively, must equal zero. The second law essentially uses Ohm's law to ...


2

When you work "fairly hard", your body can produce about 200 W of power - enough for two incandescent bulbs. Top athletes can produce more - in short bursts. Your body is roughly 25% efficient in converting "calories" (which are actually kilo calories) to Joules - meaning that if you work out hard enough to burn 600 kcal per hour, then you actually produced ...


2

A (kilo)calorie is a unit of energy, while a watt is a unit of power, which describes the rate at which energy is expended. So a 100W bulb is using 100 joules a second. A kcal is about 4184 joules, so a 100W bulb takes about 42 seconds to consume (really: convert into light and heat) a kcal. The joule is the SI (derived) unit of energy. Units of energy ...


4

An average person uses approx. 1500-2500kcal/day. Since one kcal equals 4148J in SI units, that's between 6.2-10.4MJ per day. A day has 86400 seconds, which brings us to an average power consumption of 72-120W... about as much as a light bulb. :-) Physical exercise varies between light (300kcal/h) at an additional 350W to very strenuous at probably six ...


1

You're not the first, nor the last, to find the phrase "power flow" somehow wrong. For example, from W J Beaty's article on electrical misconceptions: ELECTRIC POWER FLOWS FROM GENERATOR TO CONSUMER? Wrong. Electric power cannot be made to flow. Power is defined as "flow of energy." Saying that power "flows" is silly. It's as silly as saying that ...


0

Well, if you think about electric power, which includes current (notion of flux), then you'll end with the conclusion that if there's no flow, there is no power.



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