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lubasha [3.4K]
3 years ago
12

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

magnitude of the electric flux througha rectangular area of 1.65 m2 in the xy-plane. N m2/C
Physics
1 answer:
zzz [600]3 years ago
8 0

Answer:

The magnitude of the electric flux is 3.53\ N-m^2/C

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

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Answer:

a) 4.31 m/s²

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Explanation:

a) According to Newton's first law of motion

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a = ?

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b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

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t = 10 s

s = distance travelled = ?

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3 years ago
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Ann [662]

Answer:

Displacement: 6 meters

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goldfiish [28.3K]

<u>Answer:</u>

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

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