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3 years ago

I need help!!!!!!!

1 answer:

4 0

One simple use of the elements of the electromagnetic spectrum that we use during our everyday lives is our daily use of microwave radiation. microwave radiation is absorbed by water molecules which heats up and cooks the food whilst killing bacteria. Another would be ultraviolet radiation which we use daily in things such as light bulbs. The sun also uses this. Lastly, we use radio waves constantly. May it be tv programs, radio, or our cell phones.

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The time taken for an object to fall from a height of h m from the earth'fs surface is t s.

telo118 [61]

Answer:

B. Longer than t s,

Explanation:

Gravitational accln on earth is 9.8 m/s^2,

and one you provided as on moon is 1.6 m/s^2

that mean on moon gr. accl. is lesser!

now the time taken on earth will be lesser cuz from the same height if you drop the object from rest!

since accln on earth is higher,the object will attain higher velocity as compare to that of on moon!

✌️:)

6 0

3 years ago

An unknown amount of a radioactive isotope with a half-life of 2.0 h was observed for 6.0 h. if the amount of the isotope remain

Grace [21]

The original amount of the radioactive isotope will be 8 grams.

<h3 /><h3>What is the half-life of radioisotopes?</h3>

The amount of time required for half of a radioisotope's nuclide to decay, or change into a different species, is known as its half-life. The conversions release either beta or alpha particles, and the response can be monitored by counting the particles released.

Given that an unknown amount of a radioactive isotope with a half-life of 2.0 h was observed for 6.0 h. if the amount of the isotope remaining after 6.0 h was 24 g.

The original amount will be calculated as below:-

( 2 / 6 ) = ( Original amount / 24 )

Original amount = 4 x 2

Original amount = 8 grams

Therefore, the original amount of the radioactive isotope will be 8 grams.

To know more about the half-life of radioisotopes follow

brainly.com/question/1783783

#SPJ4

5 0

2 years ago

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

Xelga [282]

Answer:

$\lambda=1282\ nm$

Explanation:

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

For transitions:

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

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

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,

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

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

So,

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

Given, $n_i=5\ and\ n_f=3$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m$

$\lambda= 1.2828\times10^{-6}$

1 m = 10⁻⁹ nm

$\lambda=1282\ nm$

6 0

3 years ago

A force of 50n is applied to an object at an angle of 50 degrees measured relative to the horizontal. Determine the horizontal a

kobusy [5.1K]

Answer:

Fx = 32.14 [N]

Fy = 38.3 [N]

Explanation:

To solve this problem we must decompose the force vector, for this we will use the angle of 50 degrees measured from the horizontal component.

F = 50 [N]

Fx = 50*cos(50) = 32.14 [N]

Fy = 50*sin(50) = 38.3 [N]

We can verify this result using the Pythagorean theorem.

$F = \sqrt{(32.14)^{2}+ (38.3)^{2}} \\F = 50 [N]$

3 0

3 years ago

If a 0.5 kg pair of running shoes would weigh 11.55 Newton’s on Jupiter, what is the strength of gravity there?

OverLord2011 [107]

The weight of an object is given by
$F=mg$
where m is the mass of the object, while g is the strength of the gravity (which corresponds to the gravitational acceleration of the planet).

In our problem, the shoes have a mass of m=0.5 kg, and their weight is F=11.55 N. So, we can re-arrange the previous formula to find the value of g:
$g= \frac{F}{m}= \frac{11.55 N}{0.5 kg}=23.1 N kg^{-1}$
and this is the strength of the gravity on Jupiter's surface.

4 0

3 years ago