Answer:
Q = 12540 J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :
Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values,
So, 12540 J of energy is used to heat the water.
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
■■■■■■■■■■■■■■■■■■■■■■■■■■
The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
