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Lostsunrise [7]

3 years ago

13

A ball is launched horizontally from the top of a cliff with an initial velocity of 20 m/s. If the ball strikes the ground after

five seconds, then (a) how high was the cliff and (b) how far from the base of the cliff will the ball land?

1 answer:

stich3 [128]3 years ago

6 0

Answer: you are the imposter you killed red then vented in admin

Explanation:

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Answer:32m/s

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The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

<u>Finally:</u>

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

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Physical science semester B exam 66 problems , please don’t delete I’m on my last leg I need this please god help me

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HI!! If you need answers to your exam, then go to https://quizlet.com/ then search for what you need!! I hope this helps I am not sure what your exam is for, and I don't have enough info to tell you all the answers, but hopefully, this will help you!!

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The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele

Novay_Z [31]

Answer:

Acceleration of the bullet will be 1778835.6 $m/sec^2$

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion $v^2=u^2+2as$

$553^2=0^2+2\times a\times 0.855$

$a=178835.6m/sec^2$

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A rock climber stands on top of a 50 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s a

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<h2><em>Answer: b) What was the initial speed of the second stone?</em></h2>

Explanation:

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