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Snezhnost [94]
3 years ago
13

Brainliest!!! A student wanted to answer the question, “How does the amount of glycerin affect the size of a bubble blown with s

oap mixture of water, dish soap and glycerin?”. She decided to test the amount of glycerin added to a soap mixture and used 1 drop, 2, drops, 3 drops, and 4 drops. She would add 10 mL of soap to 50 mL of water for each bubble solution. She measured the size (cm) of the bubble produced for each solution.
What is the independent variable?
Physics
2 answers:
trapecia [35]3 years ago
7 0
It’s Glycerin because an independent variable is defined as the variable that is changed or controlled in a scientific experiment. It represents the cause or reason for an outcome. A change in the independent variable directly causes a change in the dependent variable. The effect on the dependent variable is measured and recorded.
laiz [17]3 years ago
5 0

Answer: water is your independent variable.

Explanation: it’s the one variable that stays consistent in the experiment

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Question 20 Unsaved The graph below represents changes in water. What process is occurring during section "B" of the graph? Ques
FinnZ [79.3K]

melting ... happpens at constant temp

4 0
3 years ago
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Describe three physical changes that occur in nature?
Alex Ar [27]

Answer: Physical changes in nature could then be erosion in a mountain, the melting of snow, and a river freezing over from the cold. Since none of these changes affect the chemical composition of the mountain, the snow, or the river, they are physical changes.

Explanation:

8 0
3 years ago
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How much momentum does a 77 kg football player have if he is running with a speed of 4 m/s
worty [1.4K]

Answer:

308 N-s

Explanation:

Momentum is given by

P= mv

P= 77 x 4

P= 308

4 0
3 years ago
Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro
blagie [28]

Answer:

21.2\times 10^{-6} T

Explanation:

i  = magnitude of current in each wire = 2.0 A

a  = length of the side of the square = 4 cm = 0.04 m

r  = length of the diagonal of the square = \sqrt{2} a = \sqrt{2} (0.04) = 0.057 m

B = magnitude of magnetic field by wires at A and C

B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )

B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )

B = 10\times 10^{-6} T

B' = magnitude of magnetic field by wire at B

B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )

B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )

B' = 7.02\times 10^{-6} T

Net magnitude of the magnetic field at D is given as

B_{net} = \sqrt{B^{2}+B^{2}} + B'

B_{net} = \sqrt{2} B + B'

B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})

B_{net} = 21.2\times 10^{-6} T

8 0
3 years ago
A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4
n200080 [17]

Answer:

Explanation:

Let v is the launch speed of the plastic ball and the angle of projection is θ.

So, in horizontal direction

v Cosθ x t = 4.8 .... (1)

In th evertical direction

1.4 = v Sin θ x t - 0.5 gt²     .... (2)

As , v Sin θ x t = 3.8      .... (3) , put in equation (2)

1.4 = 3.8 - 4.9 t²

t = 0.7 s

Put in (1) and (3)

v Cosθ x 0.7  = 4.8

v Cosθ = 6.86

and v Sinθ x 0.7 = 3.8

v Sinθ = 5.43

Now

v^{2}\left ( Sin^{2}\theta +Cos^{2}\theta  \right )=5.43^{2}+6.86^{2}

v = 8.75 m/s

3 0
3 years ago
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