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What else can I help you with?
8w-12 equals -4 what is w?
8w - 12 = -48w = 8w = 1
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.
Does a fluorescent light fitting with 4 x T8 2' 18W tubes and a 48W ballast use all 120W when running?
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.
How much electricity would a 48w appliance use in a day?
A 48 watt appliacne would use 48 x 24 or 1152 watt hours, or 1.152 kilowatt hours in one day.
What city is located 1s 48w?
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.
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
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.
