Each capacitor carry the same charge 'q'.
Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:
= V₁ +V₂ +V₃
This voltage may also be calculated using capacitance and charge;
V = Q/ C
= V₁ +V₂ +V₃
Provided that the total charge is 'q', hence the total voltage can be expressed as:
= (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)
Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.
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Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
Answer:
The frequency of wave is 160Hz.
Explanation:
Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.
So first thing, you have to make frequency the subject by dividing wavelength on both sides :



Next you have to substitute the value of v and f into the formula :
Let λ = 2.5m,
Let v = 400m/s,


Answer:
β = 114 db
Explanation:
The intensity of sound in decibles is
β = 10 log 
in most cases Io is the hearing threshold 1 10-12 W / cm²
let's calculate the intensity of each instrument
I / I₀ = 10 (β / 10)
I = I₀ 10 (β / 10)
trumpet
I1 = 1 10⁻¹² 10 (94/10)
I1 = 2.51 10⁻³ / cm²
Thrombus
I2 = 1 10⁻¹² 10 (107/10)
I2 = 5.01 10-2 W / cm²
low
I3 =1 1-12 (113/10) W/cm²
I3 = 1,995 10-1 W / cm²
when we place the three instruments together their sounds reinforce
I_total = I₁ + I₂ + I₃
I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1
I_total = 0.00251 + 0.0501 + 0.1995
I_total = 0.25211 W / cm²
let's bring this amount to the SI system
β = 10 log (0.25211 / 1 10⁻¹²)
β = 114 db
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred