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3 years ago

The number of energy levels to which an electron can jump depends on the

1 answer:

8 0

C. The amount of energy it absorbs.

Electrons can only transition between energy levels when they absorb or release certain quanta of energy.

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The size of a balloon increases when the pressure inside it increases. The balloon gets bigger when it is left in the heat from

Masteriza [31]

Explanation:

This happens because the gas inside tend to expand because its temperature gets higher.

This is why the balloon that is put in a freezer for too long tend to gets smaller, because the gas temperature that is inside the balloon decreases.

(you can try it at home)

It is related to the temperature of the gas.

5 0

3 years ago

Help plzzz!!! I only have 10 minutes to turn in

Lostsunrise [7]

The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

Kinetic energy = (1/2) (mass) (speed²)

And there we have it

The car's mass is 3,600 kg.
Its speed is 'v' m/s .
(1/2) (mass) (v²) = 5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

v² = (5406/1800) (Newton-meter/kg)

v² = (5406/1800) (kg-m²/s² / kg)

v² = (5406/1800) (m²/s²)

= = = = =

v = √(5406/1800) m/s

4 0

3 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0

3 years ago

Read 2 more answers

What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = 7.6 μc, twice the charge o

Anna [14]

Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.

dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m

I answeared your question can you answear my question pleas

6 0

3 years ago

A circular loop of radius 0.0400 m is

Alenkinab [10]

Let's see

Find enclosed area of loop

$\\ \rm\dashrightarrow \pi r^2$