The same mass dose not change but weight does
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
![V_{1}=\dfrac{Q}{C_{1}}](https://tex.z-dn.net/?f=V_%7B1%7D%3D%5Cdfrac%7BQ%7D%7BC_%7B1%7D%7D)
Put the value into the formula
![V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}](https://tex.z-dn.net/?f=V_%7B1%7D%3D%5Cdfrac%7B10.1%20%5Ctimes10%5E%7B-6%7D%7D%7B1.10%5Ctimes10%5E%7B-6%7D%7D)
![V_{1}=9.18\ V](https://tex.z-dn.net/?f=V_%7B1%7D%3D9.18%5C%20V)
The potential on the second plate
![V_{2}=V-V_{1}](https://tex.z-dn.net/?f=V_%7B2%7D%3DV-V_%7B1%7D)
![V_{2}=51.5 -9.18](https://tex.z-dn.net/?f=V_%7B2%7D%3D51.5%20-9.18)
![V_{2}=42.32\ v](https://tex.z-dn.net/?f=V_%7B2%7D%3D42.32%5C%20v)
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
![C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}](https://tex.z-dn.net/?f=C%3D%5Cdfrac%7BC_%7B1%7D%5CtimesC_%7B2%7D%7D%7BC_%7B1%7D%2BC_%7B2%7D%7D)
Put the value into the formula
![C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}](https://tex.z-dn.net/?f=C%3D%5Cdfrac%7B1.10%5Ctimes10%5E%7B-6%7D%5Ctimes1.92%5Ctimes10%5E%7B-6%7D%7D%7B%281.10%2B1.92%29%5Ctimes10%5E%7B-6%7D%7D)
![C=6.99\times10^{-7}\ F](https://tex.z-dn.net/?f=C%3D6.99%5Ctimes10%5E%7B-7%7D%5C%20F)
![C=0.69\ \mu F](https://tex.z-dn.net/?f=C%3D0.69%5C%20%5Cmu%20F)
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Answer:
<h3>The answer is 256.52 mL</h3>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
![P_1V_1 = P_2V_2](https://tex.z-dn.net/?f=P_1V_1%20%3D%20P_2V_2)
Since we are finding the new volume
![V_2 = \frac{P_1V_1}{P_2} \\](https://tex.z-dn.net/?f=V_2%20%3D%20%20%5Cfrac%7BP_1V_1%7D%7BP_2%7D%20%20%5C%5C)
From the question we have
![V_2 = \frac{59000 \times 600}{138000} = \frac{35400000}{138000} \\ = 256.521739...](https://tex.z-dn.net/?f=V_2%20%3D%20%20%5Cfrac%7B59000%20%5Ctimes%20600%7D%7B138000%7D%20%20%3D%20%20%5Cfrac%7B35400000%7D%7B138000%7D%20%20%5C%5C%20%20%3D%20256.521739...)
We have the final answer as
<h3>256.52 mL</h3>
Hope this helps you
Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²