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sladkih [1.3K]
3 years ago
6

If, in general, r were calculated as r =v/i, which circuit arrangement in part a of the experiment would have the smallest error

?

Physics
1 answer:
ladessa [460]3 years ago
4 0

Concept: According to Ohm's Law, the flow of electric current through a conductor is directly proportional to the potential difference across it, provided physical conditions (like temperature, pressure, volume etc.) remains same.

v = ir

or, r = v / i

Here, current (i) is measured by Ammeter which should be connected in series of any electrical circuit.

voltage (v) is measured by Voltmeter which should be connected parallel to the external resistance (r).

In the given experiment, the first arrangement of the circuit will show the smallest error because the voltmeter is connected exactly parallel to the external resistance.

In the second arrangement, the voltmeter is connected across external resistance (r) and Ammeter (A) and in this case, the voltmeter will not measure the exact potential drop across the external resistance (r). So, there would be more error.

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A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g
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Answer:

"Apparent weight during the "plan's turn" is  519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is \mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}

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It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vectorW_{\mathrm{app}}=m(\text { vector } g \text { -vector a })

In magnitude,

| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}

Which is quite heavier than his/her true weigh of 519.4 N

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