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17

The work done by a force is not defined by $W=\mathbf F\cdot\mathbf r$. Work is instead defined in terms of a line integral over a path (your equation just assigns a work for a force and position, which does not match what we mean by the work done by a force). We have $$W\equiv\int\mathbf F\cdot\text d\mathbf r\to\text dW=\mathbf F\cdot\text d\mathbf r$$ So ...


11

You can set up $$R=\frac VI$$ and still say that $R$ is constant and independent from $V$ in the same way that you can set up Newton's 2nd law, $$m=\frac Fa, $$ and still say that the mass $m$ is constant and independent from the force. Surely you don't suddenly get heavier just because someone pushes you with a larger force; likewise $R$ doesn't change (for ...


8

Here is the challenge you have to overcome. To actually deliver 200 watts of acoustic power into a solid block of granite requires that the transducer be impedance-matched to the load. If you press on a piece of granite, it presses back really hard and experiences almost no compression in response. This makes it an ultrahigh impedance load, and if your ...


6

When you connect the bulb to 110v, the current, I, will not be 5/11. We know $I=V/R$, so when voltage changes, amps will consequently change. It is better, for this problem, to use $P=V^2/R$ because it has two variables ($P and V$), while in $P=VI$, we have three variables.


4

As long as your entire apparatus is sitting in a gravitational field, there is no way to devise a "clever cheat" that turns gravitational potential energy into limitless mechanical work. This is not because we humans are simply not clever enough to think up some way to do this, it's because of the way that gravity works: the relationship between ...


4

Work is defined as $W = \int_{}^{} \vec F \cdot d \vec r = \int_{}^{} \vec F \cdot \vec v \enspace dt$. Power, P, is dW/dt = $\vec F \cdot \vec v$. Your relationship for work is incorrect, so your relationship for power (boxed-in relationship in your question) is not correct.


4

If the two forces were equal in absolute value, the resulting force would be null and the bucket would stay still. The key in understanding this lies in the stipulation that: if the bucket rises at a constant speed, [...] Both buckets move at the same velocity, so by Newton's Second Law the net force acting on them must be zero (null). They would not &...


4

If the bulb obeyed Ohm's law, the resistance would not change when the voltage is changed. So when the voltage is halved the current halves. Method 2 would then be correct, as would use of $P=IV$ with both $I$ and $V$ half their value at 220 V. If your bulb is a filament lamp, the resistance increases with applied voltage, as the filament gets hot. [The ...


3

Not all things are resistors, therefore not all things follow Ohm's Law. Motors, for example, are resistors, inductors, voltage sources all rolled into one. Some of these things read zero when no current is passing through the motor or if the motor is not spinning. Therefore when you try to measure the resistance of the motor at stand-still, you are ...


3

The well-known formula for the period of a simple pendulum performing small oscillations doesn't involve the mass of the bob (because the gravitational force is in the ratio $m_1/m_2$, but so too is the inertia.) However for a real pendulum there are at least three small mass-dependent effects, owing to the bob of larger mass having a larger volume ... (a) ...


3

In the equation $P = V \cdot I$, the $V$ stands for voltage, not velocity. The units of voltage are joules/coulomb, and the units of $I$ (current) are coulombs/second, so you should get the same answer.


3

Your discussion is valid till $\displaystyle{F\frac{dv}{dt}=-v\frac{dF}{dt}}$. I think the only way to reach $F(v-v_o)=v(F-F_o)$, is by assuming $F$ is constant while integrating on the left side and by assuming $v$ as constant while integrating on the right side. This is inconsistent because on the left side $F$ is being treated as a constant and on the ...


3

Complex Power is composed of a Real component (dissipated power in Watts) and an Imaginary component (Reactive stored energy in Volt-Amperes Reactive (VARs)). The total is combined as a vector quantity (Volt-Amperes = VA). A common quantity to be concerned with is Power Factor (PF). PF is the cosine of the relative phase angle between the AC Voltage ...


3

In ideal resistor the resistance is constant. However, when you derive the formula V=RI (i.e. see the Drude model) you assume that the electric potential is not too high. The non approximated relationship between current $I$ and the electric potential $v$ is not linear and in this case you can't define R to be constant. Note that when using resistors, people ...


