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schepotkina [342]
3 years ago
12

Please helppp nowwwwww

Physics
1 answer:
kogti [31]3 years ago
5 0

Answer:

4.) the dependent variable is the french fry

5.) the independent variable is the jar; whether it has a lid on or not

6.) there are plenty of control variables in this example, the type of jar (size, material of the jar, etc.), the type of french fry, the room that the jars are in, the air pressure in the room, how long the experiment is, etc.

7.) My conclusion would be that the french fry that has the most exposure to the air.

The moisture in the french fry is leaving the jar because it is exposed to the open air. It becomes stale. But because the other french fry is in the closed jar, it is not exposed to the air or any new type of bacteria that could've entered the open jar. It'll stay fresher longer because it is not exposed the way the fry in the open jar is.

Explanation:

The dependent variable is the variable that is being tested. It is the one that changes due to whatever is being placed upon it.

  • i.e. the dependent variable in an experiment that tests the growth rate of lima bean plants in different colored lighting is the lima bean plant. <em>Depending</em> on the color of the lighting, the lima bean plant will grow at different rates.

The independent variable is the variable that is changed or controlled by an experimenter

  • i.e. the independent variable in an experiment that tests the growth rate of lima bean plants in different colored lighting is the different colored lighting. The color of the lighting helps prove the hypothesis that the growth rate of a lima bean plant is dependent on the lighting. In other words, the color of the light is not dependent on the growth of the plant.

A control variable is a variable that cannot be changed and is constant throughout the entire experiment

  • i.e. temperature, gravity, pollutants in the air, air pressure, humidity, etc.
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According to the Natural Resources Defense Council, what is the largest contributor to land pollution?
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Answer: Food

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How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half
beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

V=RI

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

I=\frac{V}{R}

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: V'=\frac{V}{2}

So, the new current will be

I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}

so, the current will drop to half of its original value.

4 0
3 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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