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Svetradugi [14.3K]

3 years ago

13

If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they ha

ve to be for the electric attraction between them to equal their 700 n weight?

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

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The magnetic field at the equator points north. if you throw a positively charged object (for example, a baseball with some elec

asambeis [7]

Recall the equation for magnetic force:

F = qv x B *x is cross product, not separate variable!

If the magnetic field points towards N and you throw E, then the magnetic force would point up, or out of the page. Use the right-hand rule. You point your finger towards the direction of the object, and curl your finger to the magnetic field. Your thumb is the direction of the magnetic force.

Hope this helps!

7 0

3 years ago

Please need help with this

NeTakaya

<u>Note that</u>:

The gravitational potential energy = $mgh$

where m: is the mass, g: the acceleration due to the gravity and h is the height from the earth surface

Then, we can increase the gravitational potential energy by increasing the mass or the height from the earth surface

<u>In our question</u>, we can increase the gravitational potential energy by

<u>A) Strap a boulder to the car so that it wights more.</u>

7 0

2 years ago

Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher

AnnZ [28]

Answer:

Hello your question has some missing parts attached below is the complete question

answer : 4 μm

Explanation:

since the scale bar works the same way as a scale on a map , each bar will therefore represent 1 μm and the mature parent cell's is about 4 times the labeled value hence the Mature parent cell diameter will approximately be : 4 μm

8 0

3 years ago

A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0

2 years ago

What is the speed of a wave with a frequency of 6 Hz and a wavelength of 12 m? No links

myrzilka [38]

72 m/s

Explanation:

Given,

Frequency ( f ) = 6 Hz

Wavelength ( λ ) = 12 m

To find : -

Speed ( v ) = ?

Formula : -

v = f x λ

v

= 6 x 12

= 72 m/s

Therefore,

the speed of a wave with a frequency of 6 Hz and a wavelength of 12 m is 72 m/s.

7 0

2 years ago