Heat transfer between two substances that are in contact is called

If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit

Oxana [17]

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%

4 0

3 years ago

: In heavy rushIn heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5

zhuklara [117]

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

8 0

3 years ago

If an astronaut had a mass of 30 kg on the moon what would his mass be on earth?

Novay_Z [31]

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

hope it help..... pls add me as brainlist.

Have a nice day

7 0

3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Dovator [93]

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>

3 0

3 years ago

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