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3 years ago

Ryan wants to identify an element. Which piece of information would best help Ryan identify the element?

Physics

2 answers:

xxMikexx [17]3 years ago

8 0

The number of protons it has.

butalik [34]3 years ago

4 0

Answer: How many protons the element has

Explanation: The atoms of each element differ from those of all other elements in the number of protons. If Ryan found out how many protons the element had, he would be able to identify the element.

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At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a con

fiasKO [112]

Answer:

e.26m/s

Explanation:

Vf=Vi+at (1)

Vf=9j+(2i-4j)t

X= X₀+at

now, in the i direction

15=O+2t or t=7.5 when x position is 15

Lets put that into the (1) equation, solve for Vf.

Vf=9j+(2i-4j)7.5

Vf= 15i - 21j

Speed= $\sqrt{xcomponent^2 + ycomponent^2}$

Vf= 25.8 m/s

5 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

3 years ago

What would occur if the index of refraction of a material was less than n=1?

Bingel [31]

The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.

The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.

n=C/V

here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.

Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.

3 0

3 years ago

Read 2 more answers

Is this statement true or false concerning squall line thunderstorm development? These often form ahead of the advancing front b

Pavlova-9 [17]

Answer: The following statement is true about squall line thunderstorm development: These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.

An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.

3 0

2 years ago

A girl is sitting in a sled sliding horizontally along some snow (there is friction present).

Over [174]

Answer:

159 N

Explanation:

The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then

Fr=0.25*636=159 N

Therefore, the force of friction is equivalent to 159 N

4 0

3 years ago