A 48 watt appliacne would use 48 x 24 or 1152 watt hours, or 1.152 kilowatt hours in one day.
8w - 12 = -48w = 8w = 1
P=VI, so P=24*2 = 48W.
No, the total power consumption of the fluorescent light fitting will be 120W, including the 4 tubes and the ballast. The ballast itself consumes some power to regulate the current to the tubes, so not all 120W will be used solely by the tubes.
48*48*54= 124416Over twelve thousand cubic centimeters (if the original pallet dimensions were in centimeters if no multiply by appropriate factor!)
The city located at 1°S, 48°W is Belém, which is a city in northern Brazil. It is the capital of the state of Pará and is known for its rich culture, history, and proximity to the Amazon rainforest.
2(L+W)=P2(W+4+W)=564W+8=564W=48W=12Length is 16
Perimeter, P = 96 ft.Area, A = 500 ft^2.A = LW500= LWL = 500/WP = 2L + 2W96 = 2L + 2W substitute 500/W for L;96 = 2(500/W) + 2W divide by 2 to both sides;48 = 500/W + W multiply by W to both sides;48W = 500 + W^20 = 500 + W^2 - 48W or,W^2 - 48W + 500 = 0W = [48 +,- square root of [48^2 - (4)(1)(500)]/2W = [48 +,- square root of (2304 -2000)]/2W = (48 +,- square root of 304)/2W = (48 + 17.436)/2 or W = (48 - 17.436)/2W = 65.436/2 or W = 30.564/2W = 32.718 ft or W = 15. 282 ftL = 500/W = 500/32.718 or L = 500/15.282L = 15.282 ft or L = 32.718 ft
width=8 length=122w-4=l2w+2l=402w+2(2w-4)=402w+4w-8=406w=48w=82(8)-4=l16-4=l12=l
(1) L = 7W + 3(2) 2L + 2W = 54Substituting (1) in (2) ...2(7W+3) + 2W = 5414W + 6 + 2W = 5416W = 48W = 3L = 7x3 + 3 = 24So, length is 24 and width is 3
(1) L = 7W + 3(2) 2L + 2W = 54Substituting (1) in (2) ...2(7W+3) + 2W = 5414W + 6 + 2W = 5416W = 48W = 3L = 7x3 + 3 = 24So, length is 24 and width is 3
We can solve this by taking our basic equations for the perimeter and area of a rectangle: a = lw p = 2(l + w) And then plugging the given values into those: 42 = lw 48 = 2(l + w) Now we can solve one of them for either variable. We'll go with solving the first one for l: l = 42/w And then we can plug that into the other one: 48 = 2(42/w + w) And solve for w: 48 = 2(42/w + w) 48 = 84/w + 2w 48w = 84 + 2w2 2w2 - 48w + 84 = 0 w2 - 24w + 42 = 0 w2 - 24w + 144 = 102 (w - 12)2 = 102 w - 12 = ± √102 w = 12 ± √102 So yes, that is indeed possible, and it's length and width will be 12 - √102 and 12 + √102 (or approximately 1.9005 by 22.0995).
If there is full light then the output power will be 48W (doh!). However, this condition requires that the optimal volt/current is set at the load (or the input of a special power converter, to maintain the optimal condition). Since W=IxV, then the current at full power at 70V would be 48/70 = 680mA. This is at full power but your solar panel may have a different optimal power point (see above) so is likely to be less than this.