<span>work = 500*3 = 1500 J
1500J in 5 sec = 300 watts
Hope that helps:)</span>
Answer:
F = 3.01 N
Explanation:
Given that,
Mass of a block, m = 2.75 kg
Force applied to the block, F = 5.11 N
It is directed 53.8° above horizontal.
We need to find the total force acting on the block. The force acting on it is given by :

So, 3.01 N of force is acting on the block.
Answer:
ΔP = 986 Kpa
Explanation:
The solution is given in the pictures below
Answer:
L = 0.48 H
Explanation:
let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax be the maximum voltage and XL be the be the reactance of the inductor.
Vrms = Vmax/(√2)
= (3.00)/(√2)
= 2.121 V
then:
XL = Vrms/I
= (2.121)/(2.50×10^-3)
= 848.528 V/A
that is L = XL/(2×π×f)
= (848.528)/(2×π×(280))
= 0.482 H
Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.
<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>