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aalyn [17]
3 years ago
5

Which statement(s) correctly compare the masses of protons, neutrons, and electrons? Check all that apply

Physics
2 answers:
Elza [17]3 years ago
8 0

Answer: A & F

__Protons and neutrons have similar mass.

__Electrons are smaller than a proton or a neutron.

Explanation:

umka21 [38]3 years ago
3 0

__Protons and neutrons have similar mass.

__Electrons are smaller than a proton or a neutron.

Explanation:

The three particles involved in this problem are:

  • Proton: it is positively charge, it is found in the nucleus of the atom, and its mass is m_p = 1.67 \cdot 10^{-27} kg
  • Neutron: it has no electric charge, it is also found in the nucleus of the atom, and its mass is approximately equal to that of the proton (just slightly larger)
  • Electron: it has negative electric charge, it orbit around the nucleus of the atom, and its mass is much smaller than that of the proton: m_e = 9.11 \cdot 10^{-31} kg

We can now analyze each of the given statement:

__Protons and neutrons have similar mass.  --> TRUE

__Protons and electrons have similar mass.  --> FALSE, the electron is much lighter

__Neutrons and electrons have similar mass.  --> FALSE, the neutron is much heavier

__Protons are smaller than a neutron or an electron.  --> FALSE, protons are similar to the neutrons

__Neutrons are smaller than a proton or an electron.  --> FALSE, neutrons are similar to the protons

__Electrons are smaller than a proton or a neutron. --> TRUE

Learn more about atoms:

brainly.com/question/2757829

#LearnwithBrainly

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3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

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