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3 years ago

If the magnetic field is constant and the magnet fixed, what could one do to generate an EMF?

Physics

1 answer:

julia-pushkina [17]3 years ago

6 0

Answer:

• Bring another magnet wound with a copper wire, at the top of the fixed magnet. Here EMF will be generated by mutual induction

Explanation:

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A 4.0-kilogram block, initially at rest, is acted upon by an 8.0 newtons force for 10 seconds. How far, in meters, will the mass

Shtirlitz [24]

Answer:

Explanation:

5 0

3 years ago

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be cr

Grace [21]

Answer: 0.313 rad/s

Explanation:

The equation that relates the velocity $V$ and the angular velocity $\omega$ in the uniform circular motion is:

$V=\omega.r$ (1)

Where $r=d/2=100m$ is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.

Isolating $\omega$ from (1):

$\omega=\frac{V}{r}$ (2)

On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration $a_{c}$ is $9.8m/s^{2}$ , and is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (3)

Isolating $V$ :

$V=\sqrt{a_{c}.r}$ (4)

$V=31.3049m/s$ (5)

Substitutinng (5) in (2):

$\omega=\frac{31.3049m/s}{100m}$ (6)

$\omega=0.313rad/s$ This is the angular velocity that would produce an “artificial gravity” of 9 $9.8m/s^{2}$ .

6 0

3 years ago

The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t

Gemiola [76]

Answer:

$\omega=7.16*10^{-13}\frac{rad}{s}$

Explanation:

The angular speed is given by:

$\omega=\frac{v}{r}$

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to $\frac{m}{s}$ :

$1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}$

Now, we calculate the angular speed:

$\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}$

8 0

4 years ago

A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai

Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air ( $\rho$ ), measured in kilograms per cubic meter, is defined by following equation:

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$M$ - Molar mass, measured in kilomoles per kilogram.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 2026.5\,kPa$ , $M = 28.965\,\frac{kg}{kmol}$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ and $T = 300\,K$ , then the density of air is: