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jolli1 [7]
2 years ago
11

A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage

across the strip?
Physics
1 answer:
Alex787 [66]2 years ago
4 0
The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip. 

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.
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Answer:

I = \frac{mvb}{6}

Explanation:

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      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

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put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

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2 years ago
Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
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We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
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2 years ago
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

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v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

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So the displacement is

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\Delta t = 2.0 s - 0 s = 2.0 s

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v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

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v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

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\Delta x = 20.8 - 5.6 = 15.2 m

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\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

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