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2 years ago

14

The graph below shows the percentage of unstable nuclei remaining of a radioactive isotope.

1 answer:

mafiozo [28]2 years ago

8 0

i have no clue, sorry. jus tryna get points

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Which of the following statements describes nucleons?

Yuliya22 [10]

The correct answer is D.

A nucleon<span> is one of either of the two types of subatomic particles (neutrons and protons) which are located in the nucleus of atoms.
</span>

The total number of nucleon in the nucleus of an atom gives you an idea about the mass of that atom. In fact, one may refer mass number as nucleon number.

Simply put, nucleons are the particles that make nucleus of an atom and are held up together inside the nucleus due to nuclear force.

7 0

3 years ago

Read 2 more answers

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

An electric heating element is connected to a 110v circuit and a current of 3.2 a is flowing through the element. How much energ

Kisachek [45]

2.580 Wh.... .........

7 0

3 years ago

Read 2 more answers

The root-mean-square speed (thermal speed) of the molecules of a gas is 200m/s at 23.0°C. At 227°C the root-mean-square speed (t

777dan777 [17]

Answer:

330 m/s approx

Explanation:

The RMS speed of a gas is proportional to square root of its absolute temperature is

V ( RMS ) ∝ √T

$\frac{V_1}{V_2} =\sqrt{\frac{T_1}{T_2} }$

Here V₁ = 200 , T₁ = 23 +273 = 300K , T₂ = 227 +273 = 500 K

Putting the values

200 / V₂ = $\sqrt{\frac{300}{500} }$

V₂ = 330 m/s approx

8 0

3 years ago