Answer:
The power dissipated is 45W.
Step-by-step explanation:
The power
varies jointly with resistance
, and the square of current
:
,
where
is the constant of proportionality.
Now we are told that when
and
,
:
![15 = \alpha (1A)^2*15\Omega](https://tex.z-dn.net/?f=15%20%3D%20%5Calpha%20%281A%29%5E2%2A15%5COmega)
solving for
we get
,
which gives
![P = I^2R](https://tex.z-dn.net/?f=P%20%3D%20I%5E2R)
With the value of
in hand, we find the power dissipated when
and ![I = 3 A:](https://tex.z-dn.net/?f=I%20%3D%203%20A%3A)
![P = (3A)^2(5\Omega )](https://tex.z-dn.net/?f=P%20%3D%20%283A%29%5E2%285%5COmega%20%20%29)
![\boxed{P =45W}](https://tex.z-dn.net/?f=%5Cboxed%7BP%20%3D45W%7D)
Thus, the power dissipated is 45W.
Answer:
Add 9 + 3
Step-by-step explanation:
you should do inside the parenthesis then dividing or multiplying (whichever comes first go from left to right) then subtract and add (again whichever comes first from left to right follow the order.)
Answer:
105:175
Step-by-step explanation:
3+5 = 8
3/8 of 280 is 105.
5/8 of 280 is 175.
the ratio is 105:175
next we have to simplify it:
divide both sides by 5:
21:35 divide both sides by 7=
3:5
Answer:
x=12
Step-by-step explanation:
4x+6y=60 y=2
4x+6(2)=60
4x+12=60
4x=60-12
4x=48
x=![\frac{48}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B48%7D%7B4%7D)
x=12
Working attached below. Hope this helps! Any questions let me know :)