This is just testing your ability to recall that kinetic energy is given by:
<span>k.e. = ½mv² </span>
<span>where m is the mass and v is the velocity of the particle. </span>
<span>The frequency of the light is redundant information. </span>
<span>Here, you are given m = 9.1 * 10^-31 kg and v = 7.00 * 10^5 m/s. </span>
<span>Just plug in the values: </span>
<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
<span>k.e. = 2.23 * 10^-19 J
so it will be d:2.2*10^-19 J</span>
Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
A physical change in something doesn't change what the it is. For example, if you break glass, it will still be glass. In a chemical change where there is a chemical reaction, a new thing is formed and energy is either given off or absorbed. For example, when you burn a log. The carbon in the log is reacting to the oxygen to create ashe and smoke
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
Answer:
Yes it would be different on Earth and the moon
