Ask question

Ask question

All categories

3 years ago

11

Q 28.7: 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

1 answer:

Zinaida [17]3 years ago

3 0

Answer:the maximum Hall voltage across the strip= 0.00168 V.

Explanation:

The Hall Voltage is calculated using

Vh= B x v x w

Where

B is the magnitude of the magnetic field, 5.6 T

v is the speed/ velocity of the strip, = 25 cm/s to m/s becomes 25/100=0.25m/s

and w is the width of the strip= 1.2 mm to meters becomes 1.2 mm /1000= 0.0012m

Solving

Vh= 5.6T x 0.25m/s x 0.0012m

=0.00168T.m²/s

=0.00168Wb/s

=0.00168V

Therefore, the maximum Hall voltage across the strip=0.00168V

You might be interested in

Answers are ONLY allowed to be ionic, polar covalent, and nonpolar covalent.

Luba_88 [7]

A) It will be 2 covalent bonds
B) covalent bonds occur when there’s 2 atoms that share electrons. In this case by sharing the 2 pairs of valence electrons each atom has a total of 8 valence electrons

8 0

2 years ago

A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1

vesna_86 [32]

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

$\rho = \frac{m}{V}$

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

$m_b = 3Kg$

$r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3$

In equilibrium, the entire system is equal to zero, therefore

$\sum F = 0$

$F_g +F_h-F_b = 0$

Where,

$F_g =$ Weight of balloon

$F_h =$ Weight of helium gas

$F_b =$ Bouyant force

Then we have,

$mg+V\rho g -V\rho_a g = 0$

$\rho = \rho_0-\frac{m}{V}$

Replacing the values we have that

$\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}$

$\rho = 0.978kg/m^3$

Now by ideal gas law we have that

$PV=nRT$

$P\frac{\rho}{m} = nRT$

$P = \rho \frac{n}{m}RT$

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

$P = \frac{\rho}{M_0} RT$

$P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K$

$P = 619995.7Pa$

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa

5 0

3 years ago

In Hooke's law, Fspring=kΔx, what does the ∆x stand for

monitta

Answer:

Change in Displacement

Explanation:

delta/triangle = change

x = displacement

formula (if needed): final x - initial x

7 0

3 years ago

While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for

Nimfa-mama [501]

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

5 0

3 years ago

A disk, initially rotating at 120 rpm, is slowed down with a constant acceleration of magnitude 4 rad s2. If the disk has diamet

andrey2020 [161]

Answer:

Explanation:

given,

ω₁ = 120 rpm

1 rpm = $\dfrac{2\pi }{60} rad/s$

$\omega_1 =120\times \dfrac{2\pi }{60}$ rad/s

= 12.56 rad/s

α = - 4 rad/s²

diameter of disk = 20 cm

final angular velocity = 0

t = $\dfrac{\omega_f-\omega_i}{\alpha}$

t = $\dfrac{0-(12.56)}{-4}$

t = 3.14 s.

2) $\theta = \omega_i t +\dfrac{1}{2} \alpha t^2$

= $12.56\times 3.14 +\dfrac{1}{2}\times (-4)\times 3.14^2$

= 19. 72 radians

3) total angular distance rotated

x = θ R

x = 19.72 × 0.1 = 1.97 m

x = 2 m

3 0

3 years ago