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Five measurements of a 10.00cm rod yield: 10.4, 10.1, 10.2, 10.1, 10.4 cm. This demonstrates: @ Low accuracy, Low precision, @ High accuracy, Low precision, @ High precision, Low accuracy, @ Systematic error only

Physics

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Part 38· 3 steps

Five measurements of a 10.00 cm10.00 \, \text{cm} rod yield: 10.1,10.1,10.2,10.1,10.1 cm10.1, 10.1, 10.2, 10.1, 10.1 \, \text{cm}.

  1. Analyze the precision of the measurements.

    Precision refers to how close repeated measurements are to each other. The measurements (10.1,10.1,10.2,10.1,10.1 cm10.1, 10.1, 10.2, 10.1, 10.1 \, \text{cm}) are very close to each other, indicating high precision.

  2. Analyze the accuracy of the measurements.

    Accuracy refers to how close the measurements are to the true value (10.00 cm10.00 \, \text{cm}). The average of the measurements is approximately 10.12 cm10.12 \, \text{cm}. Since all measurements are consistently higher than the true value, they are not close to the true value, indicating low accuracy. This consistent deviation suggests a systematic error.

  3. Match with the given options.

    The measurements demonstrate high precision and low accuracy.

    The correct answer is c) High precision, Low accuracy\boxed{\text{c) High precision, Low accuracy}}.

Part 39· 4 steps

Three molecules have speeds in m/s\text{m/s} as 2.00,4.002.00, 4.00 and 5.00.

  1. Recall the formula for root mean square (RMS) speed.

    The RMS speed vrmsv_{rms} is given by:

    vrms=v12+v22+...+vn2nv_{rms} = \sqrt{\frac{v_1^2 + v_2^2 + ... + v_n^2}{n}}

    where viv_i are the individual speeds and nn is the number of molecules.

  2. Substitute the given speeds into the formula.

    vrms=(2.00)2+(4.00)2+(5.00)23v_{rms} = \sqrt{\frac{(2.00)^2 + (4.00)^2 + (5.00)^2}{3}} vrms=4.00+16.00+25.003v_{rms} = \sqrt{\frac{4.00 + 16.00 + 25.00}{3}} vrms=45.003v_{rms} = \sqrt{\frac{45.00}{3}} vrms=15.00v_{rms} = \sqrt{15.00}
  3. Calculate the RMS speed.

    vrms≈3.8729 m/sv_{rms} \approx 3.8729 \, \text{m/s}
  4. Compare with the options.

    The calculated value is approximately 3.87 m/s3.87 \, \text{m/s}.

    The correct answer is c) 3.87\boxed{\text{c) } 3.87}.

Part 40· 4 steps

A 1000 W1000 \, \text{W} heater is 75%75\% efficient. The heater is used to increase the temperature of some water from 10∘C10^\circ\text{C} to 85∘C85^\circ\text{C} in 7 hours7 \, \text{hours}. The specific heat capacity of water is 4200 J/kgK4200 \, \text{J/kgK}.

  1. Calculate the total energy supplied by the heater.

    Power P=1000 WP = 1000 \, \text{W}.
    Time t=7 hours=7×3600 s=25200 st = 7 \, \text{hours} = 7 \times 3600 \, \text{s} = 25200 \, \text{s}.
    Energy supplied Esupplied=P×t=1000 W×25200 s=25 200 000 JE_{supplied} = P \times t = 1000 \, \text{W} \times 25200 \, \text{s} = 25\,200\,000 \, \text{J}.

  2. Calculate the useful energy transferred to the water.

    Efficiency η=75%=0.75\eta = 75\% = 0.75.
    Useful energy Euseful=η×Esupplied=0.75×25 200 000 J=18 900 000 JE_{useful} = \eta \times E_{supplied} = 0.75 \times 25\,200\,000 \, \text{J} = 18\,900\,000 \, \text{J}.

  3. Calculate the change in temperature of the water.

    Initial temperature Ti=10∘CT_i = 10^\circ\text{C}.
    Final temperature Tf=85∘CT_f = 85^\circ\text{C}.
    Change in temperature ΔT=Tf−Ti=85∘C−10∘C=75∘C\Delta T = T_f - T_i = 85^\circ\text{C} - 10^\circ\text{C} = 75^\circ\text{C}. (Note: A change of 1∘C1^\circ\text{C} is equivalent to a change of 1 K1 \, \text{K}).

  4. Calculate the mass of water heated using the heat transfer formula.

    The heat absorbed by the water is Q=mcΔTQ = mc\Delta T. Here, Q=EusefulQ = E_{useful}.

    m=EusefulcΔTm = \frac{E_{useful}}{c\Delta T} m=18 900 000 J4200 J/kgK×75 Km = \frac{18\,900\,000 \, \text{J}}{4200 \, \text{J/kgK} \times 75 \, \text{K}} m=18 900 000315 000 kgm = \frac{18\,900\,000}{315\,000} \, \text{kg} m=60 kgm = 60 \, \text{kg}

    The correct answer is c) 60 kg\boxed{\text{c) } 60 \, \text{kg}}.

Part 41

What causes drift current in a p-n junction diode?

Drift current is the movement of charge carriers (electrons and holes) in a semiconductor due to the presence of an electric field. In a p-n junction, an electric field exists across the depletion region, which causes minority carriers to drift across the junction.

The correct answer is a) Electric field\boxed{\text{a) Electric field}}.

Part 42

Which of the following is operated in forward bias?

  • LED (Light Emitting Diode): An LED is specifically designed to emit light when current flows through it, which occurs when it is operated in forward bias.
  • Solar cell: Solar cells typically operate in photovoltaic mode (zero bias, generating current from light) or can be reverse biased for certain applications.
  • Photodiode: Photodiodes are usually operated in reverse bias (photoconductive mode) to detect light, as this increases the depletion region width and improves response time and sensitivity.
  • Junction diode: This is a general term for a diode, which can be forward or reverse biased. However, among the specific devices listed, the LED's primary function relies on forward bias.

The correct answer is a) LED\boxed{\text{a) LED}}.

Part 43

The main composition of biogas is?

Biogas is a mixture of gases produced by the anaerobic digestion of organic matter. Its primary components are methane (CH4\text{CH}_4) and carbon dioxide (CO2\text{CO}_2). Methane typically constitutes 50−75%50-75\% of biogas and is the main combustible component, making it the most significant part of its composition for energy purposes.

The correct answer is b) Methane\boxed{\text{b) Methane}}.

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