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3 years ago

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A new school is being built in a field of wildflowers. A class wanted to develop a research project to predict the effects of th

e new school on wildflower species. When would be the best time for the class to conduct the study?

1 answer:

kotykmax [81]3 years ago

4 0

A new school is being built in a field of wildflowers. A class wanted to develop a research project to predict the effects of the new school on wildflower species. Given the background of the class, then probably, the best time for them to do the study is during right after the school is being built. Through that, they can get some data how the school affects the flowers. So, this helps the flowers in the future.

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A research submarine has a 20-cm-diameter window that is 9.0 cm thick. The manufacturer says the window can withstand forces up

Ratling [72]

Answer:

The maximum safe depth in salt water is 3758.2 m.

Explanation:

Given that,

Diameter = 20 cm

Radius = 10 cm

Thickness = 9.0 cm

Force $F = 1.2\times10^{6}\ N$

Inside pressure = 1.0 atm

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

Put the value into the formula

$A=\pi\times(10\times10^{-2})^2$

$A=0.0314\ m^2$

We need to calculate the pressure

Using formula of pressure

$P=\dfrac{F}{A}$

Put the value into the formula

$P=\dfrac{1.2\times10^{6}}{0.0314}$

$P=38216560.50\ Pa$

$P=3.8\times10^{7}\ Pa$

We need to calculate the maximum depth

Using equation of pressure

$P=P_{atm}+\rho gh$

$h=\dfrac{P-P_{atm}}{\rho g}$

Put the value into the formula

$h=\dfrac{3.8\times10^{7}-101325}{1029\times9.8}$

$h=3758.2\ m$

Hence, The maximum safe depth in salt water is 3758.2 m.

4 0

3 years ago

a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

$x_1 = 50 cm$ represents an actual distance of

$d_1 = 1.0\cdot 10^5 km$

The actual distance between Mars and the Sun is 228 million km, therefore

$d_2=228\cdot 10^6 km$

On the scale model, this would corresponds to a distance of $x_2$ .

Therefore, we can write the following proportion:

$\frac{x_1}{d_1}=\frac{x_2}{d_2}$

And solving for $x_2$ , we find:

$x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m$

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

4 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

5. How much time does it take for a bird flying at a speed of 45 kilometers per hour to travel a

Lyrx [107]

Answer:

$40h$

Explanation:

Use the velocity formula to solve

$v = \frac{d}{t}$

In this question, you are given velocity $v = 45km/h$ , and you are given a distance, $d = 1800km$ . Time in this question is what you'll need to find.

Start by rearranging the velocity formula, to isolate for t.

$v = \frac{d}{t}$

Start by multiplying both sides by t

$v(t) = \frac{d}{t}(t)\\\\vt = d$

Then divide both sides by v.

$vt\frac{1}{v} = d/v\\ \\t = \frac{d}{v}$

Now that you've isolated for time, sub in your values and calculate.

$t = \frac{d}{v} = \frac{1800km}{45km/h} = 40 h$

8 0

2 years ago

19 dm expressed in millimeters

alina1380 [7]

1900 millimeters thats wht i got

6 0

1 year ago