Ayren purchased a 1000W (120V) hair dryer in the USA,

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Ayren's 1000W hair dryer, designed for 120V, cannot be safely used in Europe where the voltage is 220V. Initially, the calculation suggested it would draw 4.545 amps, which is under the 15A outlet limit. However, further analysis revealed that the dryer’s resistance is 14.4 ohms, leading to a current draw of 15.28 amps at 220V. This exceeds the outlet's 15A rating, indicating that using the dryer in Europe could be unsafe. Therefore, it is not advisable to operate the hair dryer in this situation.
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Ayren purchased a 1000W (120V) hair dryer in the USA, and she took it to Europe where the standard voltage is 220V. Assuming that the hair dryer can operate safely at the higher voltage, can you actually use it if the outlet is rated for 15A, or will it draw too much current (How much current)?

I went about trying to solve this with P=IV , power=current x Voltage

1000W= I (220V)

I = 1000W/220V = 4.545 Amps,

this is less than the 15A that the outlet is rated for so it should be able to, but i guess i missed something since this is incorrect. Please help.
 
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The power drawn isn't the same in Europe as in the USA.

Find the resistance of the dryer.
 
Ok i see.

1000W= I x 120V

I= 1000/120 = 8 1/3 amps

V=IR R= V/I R= 120V/8.333Amps R=14.4 ohms

then in Europe

220V= I (14.4ohms)

I= 220V/14.4ohms I=15.2778amps

this is greater than the 15 amps the outlet is rated for so it would not work.
 
Yup. Looks right.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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