Answer:
Option C
Crimp terminals
Explanation:
It's possible to crimp terminals using a multipurpose wiring tool. Since the tool selected for use during crimping also depends on the volume of work, the multipurpose wiring tool is recommended for use when the volume is small to medium. Basically, crimping tools are sized according to the wire gauge that they can fit. Since multipurpose has different sizes, that's why it's used for crimping tools.
Here I come and we wanna go home!!!
Answer: The volume is decreasing at a rate of 80 cm3/min
Explanation: Please see the attachments below
Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
Answer:
The x component of the electric field at y=2m is 
Explanation:
For a linear charge, using <u>Gauss Law</u>, we get that the <em>Electric field (radial) has the following form</em>

<em>where λ is the charge for longitud unit given in the problem, r is replaced by the y coordinate, and there are two known more data</em>. So

is the x component of the Electric field at y=2m on the y axis, which is what we wanted to know.