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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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4 years ago
Suppose that a car traveling to the west (the - x direction) begins to slow down as it approaches a traffic light. Which stateme
Amanda [17]

Answer:

(A) it's acceleration is negative but but it's velocity is positive

Explanation:

In the question it is given that begins to slow down so its speed is decreasing it is does not means that its speed is negative

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So it is not negative at all

Now the acceleration is the rate of change of velocity

a=\frac{dv}{dt}

a=\frac{v_2-v_1}{dt}=\frac{20-30}{5}=-2 m/sec^2

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3 years ago
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8 0
3 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
3 years ago
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3 years ago
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