What happens when light passes through a prism or diffraction grating?

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

mario62 [17]

Answer:

a . 0.35cm

b. 11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

$\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm$

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

$\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm$

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0

3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

6 0

3 years ago

Read 2 more answers

PLS HELP ITS WORTH SO MANY POINTS AHH<br> LONGITUDINAL <br> SURFACE<br> TRANSVERSE<br> HEAT

mash [69]

Answer:

C. transverse

7 0

2 years ago

Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determi

DerKrebs [107]

Answer:

Weight of the woman in Newton

$x = 698.6 \ N$

Mass of the woman in slug

$Mass = 4.86 \ slug$

Mass of the woman in kg

$Mass = 16 \ kg$

My weight in Newton

$W = 784 \ N$

Explanation:

From the question we are told that

The weight of the woman in pounds is $W = 157 \ lb$

Converting to Newton

1 N = 0.22472 lb

x N = 157

=> $x = \frac{157 * 1}{0.22472}$

=> $x = 698.6 \ N$

Obtaining the mass in slug

$Mass = \frac{W}{g}$

Here $g = 32.2 ft/s^2$

So

$Mass = \frac{157 }{32.2}$

$Mass = 4.86 \ slug$

Obtaining the mass in kilogram

$Mass = \frac{W}{g}$

Here $g = 9.8 \ m/s$

So

$Mass = \frac{157 }{9.8}$

$Mass = 16 \ kg$

Generally weight is mathematically represented as

$W = m * g$

Given that my mass is 80 kg then my weight is

$W = 80 *9.8$

$W = 784 \ N$

6 0

3 years ago

Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the charge of object 1 is doubled AND the charge

lbvjy [14]

Answer:

432 units

Explanation:

Let the charges be q and Q separated by a distance r. The electrostatic force , F = kqQ/r² = 72 units. If q = 2q and Q = 3Q, then the new electrostatic force is

F = k × 2q × 3Q/r² = 6kqQ/r² = 6 × 72 = 432 units

5 0

3 years ago