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tia_tia [17]
3 years ago
13

Describe the results of Ernest Rutherford's gold-foil experiment and explain how his results changed ideas about the distributio

n of positive charge within the atom.
Physics
1 answer:
DochEvi [55]3 years ago
4 0
In Rutherford's gold foil experiment, some of the positive particles would pass through the foil and some would bounce off. This led to a new theory that all of the positive subatomic particles were in the center of the atom instead of evenly spread throughout.
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The newest generation of smart phone uses three different batteries connected in a series simultaneously to extend battery life
azamat

Answer:

7.2 V

Explanation:

The three batteries are connected in series to the terminals of the phone: it means that they are connected along the same branch, so the current flowing through them is the same.

This also means that the potential difference across the phone will be equal to the sum of the voltages provided by each battery.

Here, the voltage provided by each battery is

V = 2.4 V

So, the overall voltage will be

V = 2.4 V + 2.4 V + 2.4 V = 7.2 V

8 0
3 years ago
If an atom is neutral, then the number of electrons is equal to the number of
Aloiza [94]

Answer:

Neutral atom means the number of protons is equal to the number of electrons.

Explanation:

4 0
3 years ago
A boy starts from point A and moves 5 units toward the east, then turns back and moves 3 units toward the west. What is the disp
Flura [38]
The displacement of the boy is 2 units
4 0
3 years ago
Why must objects be cooled before their mass is determined on a sensitive balance?
zubka84 [21]
Objects should be cooled before their mass is determined on a sensitive balance because it could damage the balance. Also, because it would give you wrong reading of the mass. Hot objects would warm the air around it. A warm air would expand and would produce convection as it rises causing to give the object a mass that is less than the actual. Another reason would be it would cause instability in the readings, the mass would fluctuate every now and then due to the convection currents around the object. It is always recommended to weigh the masses of objects that are in room temperature.
5 0
3 years ago
Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

From the steam table,

S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0
3 years ago
Read 2 more answers
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