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sergejj [24]
3 years ago
6

Liam did an investigation to see how water temperature affects the amount of salt that will dissolve in the water. He filled 4 b

eakers with exactly 100 milliliters of water each. He then heated each beaker to a different temperature and tested salt solubility at each different temperature. What was Liam's independent variable?
amount of water in each beaker

the number of beakers

the amount of salt that dissolved in each beaker

the temperature of the water in each beaker
Physics
2 answers:
8_murik_8 [283]3 years ago
6 0

Answer:

The correct answer is the temperature of the water in each beaker.

Explanation:

The independent variable is understood as one that can vary from one case to another, so that the researcher can obtain different results from the dependent variable being studied.

In this case, the dissolution of salt in water is the study variable. The temperature varies from one beaker to another, which is why it is our independent variable.

The amount of water in each beaker is the same, the beaker number is fixed, and the amount of salt added to each beaker is the same, this is to see how much salt dissolves at different temperatures.

Have a nice day!

docker41 [41]3 years ago
3 0

Answer:

the temperature of the water in each beaker ( last choice)

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Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

5 0
2 years ago
A plane has a takeoff speed of 88.3 m/s and can accelerate at a rate of 3m/s to reach that speed. How long does the plane take t
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v = \frac{d}{t}

and

a = \frac{v}{t}

We have acceleration and velocity so:

3 = \frac{v}{t}

88.3 = \frac{d}{t}

In the acceleration equation we can isolate for v and then plug it back into the other equation to solve...

So...

3t = v

88.3 = 3t

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E=Fe/q                                                      
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Fe=(5.8x10^5N/C)(1.5x10^-9C)
Fe=8.7x10^-4N

Fe=kq1q2/r²
8.7X10^-4N=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/r²
r²=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/(8.7x10^-4N)
√r²=√0.00002325
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nirvana33 [79]
My freind the correct anwer is C
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