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steposvetlana [31]
2 years ago
11

What is light intensity? ​

Physics
2 answers:
soldi70 [24.7K]2 years ago
7 0

Light intensity is the "amount of light given off a light source." Also, is the measure the wavelength of the amount of light given off by a light source.

Hope this helps.

Vika [28.1K]2 years ago
4 0

It's the brightness of something, and is measured as the rate at which light energy is delivered to a unit of surface, or energy per unit time per unit area.

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According to Newton's 3rd law of motion, if you hit a baseball with a bat it will _____.
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Make a claim for the behavior of charged objects with other charged objects. Make a claim for the behavior of charged objects wi
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A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ
ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

\Delta U = q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

For a proton,

q=+e=1.6\cdot 10^{-19}C

And since \Delta V=100 V

The energy gained by the proton is

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

For an alpha particle,

q=+2e=3.2\cdot 10^{-19}C

Therefore, the energy gained is

\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

q=+e=1.6\cdot 10^{-19}C

So the energy gained is the same as the proton:

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

6 0
3 years ago
How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),
fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0
3 years ago
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