Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.
They attract and stick together
Answer:
b and d
a, c, e, and f
Explanation:
Ideal gas law:
PV = nRT
Solving for temperature:
T = PV / (nR)
Therefore, temperature is directly proportional to pressure and volume, and inversely proportional to the number of molecules.
T = k PV / N
Let's say that T₀ is the temperature when P = 100 kPa, V = 4 L, and N = 6×10²³.
a) T = k PV / N = T₀
b) T = k (2P) V / N = 2T₀
c) T = k (P/2) (2V) / N = T₀
d) T = k PV / (N/2) = 2T₀
e) T = k P (V/2) / (N/2) = T₀
f) T = k (P/2) V / (N/2) = T₀
b and d have the highest temperature,
a, c, e, and f have the lowest temperature.
I believe it's magnetic stripping. I hope that helps
