Answer:
The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.
Explanation:
The application of coulonb's law is used to approach the question as shown in the attached file.
The mechanical advantage of an inclined plane is
(Length of the incline) / (its height)
= (10m) / (1m)
= 10 .
It's the same for any load, and doesn't depend on the mass that you're trying to move up or down the ramp.
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 ×
s
speed = 0.8 c = 0.8 × 3 ×
to find out
expected distance over
solution
we know c is speed of light in air is 3 ×
m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 ×
× 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
Answer:
ºC
Explanation:
First, let's write the energy balance over the duct:

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

So, let's isolate
:

The Cp of the air at 27ºC is 1007
(Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are
and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:

Surface area:

Then, the heat Q is:

Finally, find the exit temperature:

=27.0000077 ºC
The temperature change so little because:
- The mass flow is so big compared to the heat flux.
- The transfer area is so little, a bigger length would be required.