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3 years ago

What force is used to make maglev trains run along their tracks?

1 answer:

8 0

Hi, thank you for posting your question here at Brainly.

Maglev trains are called as such, since they are derived from "magnetic levitation". The force that makes them run along their tracks is through attraction and repulsion forces brought about by the magnets.

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My name is Ann [436]

The crate is moving at constant velocity when the forces acting on it are

balanced.

The value of the force <em>F</em> required to pull the crate with constant velocity is; $\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}$

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = $\mathbf{F_N}$

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

$F_N$ ·j = $\mathbf{F_N}$ × sin(θ)

The sum of the vertical component = $F_N$ ·j + $F_N$ ·j = 2· $F_N$ ·j = 2· $F_N$ ×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2· $F_N$ ×sin(θ) = m·g

θ = 45°

Therefore;

2· $F_N$ ×sin(45°) = m·g

$\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}$

Therefore;

$\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N$

$\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}$

Which gives;

$\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}$

$\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}$

Force required to pull the crate at constant velocity, <u>F = √2·μ·m·g</u>

Learn more about force of friction here:

brainly.com/question/6561298

8 0

2 years ago

Read 2 more answers

What in earth's atmosphere helps most to absorb uvb ultraviolet light from the sun?

Whitepunk [10]

The Ozone layer in the atmosphere absorbs the UV rays

Hope it helps!

4 0

3 years ago

Why dont we put (-ve) sign when calculating charge in electrostatics ?

Levart [38]

Answer:

The negative sign represents the flow of charge in an opposite direction relative to point of action.

5 0

3 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago

A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci

larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 = e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

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6 0

2 years ago