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1 year ago

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .

Physics

1 answer:

fenix001 [56]1 year ago

5 0

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)

do =original position = 2 m

vo = original <u>VERTICAL</u> velocity = 0

a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t - 1/2 ( 9.81)t^2

to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

10 m/s * .639 s =<u> 6.39 m </u>

You might be interested in

A pendulum has a length of 2 m and a 30 kg mass hanging on the end. What is the period of the

anastassius [24]

Answer:

T = 2.83701481512 seconds

Explanation:

Hi!

The formula that you will want to use to solve this question is:

$T = 2\pi *\sqrt{\frac{L}{g} }$

T--> period

L --> length of the pendulum

g --> acceleration due to gravity (9.8m/s^2)

since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)

so now we can plug in our given info into the formula above and solve!

T = 2*pi * sqrt(2/9.8)

T = 2.83701481512 seconds

*Note*

- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different

I hope this helped!

7 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers

Which is the best definition of a parallel

Oksanka [162]

A parallel circuit exists when an electric charge flows in more than one path best describes it.

<h3>What is a Parallel circuit?</h3>

This type of circuit has branches in which the current divides and only part of it flows through any of the branch.

Parallel circuit having more than one branch therefore means that electric charge will flow in more than one path thereby making option A the most appropriate choice.

Read more about Parallel circuit here brainly.com/question/12069231

5 0

2 years ago

While hovering motionless 3.0 meters an asteroid, a small

AnnZ [28]

I have no idea what you are trying to ask sorry

4 0

3 years ago

What is the altitude of the Sun at noon on December 22, as seen from a place on the Tropic of Cancer?

scZoUnD [109]

Answer:

108.217 °

Explanation:

Day of year = 356 = d (Considering year of 365 days)

Latitude of Tropic of Cancer = 23.5 °N

Declination angle

δ = 23.45×sin[(360/365)(d+284)]

⇒δ = 23.45×sin[(360/365)(356+284)]

⇒δ = 5.2832 °

Altitude angle at solar noon

90+Latitude-Declination angle

= 90+23.5-5.2832

= 108.217 °

∴ Altitude angle of the Sun as seen from the tropic of cancer on December 22 is 108.217 °

4 0

4 years ago

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .