Answer:
Explained
Explanation:
public int dimension(int [][]a2d,int nElements)
{
int count = 0;
for(int i = 0;i < a2d.length ; i++)
{
count = count + a2d[i].length;
}
return count;
}
Answer:
B :)
Explanation:
:) JUST TRUST ME I GOT IT CORRECT
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.
Answer:
2,54 cm are equal to 1 inch
Explanation:
Doing the conversion:
![55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]](https://tex.z-dn.net/?f=55%5Bcm%5D%2A%5Cfrac%7B1%5Binch%5D%7D%7B2%2C54%5Bcm%5D%7D%20%3D21%2C65%5Binch%5D)
Answer:
B : It will avoid the distortion from the atmosphere and give scientists more accurate data
Explanation: