(5w-6)2
25w2-100 = (5w+10)(5w-10) when factored
25w - 16w3 can be factored into w(25 - 16w2), and {25 - 16w2} is in the form {a2 - b2= (a+b)(a-b)}, so {a=5, b=4w}, and the final factorization is:w(5 + 4w)(5 - 4w)
49w2 - 25w = 0 factor out a w w(49w - 25) = 0 set the expression in the parenthesis to that 0 as w already = 0 49w - 25 = 0 49w = 25 w = 25/49 ============so w = 0 -------- w = 25/49 --------------
The 50W lamp bulb has higher resistance. The resistance of a bulb is inversely proportional to the power it consumes, so a 50W bulb would have half the resistance of a 25W bulb.
Can a triode tube PET 25W be damaged due to electric fluctuation
To calculate the battery backup time, you need to divide the battery capacity (7.2Ah) by the load power (25W). In this case, the calculation would be 7.2Ah / 25W = 0.288 hours or approximately 17 minutes. This means that the battery can power a 25W load for approximately 17 minutes.
the answer is 2-5w
When the electric bulb is operated at 110V instead of its rated 220V, the power consumed will be reduced to 25W. This is because power is proportional to the square of the voltage, so halving the voltage will quarter the power consumed (P = V^2/R, where R is constant for the bulb).
No!
A power output of 25W for one second is 25 joules. It is also 0.03 horsepower (electric motor scale)
If you put the 25w behind the 100w then it might not work at the full 25w and vice versa