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jek_recluse [69]
3 years ago
10

A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor

m linear charge density λ2 is located on the y-axis with their centers crossing at the origin. In what direction is the electric field at point z = a on the positive z-axis if λ1 and λ2 are positive?
A-) The positive z-direction
B-) All directions are possible parallel to the x-y plane
C-) The negative z-direction
D-) Halfway between the x-direction and the y-direction
C-) Cannot be determined
Physics
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is  located on the x-axis=\lambda_1

Linear charge density of another long line which is  located on the y-axis=\lambda_2

We have to find the direction of electric field at z=a on the positive z-axis if \lambda_1 and \lambda_2 are positive.

The direction of electric field  at z=a on the positive z-axis  is positive z-direction .

Because \lambda_1 and \lambda_2 are positive and the electric field is  applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

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Examine the given choices.
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c. 0.025 g.
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3 years ago
What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction
Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

f'=f(\dfrac{v+v_{p}}{v-v_{s}})

Where, f = frequency

v = speed of sound

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v_{s} = speed of source

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A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.​
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will

Explanation:

3 0
2 years ago
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

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x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

Magnitude of displacements,

r=\sqrt{x^2+y^2}=41.32km

Direction,

\alpha =tan^{-1}(\frac{y}{x})=24.23^o

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