suppose both the dog and the girl run at a speed of 2 m/s.Calculate both of there kinetic energies. kinetic energy of dog =

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a

vodka [1.7K]

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The final velocity depends upon initial velocity and initial height.

So, the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

3 0

3 years ago

A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

Newton’s third law in football?

trapecia [35]

Newtons third law of motion applies to football as when the ball is thrown, force is applied to the football thrown forward and also force back to the player. Therefore it states that every action has a reaction. The player does not get a huge force back from throwing the football due to the player having a bigger mass than the football being thrown.

In short, when the football is throw, there is a reaction with the ball and the player accordingly with the force put into the ball.

6 0

3 years ago

The magnetic field 0.100 m from a

Aneli [31]

Answer:

Your answer is given below:

Explanation:

4 0

3 years ago