Answer:
i think its D but i could be wrong im sorry if i am
Explanation:
edge 2020
Potential energy is highest at the top of the loop, and kinetic energy is highest at the bottom of the loop.
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
Answer:
<em>The first law states that</em> every planet describes an elliptical path about the sun as a single focus.
<em>The</em><em> </em><em>second</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The line joining the planet to the sun sweeps out equal areas in equal time intervals.
<em>The</em><em> </em><em>third</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The squares of the period of revolution is proportional to the cubes of the mean distance between the planet and the sun
