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Salsk061 [2.6K]
3 years ago
14

Which activity best demonstrates Ernest Rutherford's creativity?A. questioning the correctness of the plum pudding modelB. desig

ning an experiment to test the plum pudding modelC. making observations about the gold foil experimentD. summarizing the conclusions from the gold foil experiment as a theory
Physics
1 answer:
Rus_ich [418]3 years ago
5 0

Answer:

C.

Explanation:

Of the following activity best demonstrates the creativity of Rutherford: is making observations about the gold foil experiment.  

Rutherford's Gold Foil Test confirmed the existence of a thin, enormous atom core that would later be recognized as an atom's nucleus. To examine the effect of alpha particles on material, Ernest Rutherford, Hans Geiger and Ernest Marsden conducted their Gold Foil Experiment.

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Which of the follow represent a chemical changes
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Answer:

C) steel turning to rust in salt air

Explanation:

The missing options are:

A) ice melting to form liquid water

B) water boiling to form steam

C) steel turning to rust in salt air

D) sugar dissolving into hot coffee

In a chemical change the atoms of the reacting compounds are reordered forming new compounds. In a chemical change, new compounds appear, but in a physical change not.

Then, change of states like ice melting and water boiling are not chemical changes.

During steel rust, components of steel, like iron, are oxidized, that is, reacts with oxygen forming oxides.

The dissolution of sugar into hot coffee is a physical change in which sugar molecules get further apart in the coffee,  but they don't change.

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3 years ago
Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that
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<span>3933 watts At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3 The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams. Rounding to 4 significant figures gives me 4705 grams of water. The heat of vaporization for water is 2257 J/g. So the total energy applied is 2257 J/g * 4705 g = 10619185 J Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds. Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So 10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3) = 3933 watts</span>
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How to find the maximum amount of a substance that will dissolve using a graph?
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8 0
3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind.
RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

The time the bird takes;

\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

c)\

The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0
2 years ago
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