Answer:
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electrons in a copper wire of ...
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Explanation:
Answer: 588 joules
Explanation:
Work is done when force is applied on an object over a distance ( whether vertical or horizontal). It is measured in joules.
Thus, Workdone = Force X distance
- Vertical distance to be moved by the brick = 12 metres
- Mass of box = 5kg
- Acceleration due to gravity when box was lifted represented by g is a constant with value of 9.8m/s^2
Now, recall that Force = Mass x acceleration due to gravity
i.e Force = 5kg x 9.8m/s^2
Force = 49 Newton
So, Workdone = Force X Distance
Workdone = 49 Newton X 12 metres
Workdone = 588 joules
Thus, 588 joules of work was done.
Answer:
Explanation:By using the sunglasses while looking at If your sunglasses are polarized, instead of only seeing the surface of a lake or river, you will suddenly be able to see through the glare and into the water below.
Another one is by comparing two sunglasses, hold your glasses and another one simultaneously and look through both pairs at the same time. Then, rotate one pair of sunglasses about 60 degrees. If both pairs of glasses are polarized, the overlapping area will darken as they filter out excess light. If your pair isn't polarized, however, you won't notice any difference.
Also it can be determined by simple computer test If you rotate your glasses sideways while looking at a computer monitor through polarized glasses, portions of your screen will become blank or go dark. The same also applicable to LCD display screens such as the ones on a gas pump.
Answer:
A) The work done by the engine is: 6.8MJ/L
B) The fuel efficiency is
Explanation:
A)
We know that the gasoline releases about 3.4*10^7 J of energy for each liter, and about 80% of that energy is lost as heat; it means that the other 20% of the energy released is taken for the engine to do work. In that sense, the work done by the engine is 20% of the 3.4*10^7 J that the gasoline releases for 1 liter, so:
This last can be seen as a conversion factor, where we multiply the energy released by the gasoline by the factor (20 J taken for do work for each 100 J released).
B) We know that the car requires 5.9*10^5 J of work <u>for each km traveled</u>. That is the energy that the car requires, but it is not the energy that you have to give to the car; take in mind that the energy that you put in the car in gasoline liters will be not taken all, but just 20%. Also we know that the work done by the engine for 1 liter of gasoline is 6.8MJ, and that is just the work taken for do work (the useful energy), so we can connect both data:
The first fraction, is the ratio or the proportion of (1 km requieres 5.9*10^5 J); and we multiply by the second fraction , which is the ratio: 6.8*10^6 J of work done for each liter of gasoline.
Answer:
Option C is the correct answer.
Explanation:
We have ideal gas equation, PV = nRT
Volume of tank, V = 1 m³
Mass of water = 10 kg
Mass of 1 mole of water = 18 g = 0.018 kg
Number of moles,
Temperature, T = 160 °C = 160 + 273 = 433 K
Ideal gas constant, R = 8.314 JK⁻¹mol⁻¹
Substituting
P x 1 = 555 x 8.314 x 433
P = 2000000 Pa = 2000 kPa
The pressure in the tank is 2000 kPa
Option C is the correct answer.