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kenny6666 [7]
2 years ago
15

Define the following solar features in detail: Solar flare: Sunspot Prominence:

Chemistry
1 answer:
Over [174]2 years ago
3 0
Solar flare: a brief eruption of intense high-energy radiation from the sun's surface, associated with sunspots and causing electromagnetic disturbances on the earth, as with radio frequency communications and power line transmissions.

Sunspot Prominence: It is a large, bright, gaseous feature extending outward from the Sun's surface, often in a loop shape. It is similar to a Solar Flare

Hope this help 
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Avoiding an accident when driving can depend on reaction time. That time, measured in seconds from the moment the driver see dan
crimeas [40]

Answer:

The answer is below

Explanation:

A normal model is represented as (μ, σ). Therefore for (1.5, 0.18), the mean (μ) = 1.5 and the standard deviation (σ) = 0.18

The z score shows by how many standard deviations the raw score is above or below the mean. It is given as:

z=\frac{x-\mu}{\sigma}

a) For x < 1.35 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.35-1.5}{0.18}=-0.83

From the normal distribution table, the percent of drivers have a reaction time less than 1.35 seconds = P(x < 1.35) = P(z < -0.83) = 0.2033 = 20.33%

b) For x > 1.9 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.9-1.5}{0.18}=2.22

From the normal distribution table, the percent of drivers have a reaction time greater than 1.9 seconds = P(x > 1.9) = P(z > 2.22) = 1 - P(z<2.22) = 1 - 0.9868 = 0.0132 = 1.32%

c) For x = 1.45

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.45-1.5}{0.18}=-0.28

For x = 1.75

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.75-1.5}{0.18}=1.39

From the normal distribution table, P(1.45 < x < 1.75) = P(-0.28 < z < 1.39) = P(z < 1.39) - P(z< - 0.28) = 0.9177 - 0.3897 = 0.528 = 52.8%

d) A percentage of 10% corresponds to a z score of -1.28

z=\frac{x-\mu}{\sigma}\\\\-1.28=\frac{x-1.5}{0.18}\\\\x-1.5=-0.2034\\\\x=1.27

e) P(z < z1) - P(z< -z1) = 60%

P(z < z1) - P(z< -z1) = 0.6

P(z < -z1) = 1 - P(z < z1)

P(z<z1) - (1 - P(z < z1)) = 0.6

2P(z<z1) - 1= 0.6

2P(z<z1) = 1.6

P(z<z1) = 0.8

From the z table, z1 = 0.85

0.85=\frac{x-1.5}{0.18}and-0.85=\frac{x -1.5}{0.18}  \\\\x=1.65 \ and\ x=1.35

The reaction time between 1.35 and 1.65 seconds

8 0
3 years ago
Each square of the periodic table shows information about one element. The square usually shows the element's
omeli [17]
D, all of the above.
7 0
3 years ago
The addition of phosphate to a chemical compound is called A. glycolysis. B. oxidation. C. reduction. D. phosphorylation.
stich3 [128]

Answer:

Option (D)

Explanation:

Phosphorylation can be simply defined as the addition of a phosphate group to an organic and inorganic molecule. This process helps in regulating the processes that occur in the cells. It leads to the growth and development of cells and this process is efficiently carried out with the help of enzymes like kinase. It also plays an important role in transferring the signals within the cells, synthesis, and functioning of proteins within the cells, and storing as well as releasing of energy.

Thus, the correct answer is option (D).

5 0
3 years ago
What is the function of a hypothesis in the scientific inquiry process?
LuckyWell [14K]
It's a question that scientists can test.
3 0
3 years ago
For the following electrochemical reaction: Al3+(aq) + 3e -&gt; Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -&gt; 2F (aq) Calculate
makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=2.87-(-1.66)=4.53V

Hence, the standard electrode potential of the cell is 4.53 V.

6 0
3 years ago
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