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3 years ago

what did taking measurements reveal about the velocity of the fireball? Was the fireball’s velocity constant or changing?

Physics

1 answer:

Salsk061 [2.6K]3 years ago

7 0

Answer:

The fireball slows down when moving upward. It speeds up when moving downward. So, its velocity is always changing.

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three letters (JET) are placed in front of a plane mirror the image formed is in what arrangement???

mario62 [17]

Answer:

TEJ as this is a thing you wont get

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3 years ago

What type of electromagnetic wave does a light bulb and a radio antenna release?

andrew-mc [135]

I believe that a light bulb releases visible light and a radio antenna releases a radio waves

4 0

3 years ago

D.) How can you demonstrate the magnetic force? Write with an example

Lena [83]

Explanation:

d) Magnetic force is the power that pulls materials together (magnet e. g iron)

an example :how magnet can pick up a coin.

e) frictional force produces when two surfaces are in contact with each other.

effects of friction : I) it produces heat

II) it causes loss in power.

4 0

3 years ago

What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?

PolarNik [594]

Answer:

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$n_i=n\ and\ n_f=n-1$

Thus solving it, we get:

$\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m$

So,

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

Also, $\Delta E=h\times \nu$

So,

$h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}$

$\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

8 0

3 years ago

In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra

pogonyaev

Answer:

267.07 km

Explanation:

We have given the radius of the earth = 6378.1 km

In 2014 the difference between the magnetic north pole and geographical north pole is 2.40°

2.40° $=2.40\times \frac{\pi }{180}=0.041866 radian$

We know that linear distance is given by $S=R\Theta =6378.1\times 0.041866=267.07km$

So we have to travel 267.07 km in going from magnetic north pole to geographic north pole

3 0

3 years ago