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A 48 watt appliacne would use 48 x 24 or 1152 watt hours, or 1.152 kilowatt hours in one day.
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8w-12 equals -4 what is w?
8w - 12 = -48w = 8w = 1
Does a fluorescent light fitting with 4 x T8 2' 18W tubes and a 48W ballast use all 120W when running?
No, a fluorescent light fitting with a 4xT8 2' 18W tube and a 48W ballast does not use all the 120W when running.
If the voltage in a circuit is 24 v and the current 2 a what is the total power in the circuit?
P=VI, so P=24*2 = 48W.
What is cubic centimeters of pallet 48L X 48W X 54H?
48*48*54= 124416Over twelve thousand cubic centimeters (if the original pallet dimensions were in centimeters if no multiply by appropriate factor!)
The perimeter of a rectangular garden is 56 ft the length is 4 ft more than the width What are the dimensions of the garden?
2(L+W)=P2(W+4+W)=564W+8=564W=48W=12Length is 16
Approximate the length and width of a rectangle whose perimeter is 96 feet and whose area is 500 square feet accurate to three decimal places?
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
The length of a rectangle is 4 inches less than 2 times the width If the perimeter is 40 inches what are the dimensions?
width=8 length=122w-4=l2w+2l=402w+2(2w-4)=402w+4w-8=406w=48w=82(8)-4=l16-4=l12=l
If a rectangle is 3 more than 7 times the width and the perimeter is 54 what is the dimension of the rectangle?
(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
If a rectangle's length is 3 more than 7 times the width and the perimeter is 54 what re the dimensions of the rectangle?
(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
If I have 70 volts coming from a 45w solar panel how many amps or watts is it creating?
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.
Can a rectangle with a area of 42 have a perimeter of 48?
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).
What is the power rating of a soldering iron which has a current of 2A when connected to a 24V supply?
It depends. The 2A current, did you measure that while the soldering iron was on? Or is it rated at 2A current consumption on the device itself (on the powercable or the stem of the soldering iron). Generaly speaking you can calculted the real power consumption by using P=V*I (thus 2A*24V = 48W). But do remeber that this power consumption is in the steady state, that is, after its switch on and all transient effects have died down. To be safe allow for 3A-4A switching currents that occurs at power on.
