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amm1812
3 years ago
11

A negative charged particle of known magnitude q is placed in a uniform electric field of known strength E. What additional info

rmation is needed to determine the magnitude of the force on the particle?
a. The distance between the charged plates producing the uniform field.
b. The distance from the nearest plate to the charged particle.
c. The magnitude of the charges producing the field.
d. The mass of the negatively charged particle.
d. No additional information is needed.
Physics
1 answer:
olga_2 [115]3 years ago
8 0

Answer:d

Explanation:

Electric field is a region around charged particle where charged particle experience a force .

Magnitude of Force experienced by charged particle is given by

F=q\cdot E

where F=force experienced by charged particle

q=charge of particle

E=strength of electric field

If q and E are given then no additional information is needed to determine the magnitude of force.      

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The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU
Ipatiy [6.2K]

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun  and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

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3 years ago
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What would happen if i put in different colors in the washer with hot water?
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PLEASE HELP.You hit a hockey puck across an empty hockey rink. At first the puck moves quickly, but then it slows and comes to a
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Hi there!

Two possible answers are air resistance and friction.

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7 0
2 years ago
Read 2 more answers
A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your
Nataly [62]

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

x = x_{0} + V_{0}t + \frac{1}{2}at^2

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m      

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

x = x_{0} + V_{0}t + \frac{1}{2}at^2

1.66 = 0 + 0t + \frac{1}{2}9.8t^2

1.66 = 4.9t^2

\frac{1.66}{4.9}  = t^2

\sqrt{0.339} = t\\ t = 0.582s

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

5 0
3 years ago
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