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a_sh-v [17]
3 years ago
11

Hypothetically, suppose our resistance in part I was 200 ohms. Quantitatively calculate the impact of a 1 Ohm ammeter resistance

and a 1 megaohm voltmeter resistance (rather than ideal circumstances). Based on the uncertainties in our experiment, are these systemic errors likely to be relevant?
Physics
1 answer:
Slav-nsk [51]3 years ago
3 0

Answer:

Yes, the errors are likely to be relevant

Explanation:

A systematic error occurs as a result of the instrument used in carrying out and experiment. These errors are a result of small fluctuations in the measurement properties of the instrument. This happens when the instrument departs from non-ideal situations, for example as a result of physical expansion or change in temperature. For instance, let the resistance be measured to be up to 10 Ω ± 1 Ω

The error of the resistance, ε = 0.01Ω

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Given the following specific heat capacities, which material was have the largest change in temperature if 10 grams of each subs
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Explanation:

Comment

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You should just develop a rule.  The rule will look like this

The greater the heat capacity the (higher or lower) the change in temperature.

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Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 tim
Oksi-84 [34.3K]

Answer:

1.117935:1

Explanation:

Since the wires are of the same material, they will have the same resistivity \rho.

The cross-sectional area of the of a wire is given by;

A=\pi\frac{d^2}{4}................(1)

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

\rho=\frac{RA}{l}..................(2)

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)

Given;

l_1=1.35l_2

\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)

We then use equation (1) to fine the ratio of the area A_1 to A_2

bearing in mind that d_1=0.91d_2

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

\frac{R_1}{R_2}= 1.35*0.8281=1.117935

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