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creativ13 [48]
4 years ago
11

Current flowing in a circuit depends on two variables. Identify these variables and their relationship to current.

Physics
2 answers:
Nuetrik [128]4 years ago
5 0
Wow ! This question reads like it might have come from one of
Faraday or Maxwell's original laboratory notebooks.

Choice-A is the correct one, when you consider what "conductance"
means. Conductance is just 1/resistance .

So when you see
"A) Current is proportionate to the conductance of the circuit and
precisely proportional to the voltage applied across the circuit."

what it's saying is
"Current is inversely proportional to the resistance of the circuit, and
directly proportional to the voltage applied across the circuit."

If you write the equation for all those words, it looks like

I = V / R

and that's correct.
Tom [10]4 years ago
4 0

Answer: C) Current is inversely proportional to the resistance of the circuit and directly proportional to the voltage applied across the circuit.

Explanation:

<u>Ohm's law</u> states that current is directly proportional to the potential difference across the circuit i.e. voltage.

I ∝ V

⇒ V = I R

where R is the resistance.

Current also depends on resistance offered by the circuit.

So, more the potential difference, more the current but more the resistance offered by the circuit, lesser current would flow.

Current is inversely proportional to the resistance and directly proportional to the voltage across the circuit. Voltage is provided by a cell or a battery. Resistance is offered by the wire and the components connected in the circuit. Sometimes the batteries also have internal resistance which adds up.

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An archer wishes to shoot an arrow at a target at eye level a distance of 50.0m away. If the initial speed imparted to the arrow
notka56 [123]
<span>Given:
Hmax (distance) = 50.0m
v</span>₀ = <span>70.0m/s

Required:
what angle should the arrow make with the horizontal as it is being shot

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Hmax = v</span>₀²sin²θ / 2g
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Irina-Kira [14]

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5 0
4 years ago
The lens of a telescope has a diameter of 25 cm. You are using it to look at two stars that are 2 × 1017 m away from you and 6 ×
vagabundo [1.1K]

Answer:

a)It is NOT possible using this telescope, to see the two stars as separate stars

b)d_{min} =28.466m

Explanation:

From the question we are told that:  

Diameter of lens,d = 25 cm \approx 0.25 m

Distance from both star D_f= 2*10^{17} m

Distance between both stars D_b= 6*10^9 m

Wavelength of light \lambda =700 nm \approx 700*10^-9 m

Generally the equation for angle subtended by the two stars at the lens  is mathematically given by

 \theta=\frac{D_f}{D_b}

 \theta=\frac{6*10^9}{2*10^{17}}

 \theta=3.0*10^{-8} rad

Generally the equation for minimum angular separation of two object is mathematically given by

\theta_{min} = 1.22*\lambda/d  

\theta_{min}= \frac{1.22*700*10^-9}{0.25}  

\theta_{min}= 3.416*10^-^6 rad

Therefore

 \theta < \theta_{min}

3.0*10^{-8} rad<  3.416*10^-^6 rad

It is NOT possible using this telescope, to see the two stars as separate stars

b)

Generally the equation for minimum diameter of the lens is mathematically given by

  d_{min} =\frac{ 1.22*\lambda}{\theta}

  d_{min} =\frac{ 1.22*700*10^{-9}}{3*10^{-8}}

  d_{min} =28.466m

3 0
3 years ago
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