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avanturin [10]
3 years ago
8

A North American tourist takes his 26.0 W, 120 V AC razor to Europe, finds a special adapter, and plugs it into 220 V AC. Assumi

ng constant resistance, what power in watts does the razor consume as it is ruined?
Physics
1 answer:
Irina-Kira [14]3 years ago
7 0

Explanation:

For North America,

Power, P = 26 W

Voltage of AC razor, V = 120 V

For Europe,

Voltage of AC razor, V' = 220 V

The power of razor in North America is given by :

P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{120^2}{26}\\\\R=553.84\ \Omega

Let P' is the power consumed by razor in Europe. Now,

P'=\dfrac{V'^2}{R}\\P'=\dfrac{220^2}{553.84}\\\\P'=87.38\ \Omega

So, the razor consumed 87.38 watts of power as it is ruined.

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers
Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0
3 years ago
What net force would be needed to accelerate the box at 2.3 m/s^2
Akimi4 [234]
The pertinent equation here is F=ma.  You haven't shared the mass of the box, so I will use M to represent that mass.

Then F = M(<span>2.3 m/s^2)        (answer)</span>
6 0
3 years ago
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QUESTION:
vampirchik [111]

Answer:

I don't know why you are asking me?

5 0
2 years ago
Would someone plz help a shister outttt
tiny-mole [99]

Yaaaas, do you watch James Charles!?

8 0
3 years ago
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