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2 years ago

Please take 50 points please help me please

Physics

1 answer:

ch4aika [34]2 years ago

6 0

I am using a calculator for this question so hopefully this helps + I added working out just in case.

a) mean - 124 (add them all together then divide by 25)

b) standard deviation - 7.776888838 (7.76 to d.p)

c) the person might have eaten foods that increase or decrease the BP so every time they took their blood pressure, it might not have been done as the same routine as others (if this makes sense). Also, some blood pressures are lower than others like for example some might go down to 110 but others go all the way up to like 132, does why I said that they might have eaten food or done activites that increases the blood pressure.

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Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro

belka [17]

Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up

4 0

3 years ago

3. Read this sentence from the text: "Most of the solar radiation is absorbed by the atmosphere

Semmy [17]

I think the answer is A

6 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

Across:

vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

welcome................

6 0

2 years ago

Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (

Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud. The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

2. A galaxy collision results tostripping of gas.

3. The protogalactic cloud rotates in a very slow motion.

5. The density of protogalactic cloud is very high.

6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

1. The protogalactic cloud has high angular momentum.

4. Most of the protogalactic gases settles down into a disk.

6 0

3 years ago

Please take 50 points please help me please ​

Please take 50 points please help me please