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ycow [4]
3 years ago
12

How does gravity affect maglev trains?

Physics
1 answer:
yuradex [85]3 years ago
3 0
Earth pulls it downward to the gravitational force
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The energy generated by the sun travels to Earth as electromagnetic waves of varying lengths. Which statement describes an elect
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Answer: 4(45)

Explanation:

It’s not equal

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2 years ago
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A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =
SashulF [63]

Answer:

The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

Explanation:

Given that,

Length of the current element dl=(0.5\times10^{-3})j

Current in y direction = 5.40 A

Point P located at \vec{r}=(-0.730)i+(0.390)k

The distance is

|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}

|\vec{r}|=0.827\ m

We need to calculate the magnetic field

Using Biot-savart law

B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

We need to calculate the value of \vec{dl}\times\vec{r}

\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

7 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

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A material that has high resistance to the flow of electric current is called an electric resistor
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On Monday, Jacques announces to his parents that he wants to be called "Jack." On Wednesday, he says he wants to drop out of sch
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Jaques seems to be in the Rebelious stage
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