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3 years ago

what is haopening to the speed and acceleration as a car travels from rest to 72 km/hr in 4.0 seconds?

1 answer:

6 0

From the given information, we don't know. All we can tell is the net effect of all the changes during those 4 seconds ... the "averages".

The average change in speed is an increase of 72 km/hr (20 m/s) during that time.

The average acceleration is a constant 18 km/hr/sec (5 m/s^2) during the same time.

But both the speed and the acceleration may have gone up or down many times during the 4 seconds.

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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

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a_sh-v [17]

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Explanation:

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