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elixir [45]
2 years ago
6

Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in series with an ideal battery supplying a voltage of ∆ = 9 Volts. Sketch t

his circuit diagram. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?
Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in parallel with an ideal battery supplying a voltage of ∆ =
9 Volts. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?

Physics
1 answer:
BlackZzzverrR [31]2 years ago
4 0

Answer:

Explanation:

Check attachment for solution

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A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
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Answer:

7 deg

Explanation:

m = mass of the rod = 1.4 kg

W = weight of the rod = mg = (1.4) (9.8) = 13.72 N

k_{L} = spring constant for left spring = 59 Nm^{-1}

k_{R} = spring constant for right spring = 33 Nm^{-1}

x_{L} = stretch in the left spring

x_{R} = stretch in the right spring

L = length of the rod = 0.75 m

\theta = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m

Using equilibrium of force in vertical direction for right spring

k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m

Angle made with the horizontal is given as

\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg

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The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a
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Answer:

147.45 Hz

\Delta f=f_{Lr}-f_{Se}

Explanation:

v = Speed of sound in water = 1482 m/s

v_w = Speed of whale = 4.95 m/s

The difference in frequency is given by

\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}

Frequency of the wave in stationary condition

f_{Lr}=f\dfrac{v+v_w}{v-v_w}

Ship's frequency which is reflected back

f_{Se}=f\dfrac{v}{v-v_w}

f_{Se}=22\ kHz

\mathbf{\Delta f=f_{Lr}-f_{Se}}

\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22\times \dfrac{1482+4.95}{1482-4.95}-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz

The difference in wavelength is 147.45 Hz

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