Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
Answer:
"1.079 kg/m3
"
Explanation:
yeah that could be the answer
Answer:
5.69755 rad/s²
Explanation:
r = Radius = 1.3 m
v = Velocity of the hammer = 22 m/s
n = Number of revolutions = 4
Angular displacement
= Initial angular speed = 0
Final angular speed
Angular acceleration
Angular acceleration is given by 5.69755 rad/s²
Answer:
here's the pdf for it
IB QuestionbankExplanation:
Answer:
(a). The average current flows through the dryer is 12.5 A
(b). The resistance of the dryer is 9.6 Ω
Explanation:
Given that,
Voltage = 120 V
Power = 1500 W
(a). We need to calculate the average current flows through the dryer
Using formula of current
Where, P = power
V = voltage
Put the value into the formula
(b). We need to calculate the resistance of the dryer
Using Ohm's law
Where, I = current
Put the value into the formula
Hence, (a). The average current flows through the dryer is 12.5 A
(b). The resistance of the dryer is 9.6 Ω