<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels.
Answer:
4.96 × 10⁵ Pa
Explanation:
F = mg
This force is evenly distributed on the three leg
radius, r = d/2
= 2.8 / 2
= 1.4 cm = 0.014 m
total cross sectional area of the three legs, A = 3*pi*r^2
Pressure due to weight,
P = Weight/A
P = 4.96 × 10⁵ Pa
Answer:
1 and 3
Explanation:
<u>1 and 3 </u>
Increasing coils increases strength
COOLING the wire would increase current flow and strength of magnet
Adding an iron core will definitely increase the strength of the electromagnet
We can use the kinematic equation
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance
we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]
