Answer:
Explanation:
The most important thing to remember about parabolic motion in physics is that when an object reaches its max height, the velocity right there at the highest point is 0. Use this one-dimensional motion equation to solve this problem:
v = v₀ + at and filling in:
0 = v₀ + (-9.8)(4.0) **I put in 4.0 for time so we have more than just 1 sig fig here**
0 = v₀ - 39 and
-v₀ = -39 so
v₀ = 39 m/s
Answer:
A.
Explanation:
When the light travels through the lenses and disperses it can create other colors around objects that aren't there.
Answer:
correct answer is Fall slide, slump, creep, flow
Explanation:
solution
we know that Movement of particle under the influence of gravity
so rock and other material move down as gravity.
first rock particle fall down because falls occur very rapidly with high slope after that they slide on the slope and after sliding they slump and it occurs when the rupture surface is curved after slump process they creep.
after creeping, it can flow particle as it occurs slowly with the low slope with water.
so correct answer is Fall slide, slump, creep, flow
I'm assuming it was to keep the data consistent? The further you are from a heat source the less heat will get to you as the temperature tries to reach equilibrium and the waves start to spread out, so you should keep everything the same distance to get consistent results. I don't have any information so this is just my assumption
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁
where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
