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Anton [14]
4 years ago
8

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

find the magnitude and direction of the force this field exerts on a charge of +6.80 μc.
Physics
1 answer:
denis23 [38]4 years ago
5 0
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
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PE: 35280 Joules

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8 0
3 years ago
A wave on a string has a wavelength of 0.90 m at a frequency of 600 Hz. If a new wave at a frequency of 300 Hz is established in
kobusy [5.1K]

Answer:

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given,

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f\ \alpha\ \dfrac{1}{\lambda}

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siniylev [52]

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