Answer:
the time taken t is 9.25 minutes
Explanation:
Given the data in the question;
The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V
now, every minute, the charge lost is 9.9 %
so we need to find the time for which the charge drops below 800 mV or 0.8 V
to get the time, we can use the formula for compound interest in basic mathematics;
A = P × ( (1 - r/100 )ⁿ
where A IS 0.8, P is 2.1, r is 9.9
so we substitute
0.8 = 2.1 × ( 1 - 0.099 )ⁿ
0.8/2.1 = 0.901ⁿ
0.901ⁿ = 0.381
n = 9.25 minutes
Therefore, the time taken t is 9.25 minutes
Force equals mass*distance
F = ma
Given m = 10 kg, F = 30 N
30 = 10a
30/10 = a
3 = a
The wagon's acceleration is 3 m/s^2
Answer:
A) the current in AC electricity varies in magnitude and direction.
C) the voltage in AC electricity varies in magnitude and direction.
Explanation:
In DC current and voltage the direction of current will not change with time and it always remains the same.
So here in DC voltage and DC current the magnitude may change with time but the direction will always remain same
While in AC voltage and AC current the direction of AC will change with time
periodically.
So here magnitude and direction both will change in AC current and AC voltage.
so the correct answer is
A) the current in AC electricity varies in magnitude and direction.
C) the voltage in AC electricity varies in magnitude and direction.
Answer:
Explanation:
given,
velocity of particle 1 = 0.741 c to left
velocity of second particle = 0.543 c to right
relative velocity between the particle = ?
for the relative velocity calculation we have formula
u_x = 0.543 c
v_x = - 0.741 c
Relative velocity of the particle is
MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s so you are left with Solution 4.78 10-19 J OR .478 aJ <span>Apex - 467 nm ^.^ hopefully thats the correct thing</span>
