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0

It's a little bit hard to follow your reasoning. Let me try to give the method I would use - simple balance of energy. If the car reaches a velocity $v$ after time $t$, the power of the engine will have been used mostly for five components: Kinetic energy of car: $\frac12 m v^2$ Rotational kinetic energy of tires: $4 \times \frac12 I_t \omega_t^2$ ...


1

There are many variables that come into play in an armwrestling match, but one answer as to why a person with a larger hand and wrist has an advantage in armwrestling has less to do with strength, and more to do with leverage. In armwrestling, you and your opponent lock hands and attempt to pin each other to a pad on either side of the table. To pin your ...


0

"Stange" harmonics cannot appear. Two 50Hz sinus waves always produce another 50Hz sinus wave : $$cos(\omega t)+cos(\omega t+\phi)=2cos(\omega t+\frac{\phi}{2})cos(\frac{\phi}{2}) $$ About synchronization, you should remember that the voltage in the network is fixed, and thus only the current changes. I don't see any problem with combining two different ...


1

Short answer: in phase radiates more in the far-field. But read the long answer. Long answer: this actually depends on the frequency and the distance between the sources. For a wavelength significantly larger than the distance between the sources, the pressures emitted by each source simply add up when in-phase; and they cancel when out-of-phase. But what ...


0

Electrical sources, like current sources are conventional electrical elements that behaves according to a definition. A current source is said to impose a current. For example, when studying the circuit, you do assume that the current in the wire is chosen by the current source only. Thus, when there are two different current sources in series, you have two ...


0

If I understand correctly, the power that power sources output can be constant, but the current and the voltage is purely dependent on what loads you have in the circuit. Say my power source is a coal burner, that burns 5 coals a minute to produce 10 watts of power. Using P=IV, the current and the voltage can both be seemingly a wide variety of variables. ...


-1

They should because light bulbs connected in parallel definitely give you more energy per unit time :). It is because in parallel you have to circuits...for example, if you have a simple circuit with one bulb, and if you have total of 4 A of current, what will the current be if you add one more, in parallel? If you remember, sum of currents in two branches ...


0

In the formula $$S_X(f)= \frac{\mathcal{N_0}}{2}$$ the "one-sided" power spectrum density is $$\mathcal{N_0} = k_BT_{eq}$$ where$k_B=1.38 × 10^{-23} \text{ [W/Hz/K] }$ is Boltzmann's constant and $T_{eq}$ is the system temperature.


0

in P=VI so when connect the the batteries in parallel then voltage is improving and its current increases because of the less of the resistance ana powrer is directle proportional to both thats why power is also increases....


0

There is indeed an optimal arrangement for extracting the most usfeul energy from moving Newtonian fluids. The highest efficiency possible is $\frac{16}{27}$ - this is known as the Betz Limit or Betz Coefficient, after Albert Betz, who rediscovered it in 1920 after Fred Lanchester first calculated it in 1915. The ratio of exit speed to entry speed is 1:3 ...


1

You've begun with this: $P_l = I^2R$ $P_l = IV$ This is correct, but the $V$ here is not the line voltage, but instead the voltage drop across the resistor under consideration. Increasing the line voltage does not increase the voltage drop. Your diagram with a single resistance and a power station implies that the current in the line depends on that ...


4

You do know that your phone will transmit only enough power to reach the nearest tower right? Most of the time that is much less than the max it is capable of. But when there are buildings between you and the tower or you are on the open road you'll be happy not to be dropping so many calls... So there are two things to notice here. First - every 3 dB ...


0

Acceleration under constant power is $$ a(v) = \frac{P}{m v}$$ Direct integration has $$ t = \int \frac{1}{a(v)}\,{\rm d} v = \int \frac{m v}{P} \,{\rm d}v = \frac{1}{2} \frac{m}{P} \left( v^2 - v_0^2 \right) $$ $$ \boxed{ v(t) = \sqrt{v_0^2 + \dfrac{2 P t}{m}} }$$


1

$P=d(mv^2/2)/dt=m/2 * d(v^2 )/dt=m/2 * d(v^2 )/dv * dv/dt=m/2 * 2v * dv/dt=mva $ $dP/dt=d/dt * (mva)=m[dv/dt * a+v * da/dt]$ $dv/dt * a+v * da/dt=0$ $dv/dt * a=-v * da/dt$ $dv/(-v)=da/a$ $ln⁡(-v)+C=ln⁡(a)$ $C=ln⁡(a/v)$ $a/v=e^C$ $v=a/e^C $ $v=dv/dt * k$ , where $k=1/e^C$ $dv/v=1/k * dt$ $ln⁡(v/v_0)=1/k * t$ $v=v_0 e^{1/k * t}$ $v=C_1 e^{C_2t}$ ...


0

I think the answer is because the book got it wrong and you got it right! Your working looks fine to me - nothing more to add.


1

Calculate the kinetic energy of the water coming out in unit time. That is the power you need. $$Power = \frac12 (\rho A v) v^2 = 31.4 kW$$



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