Answer:
Where is the rest of the article to read form at?
Explanation:
Answer:
The sign of the charge is negative
The magnitude of the charge is 3.33 x 10⁻¹³ C
Explanation:
Given;
potential difference, V = -1.5 V
distance of the point charge, r = 2 mm = 2 x 10⁻³ m
The magnitude of the charge is calculated as follows;
![V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\where;\\\\k \ is \ coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2\\\\q = \frac{-1.5 \times 2\times 10^{-3}}{9\times 10^9 } \\\\q = -3.33 \times 10^{-13} \ C\\\\Magnitude \ of \ the\ charge, q = 3.33 \times 10^{-13} \ C](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bkq%7D%7Br%7D%20%5C%5C%5C%5Cq%20%3D%20%5Cfrac%7BVr%7D%7Bk%7D%20%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5Ck%20%5C%20is%20%5C%20coulomb%27s%20%5C%20constant%20%3D%209%5Ctimes%2010%5E9%20%5C%20Nm%5E2%2FC%5E2%5C%5C%5C%5Cq%20%3D%20%5Cfrac%7B-1.5%20%5Ctimes%202%5Ctimes%2010%5E%7B-3%7D%7D%7B9%5Ctimes%2010%5E9%20%7D%20%5C%5C%5C%5Cq%20%3D%20-3.33%20%5Ctimes%2010%5E%7B-13%7D%20%5C%20C%5C%5C%5C%5CMagnitude%20%5C%20of%20%5C%20the%5C%20%20charge%2C%20q%20%3D%203.33%20%5Ctimes%2010%5E%7B-13%7D%20%5C%20C)
The maximum diffraction order seen is 3.
<h3>What is the maximum diffraction order seen?</h3>
We know that the maximum angle of diffraction Q_m of the furthest bright fringe < Q = 90 degrees.
Here we need to compute the nth bright fringe for which is approximated to 90 degrees.
The angle of nth bright fringe is given by;
sin(Q_m) = n(λ)N
Approximating Q_m ≈ 90 degrees.
sin (90) = nλN
n = sin (90) / (λN)
n = 1 / ((580 x 10⁻⁶)500)
n = 3.5 orders
Since, we knew that Q_m < 90 degrees, we will choose n = 3 as the maximum number of orders.
Thus, the maximum diffraction order seen is 3.
Learn more about maximum diffraction here: brainly.com/question/14703089
#SPJ4
Answer: touch the pan to the burner
Explanation:
There are three modes of heat transfer:
conduction, convection and radiation.
For conduction, the heat transfers from a hot object to a cold object when the two are in contact.
For convection there is bulk motion of fluid occurs which transfers the heat.
For heat transfer by radiation, medium is not required.
Thus, to demonstrate conduction between pan and burner, the pan must touch the burner.