Ask question

Ask question

All categories

4 years ago

10

Randy wants to know whether a soil's porosity affects how easily seedlings grow in it.

2 answers:

telo118 [61]4 years ago

4 0

Since the main objective of this experiment is to determine the effect of porosity on seedling growth that should be the only independent variable. In short, that is the only variable that should be different to ensure fair testing.

The answer should be B:

he plants seedlings in soils with different levels of porosity and equal levels of permeability.

Permeability is not what needs to be tested. If it changes, you may not be able to determine whether it was the porosity or permeability that cause changes.

Masteriza [31]4 years ago

3 0

Answer: B: different, equal

Explanation:

Since Randy wants to know how porosity affects seedling growth, he needs to design an experiment in which he uses soils of different porosity but keeps everything else the same. This means that his experiment should involve soils with equal levels of permeability, since this factor can also affect seedling growth. Randy is likely to find that most seedlings succeed in porous soils, because the open spaces between particles give them room to grow.

You might be interested in

An arrow is moving at 35 m/s and travels for 5 seconds. How far did the arrow travel?

noname [10]

Answer:

175m

Explanation:

So it’s travelling at 35metres/1second so if we want to find 5 seconds we can do this…

$\frac{35}{1} \frac{x}{5}.... so$ 35x5 ÷ 1 = 175m/5 seconds

6 0

3 years ago

Read 2 more answers

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

8 0

3 years ago

What is the origin of cold, wet air masses.

Ber [7]

Answer:

C. over land in polar regions.

7 0

3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

4 years ago

Please help very easy 5th grade work giving brainliest

zhenek [66]

Answer:

the answer is option B because opposit sides of the magnets attract each other

8 0

3 years ago