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Helen [10]
2 years ago
15

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

second dark lines of the interference pattern when the slits are illuminated with coherent light with l
Physics
1 answer:
Ilya [14]2 years ago
5 0

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

d= \lamda*R/d

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

brainly.com/question/4931057

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Sergeeva-Olga [200]

Answer:

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5 0
2 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

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

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

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

5 0
2 years ago
While in empty space, an astronaut throws a ball at a velocity of 11 m/s. what will the velocity of the ball be after it has tra
Aliun [14]
In empty space probably means, there is no force on the ball.

(This assumption is not quite correct since there is still the force of gravity between the ball and the astronaut, but this force is very very small and can be neglected.)

Assuming there is no force on the ball, Newtown's 1st law says: When viewed in an internal frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.
 
This means:
If there is no force on the ball, there will be no acceleration on the ball either.
If the acceleration is zero, the velocity of the ball never changes.
3 0
3 years ago
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the s
Nostrana [21]

<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

\displaystyle P_E=\frac{Kx^2}{2}

When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as

U=mgh

We have then,

U=P_E

\displaystyle mgh=\frac{Kx^2}{2}

Solving for h

\displaystyle h=\frac{Kx^2}{2mg}

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

\displaystyle h=\frac{(100)0.2^2}{2(2)(9.8)}

\displaystyle h=\frac{4}{39.2}=0,10\ m

The maximum height is 0.10 meters

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