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Aleks04 [339]
3 years ago
15

Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the po

tential difference taken over the three resistors together
Physics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

\Delta V=\Delta V_1+\Delta V_2+\Delta V_3

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across R_1=\Delta V_1

Potential difference across R_2=\Delta V_2

Potential difference across R_3=\Delta V_3

We know that in series  combination

Potential difference ,V=V_1+V_2+V_3

Using the formula

\Delta V=\Delta V_1+\Delta V_2+\Delta V_3

Hence, this is required expression for potential difference.

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A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict
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Answer:

0.076 m/s

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Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
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Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

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