With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x
N and 66 degrees
ELECTRIC FORCE (F)
F = 
Where K = 9 x
N
/
The distance between
and
can be calculated by using Pythagoras theorem.
d = 
d = 46.7 cm = 0.467 m
For force
, substitute all the parameters into the formula above
= (9 x
x 3 x 1)/
= 2.7 x
/0.218
= 1.24 x
N
For force
, substitute all the parameters into the formula above
= (9 x
x 3 x 4)/
= 1.08 x
/0.1089
= 9.92 x
N
For force
, substitute all the parameters into the formula above
= (9 x
x 3 x 2)/
= 5.4 x
/0.1089
= 4.96 x
N
Summation of forces on Y component will be
=
-
Sin 45
= 9.92 x
- 1.24 x
Sin 45
= 9.04 x
N
Summation of forces on X component will be
=
-
Cos 45
= 4.96 x
- 1.24 x
Sin 45
= 4.08 x
N
Net Force = 
Net force = 
Net force = 9.9 x
N
The direction will be
Tan ∅ =
/
Tan ∅ = 9.04 x
/ 4.08 x 
Tan ∅ = 2.216
∅ =
(2.216)
∅ = 65.7 degrees
Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x
N and 66 degrees approximately.
Learn more about electric Force here: brainly.com/question/4053816
Answer:
B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.
Explanation:
B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.
<u>Answer
</u>
A. 1 and 2
<u>Explanation
</u>
At point 1 we have the highest potential energy and the kinetic energy is zero.
At 2 the potential energy is minimum and the kinetic energy is maximum.
The law of conservation of energy says that energy cannot be created nor destroyed. So, the change in P.E = Change in K.E.
P.E = height × gravity × mass. The height referred here is the perpendicular height. Gravity and mass are constant in this case.
From the diagram it can be seen clearly that the vertical height from 2 to 1 is much greater than from 4 to 3.
This shows that the change in P.E is greater between 1 and 2 and so is kinetic energy.
Inelastic collision happens when two objects joined and move together after the collision