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telo118 [61]
3 years ago
11

“Plant growth is dependent on the amount of fertilizer applied.” That was the hypothesis Mel and Bill decided on for their scien

ce project. They planted identical seeds in the same potting mix, and they used identical pots for each plant. Each pot received the same amount of water, and all the pots were placed in the same location in the greenhouse. After three weeks, Mel and Bill were able to summarize their data. Based on the data collected, what would be the MOST appropriate research question for further study?
Physics
2 answers:
Katena32 [7]3 years ago
6 0
<span>The most appropriate research question for further study would be regarding how the environmental conditions affect the growth. If the plans have the same type and amount of fertilizer and placed in different environments, the results will either prove or disprove their hypothesis.</span>
Finger [1]3 years ago
5 0

D) What is the optimum amount of fertilizer for continued plant growth?

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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of
STatiana [176]

Answer:

2.27\cdot 10^{49}

Explanation:

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F_G=G\frac{m_p m_e}{r^2}

where

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Substituting numbers into the equation,

F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

The electrical force between the proton and the electron is given by

F_E=k\frac{q_p q_e}{r^2}

where

k is the Coulomb constant

q_p = q_e = q is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N

So, the ratio of the electrical force to the gravitational force is

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3 years ago
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Answer:

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Explanation: HOPE IT HELPED

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Zigmanuir [339]
Radial acceleration is given by

a_{rad}= \frac{v^2}{r}
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v=r \omega
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a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2

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70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

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