What are the dimensions of F = ma?

The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne

aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ $V_{H} =\frac{I}{nqt} B$

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ $\frac{V}{T}$

b = —0.02 μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH = $100X10^{-6}$ V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

$a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm$

To learn more about the hall voltage , Visit: brainly.com/question/19130911

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8 0

1 year ago

A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces

snow_lady [41]

Answer:

The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds.

Initial velocity = 33 m/s

Final velocity = 0 m/s

Average velocity = (33 + 0) / 2 m/s = 16.5 m/s

Explanation:

First, how long does it take the truck to come to a complete stop?

( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds

Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.

4 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago

What is one use of the mineral gypsum fertilizer medicine wallboard electrical insulator

Semmy [17]

Answer:

The main use for gypsum is in the production of Plaster of Paris and drywall products. Gypsum is also used as a soil conditioner.

Explanation:

7 0

2 years ago

Approximating the eye as a single thin lens 2. 75 cmcm from the retina, find the eye's near-point distance if the smallest focal

RSB [31]

The near-point distance of the eye is 11 cm.

<h3>What is the eye's near-point distance?</h3>

The near-point distance of the eye is the closest possible distance an object can be from the eye in order for its image to be formed on the retina. It can also be termed the closest distance of accommodation.

The near-point distance of the eye in the given scenario can be calculated using the lens formula given below:

1/f = 1/v + 1/u

where;

f = focal length

v = image distance

u = object distance

From the data provided;

f = 2.20 cm

v = 2.75 cm

u = ?

Solving for u:

1/u = 1/f - 1/v

1/u = 1/2.20 - 1/2.75

1/u = 0.91

u = 11 cm

In conclusion, the lens formula is used to determine the eye's near-point distance.

Learn more about eye's near-point distance at: brainly.com/question/16391605

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5 0

2 years ago