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4 years ago

In a light interference pattern, alternate dark and bright fringes are seen. Why are the bright fringes formed?

Physics

2 answers:

murzikaleks [220]4 years ago

8 0

The correct choice is

d) constructive interference

bright fringes are formed as a result of constructive interference between the two waves. Constructive interference takes place between the two waves when crest of one wave falls on crest of another wave or trough of one wave falls on trough of another wave. the new wave thus produced has higher amplitude than any of the two waves. that is how we get bright fringes.

madreJ [45]4 years ago

7 0

The light interference pattern is a series of light and dark stripes.

The bright stripes are areas where there is CONstructive interference.

The dark stripes are areas where there is DEstructive interference.

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4 years ago

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The region outside the nucleus where an electron can most probably be found is the

solong [7]

Answer:

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One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing co

Kruka [31]

Answer:

$v_{1}=1.886 \frac{m}{s}$

β= 57.99 south of east

Explanation:

$v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=$ ?

Velocity in axis x the two balls come one from east and west

$m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\ v_{fx1}=1 \frac{m}{s} \\$

Velocity in axis y initial is zero so:

$v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}$

$v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}$

Angle is find using:

tan(β)= $\frac{v_{fy}}{v_{fx}}$

$\beta =tan^{-1}*\frac{1.6}{1}=57.99$

5 0

4 years ago

Someone please help!!

mart [117]

The answer is a.
The graph's line is higher if velocity is higher, and lower if velocity is lower. This is because the y axis is velocity. Thus, the fastest (highest velocity) that the car went is when the line was highest, and that is when x is 1.

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4 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

$d=v_{0}.t + \frac{1}{2} .a.t^{2}$ , where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + $\frac{1}{2}.10.t^{2}$

h = 5t²

$t_{T}$ = $\sqrt{\frac{h}{5} }$

On the other planet:

h = 0.t + $\frac{1}{2}.30.t^{2}$

h = 15.t²

$t_{P}$ = $\sqrt{\frac{h}{15} }$

Comparing the 2 planets:

$\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }$

$\frac{t_{T} }{t_{P} }$ = $\sqrt{3}$ or $t_{T} = \sqrt{3}.t_{P}$

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

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3 years ago