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Lelu [443]
3 years ago
15

A researcher is analyzing data from a high-energy particle collision. the result of the analysis gives a value of 8.8 x 10-19 c

± 0.1 x 10-19 c for the charge of one of the emitted particles. should the researcher accept the value?
Physics
1 answer:
Harrizon [31]3 years ago
5 0
Elementary charge used to determine charges of other objects is equal to a charge of electron or proton. It's value is roughly 1.6* 10^{-19} C. All other charges are whole-number multipliers of this elementary charge, meaning that we multiply elementary charge by {...,-2,-1,0,1,2,...}.

To find out if the measured charge can be accepted we need to divide it with elementary charge to see if we get whole number as result.
There are three possible values of measured charge: 8.7* 10^{-19} C\\ 8.8* 10^{-19} C \\ 8.9* 10^{-19} C

N_{1} = \frac{8.7* 10^{-19} C }{1.6* 10^{-19} C } =5.44 \\ N_{2} = \frac{8.8* 10^{-19} C }{1.6* 10^{-19} C } =5.5 \\ N_{3} = \frac{8.9* 10^{-19} C }{1.6* 10^{-19} C } =5.56

As we can see none of the possible values of a measured charge is whole-number multiplier of elementary charge so the researcher should not accept the value.
This charge can be achieved by using quarks which have value of 1/3 of elementary charge but they do not remain stable for long enough.
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An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)
weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767

T₂ = 290.15 × 3.17767 = 922.00139

\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3}  = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

Therefore;

T_4 = \frac{1383.002}{2.702} =511.859 \ k

Q_1 = c_p(T_3-T_2)

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

c_v(T_4-T_1)  = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1  \right )r_{v}^{k-1}}

Where:

β = Cut off ratio

Plugging in the values, we get;

\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1  \right )18^{1.4-1}}= 0.5191

Therefore;

\eta = \frac{\sum Q}{Q_1}

\therefore 0.5191 = \frac{150}{Q_1}

Heat supplied = \frac{150}{0.5191}  = 288.978 \ hp

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

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Answer:

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Explanation:

From Newton's second Law of Motion

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