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Airida [17]
4 years ago
10

Blue light from a tunable laser is shined on a pair of closely space slits. If the laser light is changed to red light on the sa

me two slits, how will the spacing between the fringes on the distant diffraction screen change
Physics
1 answer:
Svetllana [295]4 years ago
5 0

Answer:

<em>The fringe spacing on a distant diffraction screen will increase.</em>

Explanation:

Blue light has a shorter wavelength than red light, so, changing from blue to red light is basically increasing the wavelength of the light involved in the experiment.

In the double slit experiment, the fringe spacing on a diffraction screen is calculated from the equation below

ω = zλ/d

where ω is the fringe spacing

z is the distance of the slit to the screen

λ is the wavelength of the light used

d is separation or distance between the slits

From the equation, one can see that if other parameters are held constant, <em>increasing the wavelength will lead to an increase in the spacing between the fringes, and hence, changing the light from blue to red light will increase the fringe spacing.</em>

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Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?
Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

x_{a}(t) = at+bt^2

x_{b}(t) = ct^2-dt^3

Constants are:

a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both x_a(t) and x_b(t) depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

at+bt^2 = ct^2-dt^3

2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0

t_1 = 4 - \sqrt{3} = 2.27 s, t_1 = 4 + \sqrt{3} = 5.73 s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0

t_1 = 1 s, t_2 = 4.33 s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67 s

3 0
4 years ago
Read 2 more answers
John and Daniel are playing tug-of-war together. John is exerting 10 N of force. Daniel is exerting 12 N of force. What is their
vovangra [49]

YEP ITS D 2 N in Daniel's direction

8 0
3 years ago
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The mass of a single gold atom is 3.27×10-22 grams. how many gold atoms would there be in 236 milligrams of gold?
blsea [12.9K]

The amount of gold atoms could be calculated by dividing the total weight of the gold with the mass of a single gold atom. Just convert the given weight to grams then divide it with 3.27x10^-22 grams. The answer would be 7.22x10^20. 

8 0
3 years ago
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A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?
pav-90 [236]
F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
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4 0
3 years ago
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PSYCHO15rus [73]

Answer:

The correct option is: Total energy

Explanation:

The Hamiltonian operator, in quantum mechanics, is an operator that is associated with the<u> total energy of the system.</u> It is equal to the sum of the total kinetic energy and the potential energy of all the particles of the system.

The Hamiltonian operator was named after the Irish mathematician, William Rowan Hamiltonis denoted and is denoted by H.

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3 years ago
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