3

Use an industrial size vibrator motor bolted to the granite. The motor shaft has an eccentric mass on it which will vibrate at the rotational frequency of the motor. Not enough power? Use more than one. Use a plate compactor if the process can be conducted outdoors. Use a jackhammer, or its smaller cousin, an electric demolition hammer. If you don't want a ...


3

The problem is that you have assumed the acceleration was constant and then calculated it using: $$ a = \frac{v_{final} - v_{initial}}{time} = \frac{8.3}{150} = 0.0553~\text{m/s}^2 $$ But suppose we calculate the distance that the bucket would have moved if the acceleration had this value. That distance is: $$ s = ut + \tfrac12 at^2 = \tfrac12 \times 0.0553 \...


3

Fast breeder reactors (FBRs) and in particular liquid metal fast breeder reactors (LMFBRs) actually require more fissile material (U-235 or Pu-239) than thermal reactors- pressurized water reactors (PWRS) or boiling water reactors (BWRs)- because the fission cross section of the fissile isotopes is smaller at the higher neutron energies in the fast reactor ...


3

Electromagnetic waves are photons, and the energy of a photon is indeed determined by its frequency $f$ according to $E = h \cdot f$. However, you can simply emit more photons to increase signal strength. In classical physics, more photons would mean an electromagnetic wave of larger amplitude.


3

$11 W$ means the bulb uses 11 watts of power when it's operating at the rated voltage. If you run it for an hour for example, it will cost $0.011 kWh$, and your power company will bill you for ___ (check your local electricity prices). You can also calculate how much current the bulb will draw since you know the voltage, as well as the resistance of the bulb ...


3

The water outside the pipe pushes the water inside the pipe up the pipe until it reaches the same level as the water outside. Then the pump only needs to pump the water from this level.


2

The majority of hydrogen (about 95%) is produced from fossil fuels. Among other ways, hydrogen can be produced by electrolysis of water, but electrolysis requires electricity to start with. In most cases it would be inefficient to go through the processes of extracting hydrogen to use as fuel for generating electricity if the fossil fuels themselves will do ...


2

This is being done in remote locations. They have hydrogen electrolysis systems connected to wind turbines. When the wind is blowing they have normal wind power. When the wind stops they have stored hydrogen to provide the power. Not very efficient but certainly worthwhile in remote locations. I am in the hydrogen electrolysis business am very familiar with ...


2

If we increase the resistance of component can't we effectively reduce power loss across wire This is the same thing as saying design the component to run at high voltage low current which is exactly what we do in high voltage power lines. However you seem muddled throughout your question. You start confusing power delivered to the load as power wasted in ...


2

There's acceleration and then there's also "jerk." Jerk is the rate at which the acceleration changes i.e. da/dt. A car that goes, say, 0 to 60 in four seconds can do so in many different ways such as a constant acceleration throughout or a big jump in acceleration at the beginning. So you're right that it's not just acceleration. It's also the ...


2

From a physics point of view, it's exactly acceleration. The larger the acceleration of the vehicle, the greater the force the seat is pushing on you to achieve it. (You could of course render acceleration as a function of mass and torque or power, but that's just complicating it) That said, our bodies are not precision instruments, and our perception of &...


2

The power of the engine is defined as the rate of work done by the engine. Similarly, the power supplied by friction is the rate of work done by the frictional force, and the power supplied by gravity is the rate of work done by the gravitational force. The net power acting on the vehicle is '$F_{net}\cdot v$' which is the reusultant of the individual powers ...


2

to enlarge upon DKNguyen's answer: The electrical power distribution buss (the main "+" cable) in a vehicle possesses only ONE voltage: either the charge setpoint of the regulator (which is about 13.8 volts for a lead-acid battery system) or the battery potential which is between 11.6 and 12 volts for a lead-acid battery, depending on its state of ...


2

There is more than one relevant potential difference. You must distinguish between the potential difference, $V_L$, across the load (i.e. the 'user' that we are aiming to supply) and the potential difference, $V_W$, across just the transmitting wires (of resistance $R_W$ taken together). The load and the wires are in series across the supply so $V_\text{...


2

Brick structures fail in tension at the brick-to-mortar interface. The older the brick building is, the weaker becomes that interface, and 100 year old brick buildings are mostly held together by gravity. Earthquakes cause old brick buildings to easily collapse into low mounds of loose bricks, crushing the contents of the building. The tongue-and-groove ...


2

(I don't have enough rep to comment) The equation that you have used is derived for normal incidence (specifically for elastic collision assuming light as a particle).


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