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2 years ago

Based on these answers how can I determine when electric field strength is affected or not

Physics

1 answer:

Stolb23 [73]2 years ago

6 0

It's not that hard. Let me explain...

There are two ways to remember this.

The first one is by using the formula for the electric strength (E)

E = k × (Q / r^2) , where:

k is a constant
Q is the "central" charge. The one producing the electric field. Not the one we use to test it.
r is the distance between the charges

As you can see, field's strength is proportional to the magnitude of Q, and inverse proportional to the square of the distance. The field does not care how much you alter the charge you positioned in ot to test it.

Although that is the only way to work out the problem, there are many ways to remember it. Given that i have difficulty in remembering formulas, i always try to derive it from something i know.

By definition: E is equal to the force acted on the test charge, divided by its value.

E = F / q

F --> Coulomb's force
q --> test charge

Since i remember that:

F = k ( Q × q) / r^2

just take out from the expression the q term, since you divided by it, and all it remains is the strength of the field.

In other words, the strength changes as long as something from the expression is altered.

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Why forces are balanced and unbalanced? need help with this the lesson is tommorow

Naya [18.7K]

"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way.

-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down. It's just as if there's nobody sitting on it.

-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move. The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.

-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her. If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.

From these examples, you can see a few things:

-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a group of forces that is either balanced or unbalanced.

-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.

-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.

4 0

2 years ago

Read 2 more answers

What is the average cost for a severe injury in a collision?

koban [17]

Answer:

Explanation:

According to research and data, the average cost for a severe injury in a collision is $247,000.

These include traumatic brain injury, severe damage to limbs which could result to loss of the limbs, spinal cord injuries that could result to partial or total paralysis as well as internal damage to organs in the collision.

8 0

3 years ago

If two objects of the same size move through the air at different speeds, which encounters the greater air resistance?a. The fas

blagie [28]

Answer:

Option A is correct.

(The faster object encounters more resistance)

Explanation:

Option A is correct. (The faster object encounters more resistance)

Air resistance depends on various factors:

Speed of the object
Cross-sectional area of the object
Shape of the object

Formula:

$F=\frac{1}{2}C_d\rho A v^{2}$

As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.

6 0

2 years ago

An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

jolli1 [7]

Answer:

$v_o=39\ m/s\\t_m=4\ s$

Explanation:

Vertical Launch Upwards

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=H+\frac{v_o^2}{2g}$