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2 years ago

5

A plane accelerates to a velocity of 240 m/s in 11 s by which time it had traveled 1,400 m down the runway what were it average

and initial velocities

1 answer:

Alex Ar [27]2 years ago

5 0

The average velocity is 127.27 m/s while the initial velocity is 14.55 m/s

According to Newton's law of motion:

s = [(v + u)/2]t

Where s is the distance moved = 1400 m, v is the final velocity = 240 m/s, u is the initial velocity, t is the time taken = 11 s, hence:

1400 = [(240 + u)/2)11

2800 = (240 + u)11

u + 240 = 254.55

u = 14.55 m/s

The average velocity = (v + u)/2 = (240 + 14.55)/2 = 127.27 m/s

Therefore the average velocity is 127.27 m/s while the initial velocity is 14.55 m/s

Find out more at: brainly.com/question/13639113

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a)

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

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b)

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The electric force exerted on a charged particle is given by

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where

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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:

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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.

b)

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:

$F_{B_y}=qvB_y$

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

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Substituting,

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

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

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