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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 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
dimaraw [331]

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

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Nasa’s goal is to help industry reduce emissions from aircraft by how much by 2050 compared to 2005?
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Answer:

50%

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when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of
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<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q​</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

or

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂                                 (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂                            

or, C₂=2X/m. T₂                                   (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂  /2X  /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

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