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3 years ago

Which of the following devices is based on the property of thermal expansion?

2 answers:

6 0

D. thermometer. The warmer it gets, the more energy the particles gain and the more they move and 'expand', rising up the thermometer.

borishaifa [10]3 years ago

5 0

The correct answer to the question is : D) Thermometre

EXPLANATION:

Thermometre is the device used for measuring the temperature of any substance.

Thermal expansion is the process in which the substance will expand when heat is given to that substance. Thermometre is based on the property of thermal expansion.

Generally, the substance that is used in thermometre is mercury. It expands at a fast rate when a little bit of heat is given to the thermometre.

When the thermometre is connected to any body, the heat will flow from body at high temperature to thermometre. The mercury level rises in the thermometre tube which gives the temperature of that substance under consideration.

Hence, it is the thermometre which is based on thermal expansion.

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If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is

expeople1 [14]

Answer:

We can use 2 g H = v2^2 - v1^2 or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

8 0

3 years ago

What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.

Stolb23 [73]

Answer:

$Momentum(p) = 20kgm/s$

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

$Momentum(p) = Mass(m) * Velocity(v)$

$p = mv$

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

$Momentum(p) = 10kg * 2m/s$

$Momentum(p) = 10 * 2 * kg * m/s$

$Momentum(p) = 20kgm/s$

Hence, the momentum of the man is $20kgm/s$

5 0

3 years ago

Salt is now added to the water in the bucket, increasing the density of the liquid. what happens to the tension in the string ?

AlexFokin [52]

The question involves a ping-pong ball that is held submerged in a bucket by a string attached to the bottom of the bucket.

The answer is the tension of the string will increase. This is because making the water salty increases its density, and consequently, increases its buoyancy. This is why sea water is more buoyant than fresh water. Therefore the ping pong is pushed more upwards by the water when salt is added than initially. This gives the string more tension.

5 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .