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MAVERICK [17]
3 years ago
14

3. A laser with a wavelength of 650 nm passes through a double slit. A pattern is observed on a wall that is 1.5 meters away fro

m the slits. The spacing between 10 bright fringes is measured to be 3 cm. Calculate the separation between the slits. Show your work.
Physics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

Slit separation, d=3.25\times 10^{-4}\ m

Explanation:

It is given that,

Wavelength of laser light, \lambda=650\ nm=650\times 10^{-9}\ m

Distance between pattern and the wall, L = 1.5 m

The spacing between 10 bright fringes is, y = 3 cm = 0.03 m

We need to find the separation between the slits. According to the equation of diffraction law :

d\ sin\theta=n\lambda

Where

d is the separation between the slits

For smaller angles, sin\theta=tan\theta=\dfrac{y}{L}

d(\dfrac{y}{L})=n\lambda

d=\dfrac{n\lambda L}{y}

d=\dfrac{10\times 650\times 10^{-9}\times 1.5}{0.03}

d = 0.000325 m

or

d=3.25\times 10^{-4}\ m

So, the separation between the slits is 3.25\times 10^{-4}\ m. Hence, this is the required solution.

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Answer: the rider’s pedal force must be greater than friction and the force of gravity

Explanation:

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2 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

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2 years ago
Describe the kinetic molecular theory
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8 0
3 years ago
the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove
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We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

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From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0
1 year ago
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