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3 years ago

Part A

Physics

1 answer:

Nutka1998 [239]3 years ago

4 0

Answer:

Which sentence best states the central idea of paragraphs 21-22 in "Energy Story"?

A. Materials that are insulators and conductors have a high

resistance to electricity.

Part B

Which sentence from "Energy Story" best supports the answer in Part A?

B. "Its resistance measures how well something conducts

electricity."

Explanation:

You might be interested in

Both atypical and traditional antipsychotics _____ levels of _____.

LiRa [457]

Answer:

Both atypical and traditional antipsychotics blocks levels of serotonin.

Explanation:

These drugs are capable of reducing or turning off positive psychotic symptoms such as hallucinations, delusions, and language and behavior disorder.

In addition to also acting as mood stabilizers, they have an impact on mania, depression, and produce low anxiety.

These anti-psychotics work on the 5-HT2A receptor, it belongs to the family of serotonin receptors and it is a G protein-coupled receptor. 5-HT is short for 5-hydroxy-tryptamine, the chemical name for serotonin.

This receptor became notorious for its importance as a target for drugs like LSD. Later it became important again due to the action of many antipsychotic drugs, especially atypical ones.

5 0

4 years ago

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think

dangina [55]

Answer:

4.44 rpm

Explanation:

$\omega$ = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = $\frac{3138000}{2}\ m$

R = Radius of arm = 6 m

The acceleration due to gravity is given by

$g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2$

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

$a_c=\omega^2R$

$a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s$

Converting to rpm

$1\ rad/s=\frac{60}{2\pi}\ rpm$

$0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm$

The angular speed of the arm is 4.44 rpm

8 0

3 years ago

A vertical spring has a mass hanging from it, which is displaced from the equilibrium position and begins to oscillate. At what

Orlov [11]

Answer:

the object has least potential energy at mean position of the SHM

Explanation:

If a block is connected with a spring and there is no resistive force on the system

In this case the total energy of the system is always conserved and it will change from one form to another form

So here we will say that

Kinetic energy + Potential energy = Total Mechanical energy

As we can say that total energy is conserved so here we have least potential energy when the system has maximum kinetic energy

So here we also know that at mean position of the SHM the system has maximum speed and hence maximum kinetic energy.

So the object has least potential energy at mean position of the SHM

5 0

3 years ago

The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c

Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0

3 years ago

Read 2 more answers

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:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

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

Read 2 more answers