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Studentka2010 [4]

3 years ago

13

A __________ score allows a researcher to compare scores across different scales.

2 answers:

cupoosta [38]3 years ago

6 0

The correct answer should be C. standard

When you have standardization developed, then it is possible to compare scores across different scales and make sure that everything works relatively fine.

german3 years ago

4 0

Answer:

C

Explanation:

took the test :)

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A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

4 years ago

The Earth orbits around the sun because the gravitational force that the sun

kotykmax [81]

<h3>Question -:</h3>

The Earth orbits around the sun because the gravitational force that the sun

exerts on the Earth:

O A. causes Earth's acceleration toward the sun.

O B. is very small because the sun is so far from the Earth.

O c. is smaller than the force the Earth exerts on the sun.

O D. pushes the Earth away from the sun.

<h3>Answer -:</h3>

O A. causes Earth's acceleration toward the sun.

<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>

5 0

3 years ago

Which is the correct definition for sea floor spreading

jok3333 [9.3K]

Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other.

6 0

3 years ago

Read 2 more answers

The energy of a photon was found to be

Vaselesa [24]

now you can yourself know to which part of electromagnetic spectrum the photon belongs....

not fitting sharply in green

4 0

3 years ago

Read 2 more answers

A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate

kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid $R=1.06 m$

Emf induced $E=8.50\times 10^{-6} V/m$

no of turns per meter n=450

we know Induced EMF is given by

$\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A$

Magnetic Field is given by

$B=\mu _0ni$

thus $\frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}$

Area of cross-section

$A=\pi R^2$ where

solving integration we get

$E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2$

where r=distance from axis

R=radius of Solenoid

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s$

4 0

3 years ago