Answer:
The sum of positive and negative charges in a unit of Al2O3 equals zero.
Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.
Al203 has 2 Al atoms and 3 O atoms.
Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)
= 2(3) + 3(-2)
= 6 - 6
= 0
Explanation:
Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.
Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.
Answer: 2200J
Explanation:
M = 44kg
V = 10m/s
K.E =?
K.E = 1/2MV2 = 1/2 x 44 x (10)^2
K.E = 22 x 100
K.E = 2200J
Answer:
(a) A = 1 mm
(b) 
(c) ![a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D606.4%20m%2Fs%5E%7B2%7D%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EDistance%20moved%20back%20and%20forth%20%3D%202%20mm%20%3C%2Fp%3E%3Cp%3EFrequency%2C%20f%20%3D%20124%20Hz%3C%2Fp%3E%3Cp%3ESo%2C%20amplitude%20is%20the%20half%20of%20the%20distance%20traveled%20back%20and%20forth.%20%3C%2Fp%3E%3Cp%3E%28a%29%20So%2C%3Cstrong%3E%20amplitude%2C%20A%20%3D%201%20mm%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%28b%29%20Angular%20frequency%2C%20%CF%89%20%3D%202%20%CF%80%20f%20%3D%202%20x%203.14%20x%20124%20%3D%20778.72%20rad%2Fs%20%3C%2Fp%3E%3Cp%3EThe%20formula%20for%20the%20maximum%20speed%20is%20given%20by%20%3C%2Fp%3E%3Cp%3E%5Btex%5DV_%7Bmax%7D%3D%5Comega%20%5Ctimes%20A)


(c) The formula for the maximum acceleration is given by


[tex]a_{max}=606.4 m/s^{2}/tex]
<span>Density is 3.4x10^18 kg/m^3
Dime weighs 1.5x10^12 pounds
The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so
4/3 pi 1.9x10^3
= 4/3 pi 6.859x10^3 m^3
= 2.873x10^10 m^3
Now divide the mass by the volume
9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3
Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3
Now to figure out how much the dime weighs, just multiply by the volume of the dime.
3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg
And to convert from kg to lbs, multiply by 2.20462, so
6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
A and b!
hope this helps!