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2 years ago

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How much in kg is 1 N

1 answer:

Nookie1986 [14]2 years ago

3 0

1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s2. 1 N (Earth) = 0.101971621297793 kg.

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You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 x

Blizzard [7]

Complete Question

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced $7.5 * 10^{-3}$ meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?

Answer:

The wavelength is $\lambda = 56250 nm$

Explanation:

From the question we are told that

The distance of slit separation is $d = 7.5 *10^{-3} \ m$

The distance of the screen is $D = 4 \ m$

The distance between the bright spot closest to the center of the interference is $k = 1.5 \ cm = 0.015 \ m$

Generally the width of the central maximum fringe produced is mathematically represented as

$y = 2 * k = \frac{ D * \lambda}{d}$

=> $2 * 0.015 = \frac{ \lambda * 4}{ 7.5 *10^{-3}}$

=> $\lambda = 56250 *10^{-9} \ m$

=> $\lambda = 56250 nm$

7 0

3 years ago

A cabbie is trying to stop when he notices a fare is whistling them over. The

liberstina [14]

K.E=18750J
Mass=m=2100kg
Velocity=v

$\boxed{\sf K.E=\dfrac{1}{2}mv^2}$

$\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2$

$\\ \sf\longmapsto 18750=1050v^2$

$\\ \sf\longmapsto v^2=\dfrac{18750}{1050}$

$\\ \sf\longmapsto v^2=17.85m^2$

$\\ \sf\longmapsto v=\sqrt{17.85}$

$\\ \sf\longmapsto v=4.1m/s$

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3 years ago

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Alenkasestr [34]

Answer:

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Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

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We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

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Rasek [7]

Answer:8 meters

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SpyIntel [72]

I think is ocean but I'm not sure

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