The graph is shown below.
---|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--/--/--|--------------
-5 -4-11/3-3 -2 -1 0 1 2 8/3 3 4 5
8/3≈ 2,6
-11/3 ≈-3,6
Answer:
The electric field generated by a point charge is given by:
where
is the Coulomb's constant
Q is the charge
r is the distance from the charge
We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.
Let's calculate first the electric field generated by the positive charge at that point:
where the positive sign means its direction is away from the charge.
while the electric field generated by the negative charge is:
where the negative sign means its direction is toward the charge.
If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
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Step-by-step explanation:
Answer:
v=3
u=6
x=2√2
y=2√2
Step-by-step explanation:
First triangle :
tanΦ=v/3√3
tan30=v/3√3
(√3)/3=v/(3√3)
Cross multiply
(√3)/3 x 3√3=v
(√3 x 3√3)/3=v
(3√9)/3=v
(3x3)/3=v
9/3=v
3=v
v=3
Sin30=v/u
0.5=3/u
Cross multiply
0.5xu=3
0.5u=3
Divide both sides by 0.5
0.5u/0.5=3/0.5
u=6
Second triangle :
sin45=x/4
Cross multiply
x=4 x sin45
x=4 x (√2)/2
x=2√2
Cos45=y/4
Cross multiply
4 x Cos45=y
4 x (√2)/2=y
(4√2)/2=y
2√2=y
y=2√2
Answer:
12x-8
Step-by-step explanation:
Added both equations and used PEMDAS to combine all equations
