Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q = 
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC
Answer:
d ≈ 7,6 g/cm³
Explanation:
d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³
V = l³ = (1.74cm)³ ≈ 5,27 cm³
Answer:0.502kg
Explanation:
F4om the relation
Power x time = mass x latent heat of vapourization
P.t=ML
1260 * 15 *60 = M * 22.6 * 10^5
M= 1134000/(22.6 *10^5)
M=0.502kg=502g
A).
It would decrease because the speed of sound and temperature are proportional.
Answer:
35mA
Explanation:
Hello!
To solve this problem we must use the following steps
1. Find the electrical resistance of the metal rod using the following equation

WHERE
α=
metal rod resistivity=2x10^-4 Ωm
l=leght=2m
A= Cross-sectional area

solving

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)
3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy
I1+I2=5
4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)
V1=V2
(0.14I1)=2000(i2)
solving for i1
I1=14285.7i2
5.Now we use the equation found in step 3
14285.7i2+i2=5
