The drop in physical activity is partly due to inaction during leisure time and sedentary behaviour on the job and at home. Likewise, an increase in the use of "passive" modes of transportation also contributes to insufficient physical activity.
Answer:
- The distance between the charges is 5,335.026 m
Explanation:
To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:
where k is Coulomb's constant, 
and 
are the charges and d is the distance between the charges.
Working a little the equation, we can take:
And this equation will give us the distance between the charges. Taking the values of the problem
(the force has a minus sign, as its attractive)
And this is the distance between the charges.
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.
This problems a perfect application for this acceleration formula:
Distance = (1/2) (acceleration) (time)² .
During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²
Pick either half, and divide each side by 0.65 m/s²:
T² = (1600 m) / (0.65 m/s²)
T = square root of (1600 / 0.65) seconds
Time for the total trip between the stations is double that time.
T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)
My guess is; The warmer ocean adds water vapor to the air mass.
I'm really sorry if I'm wrong
