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2 years ago

Definition of phosphorescence

1 answer:

4 0

Light emitted by substance without combustion or perceptible heat emission of radiation in a similar manner to florescence but on a longer time scale so that emission continues after excitation ceases.

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A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

To Calculate Velocity and Average Velocity you can use the same formula as both.

Elanso [62]

Answer:

I believe it's True! Brainliest??

Explanation: Hope you have a great day :)

4 0

3 years ago

The orbit of the planets in our solar system is generally due to which characteristic of the Sun?

larisa [96]

The answer to your question is C. <span> the Sun's strong gravitational field . This is correct because i took the test :D</span>

6 0

3 years ago

Read 2 more answers

Which type of galaxy is a spherical cluster of stars with extraordinary brightness and little dust and gas?

Elis [28]

Answer:

Elliptical galaxies

Explanation:

Edwin Hubble classified galaxies into three categories

Elliptical

Spiral

Lenticular

The elliptical galaxies have an elipsoidal shape roughly. They have stars which are old and the primary light source of the galaxy. The formation of new stars is very limited. This increases the brightness of the galaxy. The mass of the stars are low. So, far the percentage elliptical galaxies is low compared to other galaxies.

3 0

3 years ago

Read 2 more answers

Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts

nignag [31]

<span>Depends on the precision you're working to. proton mass ~ 1.00728 amu neutron mass ~ 1.00866 amu electron mass ~ electron mass = 0.000549 amu Binding mass is: mass of constituents - mass of atom Eg for nitrogen: (7*1.00728)-(7*1.00866)-(7*0.000549) -14.003074 = 0.11235amu Binding energy is: E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So: 0.11235 * 931.5 = 104.6MeV Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV This is probably about right; it sounds like the right size! Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :) 1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules). It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.</span>

7 0

3 years ago