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Arte-miy333 [17]

3 years ago

13

The atomic mass of an atom can be found by

1 answer:

Anton [14]3 years ago

5 0

A. Adding the number of protons and number of neutrons in the atom

Electrons are so incredibly small that the number of electrons makes almost no difference in the atomic mass. So b and c are not the answer.

D doesn't really make sense because like if you want to know the mass of a basket of apples and oranges, you'd add up the apples' and oranges' masses, not subtract.

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The relationship between frequency and period is...

Kazeer [188]

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Frequency, f, is how many cycles of an oscillation occur per second and is measured in cycles per second or hertz (Hz). The period of a wave, T, is the amount of time it takes a wave to vibrate one full cycle. These two terms are inversely proportional to each other: f = 1/T and T = 1/f.

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7 0

3 years ago

Wile E. Coyote stands on top of a 90-meter high cliff, looking down at the Roadrunner. If

Lina20 [59]

Answer:

33.516 kJ

Explanation:

Potential energy is given by:

PE = mgh

Where m is the mass, g is acceleration due to gravity, and h is the height. In this case:

PE = 38kg x 9.8m/s^2 x 90m = 33516 kg m^2/s^2 = 33516 J = 33.516 kJ

6 0

3 years ago

Which of these is the unit of measure for work?

Anastasy [175]

The SI unit of measure for work, as well as
for all other kinds of energy, is the "joule".

7 0

3 years ago

Read 2 more answers

What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

or

$\frac{Q_2}{3.78\times 10^3}=0.72$

or

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

or

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

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3 years ago

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3 years ago