<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
Answer:
A) a = 2.31[m/s^2]; B) t = 14.4 [s]
Explanation:
We can solve this problem using the kinematic equations, but firts we must identify the data:
Vf= final velocity = take off velocity = 120[km/h]
Vi= initial velocity = 0, because the plane starts to move from the rest.
dx= distance to run = 240 [m]
![v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bi%7D%20%5E%7B2%7D%2B2%2Ag%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%3D120%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A%5Cfrac%7B1hr%7D%7B3600sg%7D%20%2A%20%5Cfrac%7B1000m%7D%7B1km%7D%20%3D33.33%5Bm%2Fs%5D%5C%5C%5C%5CReplacing%5C%5C33.33%5E%7B2%7D%3D0%2B2%2Aa%2A%28240%29%5C%5C%20a%3D%5Cfrac%7B11108.88%7D%7B2%2A240%7D%5C%5C%20%20a%3D2.31%5Bm%2Fs%5E2%5D%5C%5C)
To find the time we must use another kinematic equation.
![v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Creplacing%3A%5C%5C33.33%3D0%2B%282.31%2At%29%5C%5Ct%3D%5Cfrac%7B33.33%7D%7B2.31%7D%5C%5C%20t%3D14.4%5Bs%5D)
You can describe the motion of an object by its position, speed, direction, and acceleration
Answer:
t = 0.28 seconds
Explanation:
Given that,
Force acting on a firework, F = 700 N
The momentum of the firework, p =200 kg-m/s
We need to find the time before it explodes. Ket the time be t. We know that, the rate of change of momentum is equal to external frce. So,
So, the required time is equal to 0.28 seconds.
So first Identify all the given Varibales so u can choose which Eqauton to use
D=200m
T=4s
Vi=10m/s
Vf=?
You should this equation
D= 0.50(Vf+Vi)T
Plug in the values
200= 0.50 (Vf+10) 4
Divide the 4 out of the right side and if you do sumthing to one side you gotta do it to the other
200 divided by 4= 0.50(Vf+10)
50= 0.50(Vf+10)
Now expand the 0.50
So 50= 0.5Vf + 5 (because 0.5 times 10 is 5)
Now get rid of the 5
50-5= 0.5Vf
45 =0.5Vf now Divide the 0.5 out
45 divided by 0.5 = Vf
And 45/0.5 is 90
So 90=Vf
Therefore the final Velocity is 90m/s
