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2 years ago

Calculate the work done by a 47N force pushing a pencil against a force of 23 N

1 answer:

8 0

I believe the answer would be 24N
Explanation: Because the forces are going against each other, in order to calculate the work done we need to subtract them. 47-23=24

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A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block attached to it by 0.5 m. The block is released from

morpeh [17]

Answer:

0.7000cm

Explanation:

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6 0

3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

A rocket sled is tested at "5 g" (5 times the acceleration due to gravity). If the sled starts from rest at position do= 0.00, h

stealth61 [152]

D=at²
441m=(5*9.81m/s²)(t²)
t²=441/(5*9.81)
t≈√8.99
t≈3 sec

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3 years ago

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devlian [24]

Explanation:

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