Answer: gravity, circuits
Explanation:
Answer:
E1 = 10.15 * 10^4 N/C
E2 = 0
E3 = 10.15 *10^4 N/C
Explanation:
Given data:
Two 13 cm-long thin glass rods ( L ) = 0.13 m
charge (Q) = +11nC
distance between thin glass rods = 4 cm .
<u>Calculate the electric field strengths </u>
electric charge due to a single glass rod in the question ( E ) = 
equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced
applying equation 1 to determine E1
E1 = 
( distance from 1 rod is 0.01 m and from the other rod is 0.03 )
= 
= 10.15 * 10^4 N/C
applying equation 1 to determine E2
E2 = 
therefore E2 = 0
E1 = E3
hence E3 = 10.15*10^4 N/C
Answer:
Options A, B, and C are all possible.
Explanation:
We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?
If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.
If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.
If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.
The resistance of the lamp is apparently 50V/2A = 25 ohms.
When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.
In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.
The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.
The power delivered by the 200V mains is (200V) x (2A) = 400 watts.
The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.
The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.
Together, they add up to the 400 watts delivered by the mains.
CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
