its c so componets arr r\very fragle im aying this because it needs more then 20 letters
The total amount of energy in a closed system must stay constant.
Answer:
Hi where is question. Hope you understand me
Answer: time taken to charge to 95%
t = -5.80[ln(1-0.95)]
t = 17.38ms
Explanation:
For an RC Charging circuit
Where Vs
Vc = Vs (1 - e^(-t/RC))
Vc/Vs = 1 - e^(-t/RC)
-t/RC = ln(1 - Vc/Vs)
t = -RC[ln(1 - Vc/Vs)] and RC = k = -t/ln(1 - Vc/Vs)
Where ;
Vc = voltage across the capacitor
Vs = voltage supply
t = charging time = 2.5ms
k = RC = time constant.
Vc/Vs = 0.35
To calculate the time constant k;
k = -t/ln(1- Vc/Vs)
k = -2.5/ln(1-0.35)
k = 5.80ms
time taken to charge to 95%
t = -5.80[ln(1-0.95)]
t = 17.38ms
Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:

where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,

<u>h = 593.50 m</u>
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,

<u>h₁₁ = 103 m</u>
c)
Now, we use first equation of motion for complete motion:

where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,

<u>vf = 107.91 m/s</u>