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2 years ago

What direction can Earth's plates move

1 answer:

3 0

The movement of the plates creates three types of tectonic boundaries: convergent, where plates move into one another; divergent, where plates move apart; and transform, where plates move sideways in relation to each other. They move at a rate of one to two inches (three to five centimeters) per year.

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When a wire loop is connected to a battery, what is produced in the loop?

Vinil7 [7]

Answer:

magnetic field

Explanation:

3 0

3 years ago

In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector

il63 [147K]

Answer:

$|B|=27.00425726m$

$\alpha =210.3781372$ °

Explanation:

Let's use the component method of vector addition:

$A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844$

Now, we know:

$C_x=A_x+B_x\\\\C_y=A_y+b_y$

So:

$B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774$

Now lets calculate the magnitude of the vector B:

$|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m$

Finally its angle is given by:

$\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344$ °

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.

8 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

What is the resultant of vectors shown

Ilya [14]

Answer:

adding to or more vectors together . When displacement vectors are added, the result is a resultant displacement. But any two vectors can be added as long as they are the same vector quantity.

Explanation:

4 0

4 years ago

When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is 2 A. What is the resistance

Romashka-Z-Leto [24]

Now we know by Ohm's law that

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Introducing the constant of proportionality, the resistance,the Ohm's law can be mathematically represented as

V=I x R

Where V is the voltage measured in volts

I is the current measured in amperes

R is the resistance measured in ohms

Given:

I = 2 A

V= 110 V

Applying Ohm's law and substituting the given values in the above formula we get

V=I x R

110 = 2 X R

R = 55 ohms

4 0

3 years ago