Explanation:
I assume the acceleration calculated in part (b) is the 3.33 m/s² from your other question.
Use Newton's second law to find the total force:
F = ma
F = (60,000 kg) (3.33 m/s²)
F = 200,000 N
Since there are 2 engines, the thrust from each is half of this:
F = 100,000 N
In reality, there are forces other than thrust. There are also drag forces (rolling friction and air resistance).
From Newton's second law, if we increase the mass and keep the force the same, the acceleration decreases. So it would take longer to reach the take-off speed.
The correct answer is
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C) 1200 g/m3. Let's see why. The relationship between liters and cube decimeters is
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Therefore,
However, we also know that
Therefore,
and
Therefore, the density of the problem
becomes
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.
Area because CM^2 is the unit used for Area
Answer:
1). Brake pedal or lever
2). A pushrod also called an actuating rod