a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q = 
is located in the center of a spherical cavity of radius , 
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%2A%5Ctimes%2010%5E%7B-4%20%7D%5B8.57375%20-%202.81011%20%5D%5Ctimes%207.35%5Ctimes%2010%5E%7B-4%7D)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)
Electric field at 
m due to shell
E1 = 
Electric field at due to 'q' at center 
E2 =
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)
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Answer:
$ 1.1
Explanation:
From the question given above, the following data were obtained:
Cost per kWh = $ 0.1
Current (I) = 10 A
Voltage (V) = 220 V
Time (t) = 5 h
Cost of operation =?
Next, we shall determine the power the electric oven. This can be obtained as follow:
Current (I) = 10 A
Voltage = 220 V
Power (P) =?
P = IV
P = 10 × 220
P = 2200 W
Next, we shall convert 2200 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
2200 W = 2200 W × 1 KW / 1000 W
2200 W = 2.2 KW
Thus, 2200 W is equivalent to 2.2 KW.
Next, we shall determine the energy consumed by the electric oven. This can be obtained as follow:
Power (P) = 2.2 KW
Time (t) = 5 h
Energy (E) =?
E = Pt
E = 2.2 × 5
E = 11 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost $ 0.1
Therefore,
11 KWh will cost = 11 × 0.1
11 KWh will cost = $ 1.1
Therefore, the cost of operating the electric oven is $ 1.1
If it is not exposed to sunlight often... then it might not be able to produce sufficient amounts
