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.
Efficiency = 45J/120J = 0.375
Or
0.375 * 100% = 37.5 %
Answer:
C) 16.3 ml
Explanation:
Density is equal to the ratio between the mass of an object and its volume:
where
m is the mass
V is the volume
In our problem, we know:
- density of aluminium: 
- mass of the aluminium foil: 
So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
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