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BlackZzzverrR [31]
3 years ago
8

Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperatu

re. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.
Physics
1 answer:
butalik [34]3 years ago
4 0

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

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A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at
GuDViN [60]

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

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Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

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3 years ago
bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc
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Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

m_A = mass of car A;

v_{A1} = the initial velocity of car A;

v_{A2} = the final velocity of car A;

and

m_B = mass of car B;

v_{B1} = the initial velocity of car B;

v_{B2} = the final velocity of car B.

Then, the law of conservation of momentum demands that

m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}

And the conservation of kinetic energy says that

\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2

These two equations are solved for final velocities  v_{A2} and v_{B2} to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

\boxed{v_{A2} = -1.05m/s}

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

\boxed{v_{B2} = 3.49m/s}

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

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Cierto volumen de gas se encuentra a 60°c de temperatura y 5atm de presión, es calentado hasta 140°c, estado en el cual ocupa un
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Answer:

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Explanation:

The system is about a ideal gas. Then you can use the equation for ideal gases for a volume V1, temperature T1 and pressure P1:

P_1V_1=nRT_1   (1)

And also for the situation in which the variables T, V and P has changed:

P_2V_2=nRT_2   (1)

R: constant of ideal gases = 0.082 L.atm/mol.K

For both cases (1) and (2) the number of moles are the same. Next, you solve for n in (1) and (2):

n=\frac{P_1V_1}{RT_1}\\\\n=\frac{P_2V_2}{RT_2}

Next, you equal these equations an solve for T2:

\frac{P_1V_1}{RT_1}=\frac{P_2V_2}{RT_2}\\\\V_1=\frac{P_2V_2T_1}{P_1T_2}

Finally you replace the values of P2, V2, T1 and T2:

V_1=\frac{(7atm)(680L)(140\°C)}{(60\°C)(5atm)}=2221.33\ L

Hence, the initial volume of the gas is 2221.33 L

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