50 g of liquid X at 10 Celcius and 200 g of liquid Y
mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx) * (ty - t) / (t-tx)
cx/cy = 200/50*(40-15)/(15-10) = 20
cx/cy = 20
Well are having freckles recessive or dominant?
The equivalent capacitance between A and B points is 2.5F.
<h3>What is parallel plate capacitor?</h3>
The two parallel plates placed at a distance apart used to store charge when electric supply is on.
The capacitance of a capacitor is given by
C = ε₀ A/d
From the given circuit C1, C2 and C3, C4 are in parallel C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F
C1, C2 = 4 +4 =8F
C3, C4 = 2 +4 =6F
Now , all capacitors are in series.
Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2
Ceq = 2.5 F
Thus, the equivalent capacitance between A and B points is 2.5F.
Learn more about parallel plate capacitor.
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Answer:
The bulbs should be connected in parallel.
Explanation:
We want to find out a way to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other bulbs stays lit.
We must connect the two bulbs in parallel so that even when one bulb is burns out, it will have no effect on the other bulb and the 2nd bulb will keep on working. The current flowing in each bulb will depend upon the resistance of each bulb and the voltage will be same across each bulb.
On the other hand, if we use a series circuit then if one bulb burns out then the there is no flow of current in the circuit and therefore, the second bulb will not be operational.
The current flowing through each bulb is given by
I = V/R
The voltage across each bulb is given by
V = IReq
Where I is the current and Req is the equivalent resistance of the two bulbs connected in parallel and is given by
Req = (R₁*R₂)/(R₁+R₂)
The connection diagram is attached where two bulbs are connected in parallel and are power with a battery.
How many stairs?
You can use this to find the work
U
W=mgh
And the power by
P=W/T