Pass through it easily is called a Metals such as iron, steel, copper, and aluminum are good conductors.
Answer:
The answer is C. Pollution
Answer:
You will need 450 cells (3 cm each) to meet the voltage/current requirement.
The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.
The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.
Explanation:
To meet the voltage:
120 [v] required voltage
0.8 [v] voltage of each cell
![\frac{120}{0.8} =150[v]\\](https://tex.z-dn.net/?f=%5Cfrac%7B120%7D%7B0.8%7D%20%3D150%5Bv%5D%5C%5C)
So we need 150 cells in series for the voltage.
To meet the current
1.0 [A] Required current
350[mA]=0.35[A] cell current
1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.
See the attached figure
<span>Answer:
1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use.
Given: m= 1200kg v initial = 95km/hr v final = 0
2nd, focus on the units - in most cases units speak for the concept
the unit of the unknown is kcal, thus its the unit of energy or work
so, W = ?
3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present
since W = delta K.E =delta P.E
W= 0.5m( vf^2 - vi^2) ---> best formula
4th, Substitute the given to the formula
since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J
we express first 95 km/hr to m/s
95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec
W= 0.5(1200kg)[(0^2- (26.39m/sec)^2]
W=600 kg(0 - 696.43m^2/s^2)
W=600kg(-696.43m^2/s^2)
W=417859.3Nm or 417859.3 J
W = 417859.3 J x 1 cal /4.19 J
W = 99,727.7 cal or 99.728 kcal</span>
Answer:
The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th its weight on the earth.
