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3 years ago

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A positively charged particle initially at rest on the ground accelerated upward to 100m/s in 2.00s. If the particle has a charg

e-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform, what are the magnitude and direction of the electric field?

1 answer:

VashaNatasha [74]3 years ago

4 0

Answer:

Explanation:

From the question we are told that

The initial velocity is $u = 100 m/s$

The time taken is $t = 2.0 s$

The charge to mass ratio is $Q/m = 0.100 C/kg$

Generally the acceleration is mathematically evaluated as

$a = \frac{u}{t }$

substituting values

$a = \frac{100}{2}$

$a = 50 \ m/s^2$

The electric field is mathematical represented as

$E = \frac{(a+g)}{Q/m}$

substituting values

$E = \frac{(50+9.8)}{0.100}$

$E = 598 \ N/C$

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