Well let's see:
A). No. A capacitor doesn't measure anything.
B). No. The power delivered to the circuit is determined by
the battery or power supply and all the things in
the circuit that dissipate energy. A capacitor doesn't
do any of these things.
C). No. If any current actually flows between its plates,
the capacitor is shot and can't do its job, and
must be replaced.
D). Yes. A capacitor stores charges on its plates, and
electrical energy in the field between its plates.
answer - d
set up equation
the velocity can be found through the change in kinetic energy
due to the conversation of energy, the change in kinetic energy is the same as the change in potential energy when the height is changed
ΔK = ![\frac{mv^{2}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7B2%7D)
ΔU = mgΔh
ΔK = ΔU
= mgΔh
= 2gΔh
v = ![\sqrt{2gh}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D)
values
g = 9.8![\frac{m}{s^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
Δh = 42 - 4.2 = 37.8 m
plug in values
v = ![\sqrt{2gh}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D)
v = ![\sqrt{2 * 9.8 * 37.8}](https://tex.z-dn.net/?f=%5Csqrt%7B2%20%2A%209.8%20%2A%2037.8%7D)
v = 27.2![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
answer choice D
Answer:
1.dr/dt=0.0096cm/s
2. dA/dt=2.19cm^2/s
Explanation:
A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rate of change of surface area at this moment?
for this question, we need to analyze the parameters we know
V=volume of the spherical balloon 1000π cm3
volume of the sphere=![\frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D)
1000π=4/3πr^3
dividing both sides by 4
250*3=r^3
r=9.08cm, the radius of the balloon
dv/dt=dv/dr*dr/dt...................................1
dv/dr ,means
V=![\frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D)
dv/dr=4*pi*r^2
dv/dt=10 cm3/s
from equ 1
10=4*pi*9.08^2*dr/dt
10=1036 dr/dt
dr/dt=10/1036
dr/dt=0.0096cm/s
2. to find the rate at which the area is changing we have,
dA/dt=dA/dr*dr/dt
area of a sphere is 4πr^2
differentiate a with respect to r, radius
dA/dr=8πr
dA/dt=8πr*0.0096
dA/dt=8*pi*9.08*0.0096
dA/dt=2.19cm^2/s
is the rate of change of the surface area
Answer:
The acceleration of the cart is 283.54g
Explanation:
It is given that,
Initial speed of the car, u = 35 km/h = 9.72 m/s
Finally, it stops, v = 0
Diameter of the dime, d = 1.7 cm
Let a is the acceleration of the car. Using third equation of motion to find it as :
![v^2-u^2=2ad](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ad)
![-u^2=2ad](https://tex.z-dn.net/?f=-u%5E2%3D2ad)
![a=\dfrac{-u^2}{2d}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B-u%5E2%7D%7B2d%7D)
![a=-\dfrac{(9.72)^2}{2\times 1.7\times 10^{-2}}](https://tex.z-dn.net/?f=a%3D-%5Cdfrac%7B%289.72%29%5E2%7D%7B2%5Ctimes%201.7%5Ctimes%2010%5E%7B-2%7D%7D)
![a=-2778.77\ m/s^2](https://tex.z-dn.net/?f=a%3D-2778.77%5C%20m%2Fs%5E2)
So, the acceleration of the car is
and it is decelerating.
Since, ![g=9.8\ m/s^2](https://tex.z-dn.net/?f=g%3D9.8%5C%20m%2Fs%5E2)
So, ![a=\dfrac{2778.77}{9.8}=283.54\ g](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B2778.77%7D%7B9.8%7D%3D283.54%5C%20g)
So, the acceleration of the cart is 283.54g. Hence, this is the required solution.