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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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With good tires and breaks, a car traveling 50 mi/hr on dry pavement can travel 400 ft when the driver reacts to something he se

Bogdan [553]

Answer: $6.72 m/s^2$

Explanation:

Given

initial velocity $u=50 mi/hr\approx 73.33 ft/s$

Distance traveled before stopping $s=400 ft$

using equation of

$v^2-u^2=2as$

where v=final velocity

u=initial Velocity

a=acceleration

s=displacement

v=0 as car stops after travelling $400 ft$

$0-73.33^2=2\times a\times 400$

$a=-6.72 m/s^2$

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3 years ago

A calculator has a resistance of 22 Ω. What is the power rating for this calculator when connected to a 1.5 V battery?

GREYUIT [131]

Answer:

Explanation:

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3 years ago

The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of

Fed [463]

Answer:

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

$\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})$ (Eq. 1)

Where:

$\Delta l$ - Expansion of the rod portion, measured in meters.

$\alpha$ - Linear coefficient of expansion for titanium, measured in $\frac{1}{^{\circ}C}$ .

$l_{o}$ - Initial length of the rod portion, measured in meters.

$T_{o}$ - Initial temperature of the rod portion, measured in Celsius.

$T_{f}$ - Final temperature of the rod portion, measured in Celsius.

If we know that $\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $l_{o} = 10\,m$ , $T_{o} = 0\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , the expansion experimented by the rod portion is: