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3 years ago

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Suppose you are traveling down the highway at 100 km/hr, and want to pass a truck. You accelerate from 100 km/hr to 120 km/hr in

3 seconds in order to do so. What is your average acceleration, in meters per second squared (m/s 2 )

1 answer:

Olenka [21]3 years ago

8 0

Answer:

1.87 ms-2

Explanation:

We need to convert the velocity from km/hr to m/s as follows;

100 × 1000/3600 = 27.7 m/s

Also

120 × 1000/3600 = 33.3 m/s

From;

v= u + at

v= 33.3 m/s

u= 27.7 m/s

t= 3 secs

v= u + at

a = v- u/t

a = 33.3 - 27.7/3

a= 1.87 ms-2

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The storage coefficient of a confined aquifer is 6.8x10-4 determined by a pumping test. The thickness of the aquifer is 50 m and

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In a TV set, an electron beam moves with horizontal velocity of 4.8 x 10^7 m/s across the cathode ray tube and strikes the scree

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A student locates a double-slit assembly 1.40 m from a reflective screen. The slits are separated by 0.0572 mm. (a) Suppose the

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Complete Question

A student locates a double-slit assembly 1.40 m from a reflective screen. The slits are separated by 0.0572 mm.

(a) Suppose the student aims a beam of yellow light, with a wavelength of 589 nm, toward the slit assembly, and this makes an interference pattern on the screen. What distance (in cm) separates the zeroth-order and first-order bright fringes (a.k.a. maxima)?

(b)

Now suppose that blue light (with

λ = 415 nm)

is used instead. What distance (in cm) will now separate the second-order and fourth-order bright fringes?

Answer:

a

The distance of separation is $z_1 - z_o = 1.44cm$

b

The distance of separation is $z_4 - z_2 = 2.031cm$

Explanation:

From the question we are told that

The distance from the screen is $D = 1.40m$

The slit separation is $d = 0.0572 mm = 0.0572 *10^{-3} m$

The wavelength of the yellow light is $\lambda_y = 598nm$

The distance of a fringe from the central maxima is mathematically represented as

$z_n = n \frac{\lambda_y D}{d}$

Where n is the order of the fringe so the distance of separation between

The distance that separates first order from zeroth order bright fringe can be evaluated as

$z_1 - z_o = (1 - 0 ) \frac{\lambda_y D}{d}$

Substituting values

$z_1 - z_o = (1 - 0 ) \frac{590*10^{-9} 1.40}{0.0572 *10^{-3}}$

$z_1 - z_o = 0.0144m$

Converting to cm

$z_1 - z_o = 0.0144m = 0.0144*100 = 1.44cm$

b

The wavelength of blue light is $\lambda _b$

So the distance that separates second order from fourth order bright fringe can be evaluated as

$z_4 - z_2 = (4 - 2 ) \frac{\lambda_y D}{d}$

Substituting values

$z_4 - z_2 = (4 - 2 ) \frac{415*10^{-9} 1.40}{0.0572 *10^{-3}}$

$z_4 - z_2 = 0.02031 \ m$

Converting to cm

$z_4 - z_2 = 0.02031m = 0.02031*100 = 2.031cm$

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3 years ago