True as the independent variable is over the dependent variable and controls it as the dependent relies on the independent.
Types of Gas? I'm not exactly sure what you're asking
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
If a negatively charged balloon is brought near one end of the rod but not in direct contact, then <u>the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon</u>, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.
The equation 
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:
Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:
When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:
Therefore, the equation represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!
