What's the momentum (in kg-m/s) of a cannonball with a mass of 21 moving with a velocity of 25 m/s?

A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0 nm and 589.6 nm. The

DIA [1.3K]

The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

L = 57.88 mm

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

Read more at; brainly.com/question/17161594

8 0

2 years ago

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S

Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic ﬁeld

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

8 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$