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MrMuchimi
3 years ago
13

Calculate the momentum of a 10 kg bowling ball rolling at 2m/s towards north.

Physics
1 answer:
densk [106]3 years ago
8 0

Answer:

momentum=mass x velocity= 10 x 2 = 20kgm/s

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Using an elastic cord, the astronaut in the black shirt pulls the astronaut with the red shirt back towards him before releasing
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3 years ago
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How do scientists use the Doppler effect to understand the universe?
professor190 [17]
There's a very subtle thing going on here, one that could blow your mind.

Wherever we look in the universe, no matter what direction we look,
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Now:  If ... say tomorrow ... a competent Physicist discovers another way
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theory along with it !


Now that our mind has been blown, come back down to Earth with me,
and I'll give you something else to think about:

It's true that when we look at distant galaxies, we do see their light
arriving in our telescopes with longer wavelengths than it should have.
And then we use the Doppler effect to calculate how fast that galaxy
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every day.                                   I mean every night.

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6 0
3 years ago
Read 2 more answers
A ball of radius 0.220 m rolls along a horizontal table top with a constant linear speed of 3.70 m/s.?
4vir4ik [10]
<span>16.82 x 0.04 = 0.67 rad
I hope I helped if you really need I can explain to you how I got that answer but Thats correct im sorry it took 2 days for me to find this answer but if you or anybody else still needs the answer for this question here it is :) have a fantastic day guys Spring Break is coming up soon :)</span>
6 0
3 years ago
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

5 0
3 years ago
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