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3 years ago

10

Meghan explains to Zach the difference between commensalism, mutualism and parasitism. Which sentence(s) did she include in her

explanation? Select all that apply.
A.) A mutualistic relationship benefits only one of the species.
B.) In commensalism, one species is unaffected by the relationship.
C.) Parasitism involves a parasite and a host.
D.) Mutualism is a relationship in which one species preys upon another.
E.) Commensalism is a relationship between two species that compete with each other.

2 answers:

vlabodo [156]3 years ago

8 0

C is the correct answer I’m sure

Lady_Fox [76]3 years ago

3 0

I believe the answer is C. and B. i apologize if its incorrect.

Hope this helps have a great day

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Whitch group of elaments shares charicteristics whith both nonmetals and metloids

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3 years ago

A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.

kenny6666 [7]

Answer:

a) $|\Delta E|=4.58\: J$

b) $F=61.90\: N$

Explanation:

a)

We can use conservation of energy between these heights.

$\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})$

$\Delta E=0.608*9.81(0.6026-1.37)$

Therefore, the lost energy is:

$|\Delta E|=4.58\: J$

b)

The force acting along the distance create a work, these work is equal to the potential energy.

$W=\Delta E$

$F*d=mgh$

Let's solve it for F.

$F=\frac{mgh}{d}$

$F=\frac{0.608*9.81*1.37}{0.132}$

Therefore, the force is:

$F=61.90\: N$

I hope is helps you!

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3 years ago

When a rope is wrapped around a wheel (or pulley) and pulled, the lever arm is the ________ Blank 1 of the wheel and the angle b

m_a_m_a [10]

Answer:

the DISTANCE between the lever arm and the force is always 90º

Explanation:

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3 years ago

What types of crust are colliding between the North American Plate and the Pacific Plate?

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Answer: G00gle got you bro

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3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

2 years ago

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