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Nesterboy [21]
2 years ago
13

5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw

ay from the central maximum. At what angle does the 18th minimum occur? A) 26.0° B) 1.94° C) 5.05° D) 0.44°
Physics
1 answer:
abruzzese [7]2 years ago
3 0

For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle  is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle   is mathematically given as

\theta=n(\lambda/d)

Therefore

\theta 1/ \theta 2=n1(\lambda/d)/ n2(\lambda/d)

In conclusion

theta/7=16/5

theta=10*7/5

theta=25.3

Read more about Angle

brainly.com/question/14362353

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A long solenoid that has 1000 turns uniformly distributed over a length of 0.400 m produces a magnetic field of magnitude 1.00 ×
yKpoI14uk [10]

0.03185A current is required in the windings for that to occur.

  • Solenoids are essentially coils of cord. These generate a magnetic subject which strives a pressure over a steel element.A solenoid is a fundamental time period for a coil of cord that we use as an electromagnet.
  • We additionally consult with the tool which could convert electric power into mechanical power as a solenoid.
  • Actually it generates a magnetic subject for developing linear movement from the electrical cutting-edge. With using a magnetic subject.
  • Magnetic field at the centre of a solenoid of length 'L' having 'N' number of turns with a current 'I' is given by B=\frac{u_0NI}{L}   ...(1)

It is given that 1000 turns uniformly distributed over a length of 0.400 m produces a magnetic field of magnitude 1.00 ×10⁻⁴T .

Putting above values in equation (1) , we get

1.00\times10^{-4}=\frac{4\pi \times10^{-7}\times1000I}{0.400} \\\\I=\frac{1.00\times10^{-4} \times0.400}{4\pi \times1000} \\\\I=0.03185A

Learn about solenoid more here :

brainly.com/question/15218721

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7 0
2 years ago
Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget
Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

P_t=m_1v_1+m_2v_2+...

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

P_1=-2mv

The second cart of mass m goes to the right at a speed v

P_2=mv

The total momentum before the impact is

P_t=-2mv+mv=-mv

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass m_2 has a negative momentum and then the sum of both masses keeps being negative

3 0
3 years ago
Read 2 more answers
2nd grade work. Anyone?
Temka [501]

Answer:

A. shadow

Explanation:

5 0
3 years ago
Read 2 more answers
Hey! please help i’ll give brainliest!
AURORKA [14]
Answer is : 2nd option!
4 0
2 years ago
A spring with spring constant of 33 N/m is stretched 0.15 m from its equilibrium position. How much work must be done to stretch
Hoochie [10]

Work done is 0.442J

<u>Explanation:</u>

Given:

Spring constant, k = 33 N/m

Distance, x₁ = 0.15m

Additional distance, x₂ = 0.072 m

Total distance = 0.15 + 0.072 m

                        = 0.222 m

Work done, W = ?

We can calculate work done by the formula

W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2

On substituting the value we get:

W = \frac{1}{2}k [(x_2)^2  - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J

Therefore, work done is 0.442J

7 0
3 years ago
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