Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Answer:19.32 m/s
Explanation:
Given
initial speed of car(u)=4.92 m/s
acceleration(a)=![3.2 m/s^2](https://tex.z-dn.net/?f=3.2%20m%2Fs%5E2)
Speed of car after 4.5 s
using equation of motion
v=u+at
![v=4.92+3.2\times 4.5=4.92+14.4](https://tex.z-dn.net/?f=v%3D4.92%2B3.2%5Ctimes%204.5%3D4.92%2B14.4)
v=19.32 m/s
Displacement of the car after 4.5 s
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![19.32^2-4.92^2=2\times 3.2\times s](https://tex.z-dn.net/?f=19.32%5E2-4.92%5E2%3D2%5Ctimes%203.2%5Ctimes%20s)
![349.05=2\times 3.2\times s](https://tex.z-dn.net/?f=349.05%3D2%5Ctimes%203.2%5Ctimes%20s)
s=54.54 m
That certain change from that ecosystem will require 50 years or longer because big ecosystems need a long time to restablize the living ecosystem.
Answer:
P=740 KPa
Δ=7.4 mm
Explanation:
Given that
Diameter of plunger,d=30 mm
Diameter of sleeve ,D=32 mm
Length .L=50 mm
E= 5 MPa
n=0.45
As we know that
Lateral strain
![\varepsilon _t=\dfrac{D-d}{d}](https://tex.z-dn.net/?f=%5Cvarepsilon%20_t%3D%5Cdfrac%7BD-d%7D%7Bd%7D)
![\varepsilon _t=\dfrac{32-30}{30}](https://tex.z-dn.net/?f=%5Cvarepsilon%20_t%3D%5Cdfrac%7B32-30%7D%7B30%7D)
![\varepsilon _t=0.0667](https://tex.z-dn.net/?f=%5Cvarepsilon%20_t%3D0.0667)
We know that
![n=-\dfrac{\epsilon _t}{\varepsilon _{long}}](https://tex.z-dn.net/?f=n%3D-%5Cdfrac%7B%5Cepsilon%20_t%7D%7B%5Cvarepsilon%20_%7Blong%7D%7D)
![\varepsilon _{long}=-\dfrac{\epsilon _t}{n}](https://tex.z-dn.net/?f=%5Cvarepsilon%20_%7Blong%7D%3D-%5Cdfrac%7B%5Cepsilon%20_t%7D%7Bn%7D)
![\varepsilon _{long}=-\dfrac{0.0667}{0.45}](https://tex.z-dn.net/?f=%5Cvarepsilon%20_%7Blong%7D%3D-%5Cdfrac%7B0.0667%7D%7B0.45%7D)
![\varepsilon _{long}=-0.148](https://tex.z-dn.net/?f=%5Cvarepsilon%20_%7Blong%7D%3D-0.148)
So the axial pressure
![P=E\times \varepsilon _{long}](https://tex.z-dn.net/?f=P%3DE%5Ctimes%20%5Cvarepsilon%20_%7Blong%7D)
![P=5\times 0.148](https://tex.z-dn.net/?f=P%3D5%5Ctimes%200.148)
P=740 KPa
The movement in the sleeve
![\Delta =\varepsilon _{long}\times L](https://tex.z-dn.net/?f=%5CDelta%20%3D%5Cvarepsilon%20_%7Blong%7D%5Ctimes%20L)
![\Delta =0.148\times 50](https://tex.z-dn.net/?f=%5CDelta%20%3D0.148%5Ctimes%2050)
Δ=7.4 mm