Answer:
The answer would be 0.04ohms.
Explanation:
Hopefully this helps
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
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.
Answer:
The answer is a for Plato users.
Explanation:
Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.
