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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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Answer:

By Gaining Electrons

Explanation:

A nuetral atom is negative when it gains electrons, and it can be positive when it loses electrons.

4 0
2 years ago
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Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resi
Bond [772]

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

7 0
3 years ago
A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical
iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

T = 2 \pi \sqrt{\frac{m}{k} } \\

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} =  \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k =  17.71N/m

Hence the force constant of the spring is 17.71N/m

4 0
3 years ago
A scientific hypothesis must be related to nature and be..
PSYCHO15rus [73]

Answer:

A scientific hypothesis must be tetable so it can become a scientific theory.

Explanation: I think

8 0
3 years ago
Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the poten
alukav5142 [94]

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

4 0
2 years ago
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