Two very narrow slits are spaced 1.80 mm apart and are placed 35.0 cm from a screen. What is the distance between the first and
1 answer:
The distance between the first and second dark lines of the interference pattern is mathematically given as
d= 0.107 m
<h3>What is the distance between the first and second dark lines of the interference pattern?</h3>Question Parameters:
Two very narrow slits are spaced 1.80 mm apart and are placed 35.0 cm from a screen
coherent light with 550 nm
Generally, the equation for the distance between the first and second dark lines is mathematically given as
Therefore
d= 550*10^-9*0.35/(1.8*10^-6)
d= 0.107 m
In conclusion, the distance between the first and second dark lines
d= 0.107 m
Read more about Distance
You might be interested in
The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
<h3>Angular Speed of the pulley </h3>
The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;
K.E = P.E
Substitute the given parameters and solve for the angular speed;
The linear speed of the block after travelling 0.7 m;
v = ωR₂
v = 35.39 x 0.03
v = 1.1 m/s
Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
Learn more about conservation of energy here: brainly.com/question/24772394
<u><em>The question doesn't provide enough data to be solved, but I'm assuming some magnitudes to help you to solve your own problem</em></u>
Answer:
<em>The maximum height is 0.10 meters</em>
Explanation:
<u>Energy Transformation</u>
It's referred to as the change of one energy from one form to another or others. If we compress a spring and then release it with an object being launched on top of it, all the spring (elastic) potential energy is transformed into kinetic and gravitational energies. When the object stops in the air, all the initial energy is now gravitational potential energy.
If a spring of constant K is compressed a distance x, its potential energy is
When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as
We have then,
Solving for h
We have little data to work on the problem, so we'll assume some values to answer the question and help to solve the problem at hand
Let's say: x=0.2 m (given), K=100 N/m, m=2 kg
Computing the maximum height
The maximum height is 0.10 meters