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poizon [28]
3 years ago
9

A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference

between x = 0.00 m plane and the x = 3.60 m plane? V(3.60 m) = kV (b) A point particle that has a charge of +2.90 µC is released from rest at the origin. What is the change in the electric potential energy of the particle as it travels from the x = 0.00 m plane to the x = 3.60 m plane?
Physics
1 answer:
EastWind [94]3 years ago
8 0

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference  ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J

ΔU =18.79 mJ

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Answer:

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A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al
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Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

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x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

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          ay = (v_{oy} -v_{y}) / t

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the sign indicates that the acceleration goes in the negative direction of the y axis

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A 15kg ball accelerates at a rate of 3m/s/s. What force was required?
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Answer:

<h2>45 N</h2>

Explanation:

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force = mass × acceleration

From the question we have

force = 15 × 3

We have the final answer as

<h3>45 N</h3>

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