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timofeeve [1]
3 years ago
9

An average human weighs about 600 N. If two such generic humans each carried 1.5 coulomb of excess charge, one positive and one

negative, how far apart would they have to be for the electric attraction between them to equal their 600 N weight?
Physics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:If two such generichumans each carried 1.0 coulomb of excess charge, one posit... ... Charge, One Positive Andone Negative, How Far Apart Would They Have To Be For The Electricattraction Between Them To Equal Their 650-N Weight? ... 21.5 An average human weighs about 650 N. If two such generichumans ...

Explanation:

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Answer:

Elements in Group 14 could lose four, or gain four electrons to achieve a noble gas structure. In fact, if they are going to form ions, Group 14 elements form positive ions. Carbon and silicon form covalent bonds. Carbon's millions of organic compounds are all based on shared electrons in covalent bonds.

Explanation:

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Which of the following Illustrates 2 resistors in a series circult? A, B, C, D.​
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Answer:

It's either B or D, I'm not positive which it is

Explanation:

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2 years ago
A kayakeris paddling 2.50 m/s at an angle of 45° (northeast) and the current is moving 1.25 m/s at an angle of 315° (southeast).
PIT_PIT [208]

The kayaker has velocity vector

<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )

<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )

and the current has velocity vector

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That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that

tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s)   →   <em>θ</em> ≈ 18.4º

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5 0
3 years ago
4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
Anon25 [30]

a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

\sum F=3.55-0.52=3.03 N

Now we can find the acceleration by using Newton's second law of motion, which states that:

\sum F=ma

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

W=mg

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the gravitational field strength

So, the gravitational force on the box is

W=(0.200)(9.8)=1.96 N

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

\sum F=N-W

According to Newton's second law,

\sum F=ma

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

a=0

This means that the net force is zero:

\sum F=0

And so, we can find the normal force:

N-W=0\\N=W=1.96 N

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Oksana_A [137]

Answer:

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Explanation:

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