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4 years ago

Which part of an atom is actively exchanged or shared in a chemical bond?

Physics

2 answers:

RideAnS [48]4 years ago

6 0

<em>The part of an atom that is actively exchanged or shared in a chemical bond is the;</em>

B. Electron

<u>Electrons are negatively charged atoms.The electron is actively exchanged in a chemical bond </u>

Naddika [18.5K]4 years ago

3 0

Which part of an atom is actively exchanged or shared in a chemical b or shared in a chemical bond

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A 10.-kilogram object, starting from rest, slides down a frictionless incline with a constant acceleration of 2.0 m/sec2 for fou

aniked [119]

Answer:

C) 100N

Explanation:

Formula for calculating the Weight of an object is expressed as;

Weight = mass × acceleration due to gravity

Given

Mass of the object = 10kg

Acceleration due to gravity = 9.81m/s²

Substitute into the formula above

Weight = 10×9.81

Weight = 98.1N

Hence the approximate weight of the object is 100N

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3 years ago

Using an example, such as the natural building of sand dunes, the process of DEPOSITION causes?

andre [41]

The correct answer is c. The process of deposition causes rock and soil to be slowly gained. Deposition is a geological process in which soil and rocks are added to a landform.

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3 years ago

Read 2 more answers

9) An electrical appliance has a resistance of 25 N. When this electrical ap-

Neko [114]

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

6 0

3 years ago

The drawing shows two long, thin wires that carry currents in the positive z direction. Both wires are parallel to the z axis. T

erica [24]

Answer:

The magnitude of the magnetic field at the origin is $2.56\times 10^{-6}\ T$ .

Explanation:

Given :

50-A wire is in the x-z plane and is 5 m from the z axis.

Also , 40-A wire is in the y-z plane and is 4 m from the z axis.

Now , since both the wire are perpendicular to each other .

Therefore , magnetic field are also perpendicular to each other .

Magnetic field at origin due to wire 1 is :

$B_1=\dfrac{\mu_oI_1}{2\pi R_1}\\\\B_1=\dfrac{(50)\mu_o}{2\pi( 5)}\\\\B_1=\dfrac{5\mu_o}{\pi}$

Magnetic field at origin due to wire 2 is :

$B_2=\dfrac{\mu_oI_2}{2\pi R_2}\\\\B_2=\dfrac{(40)\mu_o}{2\pi( 4)}\\\\B_2=\dfrac{4\mu_o}{\pi}$

Now , therefore net magnetic field is :

$B=\sqrt{B_1^2+B_2^2}\\\\B=\sqrt{(\dfrac{5\mu_o}{\pi})^2+(\dfrac{4\mu_o}{\pi})^2}\\\\B=\dfrac{\sqrt{41}\mu_o}{\pi}$

Putting value of $\mu_o=4\pi \times 10^{-7}\ H/m$

We get ,

$B=\sqrt{41}\times 4\times 10^{-7}\\B=2.56\times 10^{-6}\ T$

Therefore, the magnitude of the magnetic field at the origin is $2.56\times 10^{-6}\ T$ .

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3 years ago

Which velocity-time graph is equivalent to the given position-time graph?

Agata [3.3K]

Answer:

Explanation:

it's hard to explain... sorry

6 0

3 years ago