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ladessa [460]
4 years ago
12

A scientist in a lab would like to determine if an object conducts electricity. Which of the following experiments would test th

is? A. Place the object on a source of heat, and measure the time it take to heat up. B. Float the object in water. C. Include the object in a circuit, and check to see if the circuit still works. D. Weight the object before and after refrigeration. ​
Physics
2 answers:
lubasha [3.4K]4 years ago
7 0

Answer: C. Include the object in a circuit

Explanation:

mr Goodwill [35]4 years ago
6 0

A scientist in a lab would like to determine if an object conducts electricity. How could he test this ?

A. <u>Place the object on a source of heat</u>, and measure the time it take to heat up.  <em>No.  This would  tell him how well the object conducts heat, but not electricity. </em>

B. <u>Float the object in water.</u>  <em>No.  This would tell him the object's density, and also maybe how well it can absorb water, but nothing about conducting electricity. </em>

<em>C. Include the object in a circuit,</em> and check to see if the circuit still works.  <em>YES !</em><em>  If the circuit still works, then the object conducts electricity.  If it doesn't then it doesn't. </em>

D. <u>Weight the object</u> before and after refrigeration.  <em>No.  This doesn't tell him much of anything ... there's no reason why the object should weigh more or weigh less when it's warm or cold.​</em>

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calculate the energy spent on spraying a drop of mercury of 1 cm radius into 10^6 droplets all of same radius. surface tension o
WARRIOR [948]

Answer:

ANS : .Energy spent on spraying =4.3542*10^{-4}J

Explanation:

<em>Given:</em>

  • <em>Radius of mercury = 1cm initially ;</em>
  • <em>split into 10^{6} drops ;</em>

Thus, volume is conserved.

i.e ,

\frac{4}{3} \pi R_{o}^{3} = 10^{6}*\frac{4}{3} \pi R_{n}^{3}\\R_{n}=\frac{R_{o}}{10^{2}} = \frac{1cm}{100} = 0.01 cm

  • Energy of a droplet = TΔA

Where ,

  • <em>T is the surface tension </em>
  • <em>ΔA is the change in area</em>

Initial energy E_{i} = T*A_{i}\\= 0.0035 * 4 *\pi *0.01^{2}\\=4.398*10^{-6}J

Final energy E_{f}=10^{6}*T*A_{f}\\=10^{6}*0.0035*4*\pi *(0.0001)^2\\=4.39823*10^{-4}

∴  .Energy spent on spraying = =E_{f}-E{i}\\=(439.823-4.39823)*10^{-6}\\=4.3542*10^{-4}J

ANS : .Energy spent on spraying =4.3542*10^{-4}J

6 0
3 years ago
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