All categories

2 years ago

light of wavelength 520 nm falls on a slit that is 3.20 um wide. Estimate how far the first bright-sh difrrection fringe is

Physics

1 answer:

Westkost [7]2 years ago

6 0

Answer:

because of the gravity of the earth

You might be interested in

As a marble rolls across the floor, it gradually slows to a stop due to friction. Which statement best describes the change in m

Olegator [25]

Answer:

I belive it would be "C"

Explanation:

If it was any of the other answers "B" it would instantly stop. "A" it would roll forever.

3 0

3 years ago

A stone on ground is zero energy

NNADVOKAT [17]

Answer:

A stone on the ground does not have zero energy…there is an internal potential in every object. Aldo is not in action or in any mechanical motion it is being acted upon by gravity and also molecular forces and energy.

Hope this helps !

7 0

3 years ago

Which of the Jovian planets have rings?

stira [4]

Jupiter , Saturn , Uranus , Neptune .

7 0

3 years ago

Read 2 more answers

Sb + Cl2 → SbCl3 Balance chemical equation

melamori03 [73]

Answer:

Cl2 + Sb → SbCl3

Cl2 + Sb → SbCl5

Cl2 + Sb → SbCl

Cl2 + Sb → SbCl2

Cl2 + Sb → SbCl3 + SbCl5

Explanation: Hope it will help

7 0

3 years ago

A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order t

NikAS [45]

Answer:

C. $\frac{25}{7}$ m

Explanation:

We are given that

Weight of board=w=10 N

Length of board=L=5 m

Tension in the string=T=3 N

Applied upward force=F=7 N

We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

The board is uniform therefore, the center of board is the mid- point of board.

Therefore, the lever arm of weight= $r_1=\frac{L}{2}=\frac{5}{2}m$

Now, the torque exerted by the weight of the board

$\tau_1=Force\times perpendicular\;distance=10\times \frac{5}{2}=25 N$

The torque exerted by applied force= $\tau_2=7\times r=7r$

In static equilibrium

The sum of rotational forces=0

$\tau_1+\tau_2=0$

The two rotational force act in opposite direction therefore,

$\tau_2=-7r$

Substitute the values

$25-7r=0$

$7r=25$

$r=\frac{25}{7}$ m

Hence, option C is true.

7 0

3 years ago