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valentina_108 [34]
2 years ago
14

As in the video, we apply a charge + to the half-shell that carries the electroscope. This time, we also apply a charge – to the

other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?
A) It deflects more than it did at the end of the video.
B) It does not deflect at all.
C) It deflects the same amount as at end of the video.
D) It deflects less than it did at the end of the video.
Physics
1 answer:
horrorfan [7]2 years ago
8 0

Note: Although the video is not provided in this question, it is not needed to answer the question.

Answer:

B) It does not deflect at all

Explanation:

Since both half shells contain opposite charges, the two shells become electrically neutral when they are brought together and the electroscope discharges. On separating the two half shells again, the needle does not deflect because the half shells have now lost their charges to become neutral.

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Which type of graph is best suited to show the breakdown of how this teacher weights each category ?
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Answer:

pie graph?

Explanation:

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A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca
UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial  energy to gravitational potencial energy, so:

E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J

then we have only gravitational potencial energy when the ball is at its maximun height.

E_g=m*g*h

because all the energy was transformed Eg=Ee

h=\frac{0.081J}{9.8m/s^2*m}

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m

6 0
3 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
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5 0
3 years ago
It is found that the most probable speed of molecules in a gas at equilibrium temperature
kaheart [24]

Answer:

\frac{T_2}{T_1} = 1

Explanation:

The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

v = \sqrt{\frac{3RT}{M} }

where,

v = root mean square velocity of molecules:

R = Universal Gas Constant

T = Equilibrium Temperature

M = Molecular Mass of the Gas

Therefore,

For T = T₁ :

v = \sqrt{\frac{3RT_1}{M} }

For T = T₂ :

v = \sqrt{\frac{3RT_2}{M} }

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

\sqrt{\frac{3RT_1}{M} }=\sqrt{\frac{3RT_2}{M} }\\\\\frac{T_2}{T_1} = 1

8 0
3 years ago
Question 17 A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off
ivanzaharov [21]

Complete Question:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

(Question attached)

Answer:

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C (rounded to 1 decimal place)

Explanation:

A calorimeter is used to measure the heat of chemical or physical reactions. The example given in the question is using the calorimeter to determine the specific heat capacity of iron.

When the system reaches equilibrium the iron and water will be the same temperature, T_{e}. The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.

Q=mc(T_{e}-T_{initial}) (Eq 1)

-Q_{iron}=Q_{water} (Eq 2)

Water:

m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C

Iron:

m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C

Substituting Eq 1 into Eq 2 and details extracted from the question:

-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})

-59.1*c_{iron}(27.6-85)=100.0*4.186(27.6-23)

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C

4 0
3 years ago
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