Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr
1 answer:
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First,
where is density,
is mass, and
is volume. We can compute the volume of the roll:
When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have
where is the given area, so
If we're taking significant digits into account, the volume we found would have been , in turn making the thickness
.
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge at a point
distance away is given by
where,
= Coulomb's constant, having value
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.