Answer:
The answer is %pearlite = 0.06%
Explanation:
according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:
%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%
The percentage of cementite is equal to:
%cementite = (1.88-0)/(2-0) = 0.94%
Guessing you want the average speed. We can multiple each speed by the time we spent going that speed, and them all together and then divide by the total time we spent in traffic to get the average speed. We spent a total of 7.5 minutes in traffic, so average speed = (12*1.5+0*3.5+15*2.5)/7.5 = 7.4 m/s
Explanation:
A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon
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<h2>
Answer:442758.96N</h2>
Explanation:
This problem is solved using Bernoulli's equation.
Let
be the pressure at a point.
Let
be the density fluid at a point.
Let
be the velocity of fluid at a point.
Bernoulli's equation states that
for all points.
Lets apply the equation of a point just above the wing and to point just below the wing.
Let
be the pressure of a point just above the wing.
Let
be the pressure of a point just below the wing.
Since the aeroplane wing is flat,the heights of both the points are same.

So,
Force is given by the product of pressure difference and area.
Given that area is
.
So,lifting force is 
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>