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alina1380 [7]
3 years ago
13

How are chargeable cells different from ordinary dry cells​

Physics
1 answer:
topjm [15]3 years ago
6 0
Ordinary cells can convert chemical energy to electrical energy only, but rechargeable cells can also store electrical energy into chemical energy and vice versa. You will study more about it in your higher classes. secondary cells can be recharged and used again but dry cells cannot be recharged.
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Which shared property belongs in the region marked "B"?
NNADVOKAT [17]
The properti that you didnt share
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3 years ago
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
What measurement is the quantity of mass per unit volume
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That's the definition of the property called "density".
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3 years ago
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What is your acceleration while sitting in your chair. the latitude of corvallis is 44.4˚.?
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 <span>You can start with the equations you know 

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If you draw a circle and have the radius going out at a 44.4 degree angle above the center you can then find the r. 

r=Rcos(44.4) 
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Now you can plug this value into the acceleration equation from above... 

a= 1.8*10^8/7.47*10^9 
a= .0241 m/s^2 </span>
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3 years ago
PLS HELP DUE AT 10 PM
Rina8888 [55]
I believe it is C hope i helped!
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