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LUCKY_DIMON [66]

3 years ago

5

2 pts.) An electron is placed in an electric field of intensity 700 N/Cj. What are themagnitude and direction of the acceleratio

n of this electron due to this field? (melectron=9.1 × 10-31 kg, qe= 1.60 × 10-19 C)

1 answer:

Gnoma [55]3 years ago

6 0

Explanation:

We have,

Electric field intensity is 700 N/Cj

It is required to find the magnitude and direction of the acceleration of this electron due to this field.

The electric force is balanced by the force due to its motion i.e.

qE =ma

a is acceleration of this electron

$a=\dfrac{qE}{m}\\\\a=\dfrac{1.6\times 10^{-19}\times 700}{9.1\times 10^{-31}}\\\\a=1.23\times 10^{14}\ m/s^2$

So, the acceleration of the electron is $1.23\times 10^{14}\ m/s^2$ and it is moving in +y direction.

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loris [4]

Answer:

27.1m/s

Explanation:

Given parameters:

Height of the building = 30m

Initial velocity = 12m/s

Unknown:

Final velocity = ?

Solution:

We apply one of the kinematics equation to solve this problem:

v² = u² + 2gh

v is the final velocity

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v² = 12² + (2 x 9.8 x 30)

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3 years ago

As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins

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Answer:

A. $w=8.3rev/s$

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Explanation:

A.

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B.

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4 years ago

irga5000 [103]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of $X = 1000 \ N$ and $Y = 1400 \ N$

Explanation:

Considering the diagram,

Taking moment about the pivot , we have

$X * 1 = 400 * 2.5$

=> $X = 1000 \ N$

Generally given that the bar is at equilibrium then the upward forces is equal to the down ward force

So

$Y = X + 400$

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3 years ago

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marta [7]

Answer:

$74.31\ \text{N}$

$16.59^{\circ}$

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$\theta$ = Angle between the vectors = $45^{\circ}$

Resultant is given by

$R=\sqrt{P^2+Q^2+2PQ\cos\theta}\\\Rightarrow R=\sqrt{50^2+30^2+2\times 50\times 30\times \cos45^{\circ}}\\\Rightarrow R=74.31\ \text{N}$

Angle of resultant

$\phi=\tan^{-1}\dfrac{Q\sin\theta}{P+Q\cos\theta}\\\Rightarrow \phi=\tan^{-1}\dfrac{30\times \sin45^{\circ}}{50+30\cos45^{\circ}}\\\Rightarrow \phi=16.59^{\circ}$

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Direction of the resultant is $16.59^{\circ}$

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kozerog [31]

A coherent, typically large body of matter with no definitive shape
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3 years ago