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4 years ago

6

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

central maximum. at what angles do the second bright spots occur?

1 answer:

o-na [289]4 years ago

3 0

The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.

(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m

Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.

(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ

θ=37.69 degrees

Hopes this helps!

P.S. I hope this is right. If not sorry in advance.

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3 years ago

A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to

shutvik [7]

k = 5.29

a = 0.78m/s²

KE = 0.0765J

<u>Explanation:</u>

Given-

Mass of air tracker, m = 1.15kg

Force, F = 0.9N

distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

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3 years ago

An object is moving to the left with a constant speed. What can be concluded about the forces acting upon the object?

MariettaO [177]

Answer: sum of forces is zero

Explanation: acording to the newton first law you can say that the sum of the forces acting over the object is zero no matter the direction of movement

good luck

mario

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3 years ago

if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?

LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

<u>The complete question is written below: </u>

<u></u>

<em>In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of $1.80 m/s^{2}$ , it would go from rest to its top speed in 85.6 s. </em>

<em>a) What was the speed of the vessel? </em>

<em> </em>

<em>b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute? </em>

<em></em>

<em>Calculate the answers in both meters per second and kilometers per hour</em>

<em></em>

a) The average acceleration $a_{av}$ is expressed as:

$a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t}$ (1)

Where:

$a_{av}=1.80 m/s^{2}$

$\Delta V$ is the variation of velocity in a given time $\Delta t$ , which is the difference between the final velocity $V$ and the initial velocity $V_{o}=0$ (because it starts from rest).

$\Delta t=85.6 s$

Isolating $V$ from (1):

$V=a_{av}\Delta t + V_{o}$ (2)

$V=(1.80 m/s^{2})(85.6 s) + 0 m/s$ (3)

$V=154.08 \frac{m}{s}$ (4)

If $1 km=1000m$ and $1 h=3600 s$ then:

$V=154.08 \frac{m}{s}=554.68 \frac{km}{h}$ (4)

b) Now we need to find the final velocity when $\Delta t=1 min=60 s$ :

<em></em>

$V=(1.80 m/s^{2})(60 s) + 0 m/s$ (5)

$V=108 \frac{m}{s}=388.8 \frac{km}{h}$ (6)

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3 years ago

An electron is accelerated from rest by a potential difference of 412 V. It then enters a uniform magnetic field of magnitude 18

Salsk061 [2.6K]

Explanation:

Given that,

Potential difference, V = 412 V

Magnitude of magnetic field, B = 188 mT

(a) The potential energy of electron is balanced by its kinetic energy as :

$eV=\dfrac{1}{2}mv^2$

v is speed of the electron

$v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s$

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

$r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m$

Hence, this is the required solution.

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3 years ago