The magnetic fields of magnets are most powerful at the <u>poles of a magnet</u>. Without any difference in the magnitude, both the poles have equal amount of Magnetic field intensity at their poles.
If you see the field lines we can see the field lines start from the North Pole run parallel and end at the South Pole. The field lines are concentrated more at the poles hence the magnetic field intensity or the magnetic field itself is more at the poles than any other part.
Answer:
A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.
Explanation:
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer
given,
SAT is 500 with a standard deviation of 100.
a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.
sample mean = 
= 
= 5
95% confidence level is achieved within +/- 1.960 standard deviations.
1.960 standard deviations x 5 is equal to +/- 9.8
confidence interval = 511 - 9.8 --- 511 + 9.8
= 501.2-----520.8
The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.
When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.
Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.
So the light of "bulb a" will be dimmer.