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3 years ago

What types of bond forms when two or more atoms share electrons?

2 answers:

8 0

D) Hydrogen is the correct/ right answer because it says in the science book in the chapter that it is on to find the correct answer as well as look in the chapter that is on and look on the pages

Kazeer [188]3 years ago

6 0

<span>A.) covalent

This type of bond happens when the electrons are shared between the atoms.
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Which substance is a greenhouse gas that is emitted by human activities, is the most widespread, and remains in the air the long

e-lub [12.9K]

Answer:

I am not really sure, but it is probably Carbon Dioxide

Explanation:

3 0

3 years ago

Read 2 more answers

Calculer l intensité du courant qui le traverse

MissTica

Answer:

Le calcul du courant se fait avec deux éléments : la tension et la valeur de la résistance. Courant (A) = tension (V) / résistance (Ohm) ce qui donne la formule I = U/R.

please mark me as brainalist

6 0

2 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

V125BC [204]

Answer:

Current in the toroid will be $I=17.32\times 10^{-3}A$

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid $E=9\times 10^{-16}J$

We have to find the current in the toroid

Magnetic energy stored is equal to $E=\frac{1}{2}Li^2$

$9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2$

$I=17.32\times 10^{-3}A$

So current in the toroid will be $I=17.32\times 10^{-3}A$

5 0

3 years ago

A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made

olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

$B=\dfrac{\mu_oI}{2\pi r}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}$

B = 0.000002 T

$B=10^{-5}\ T$

(2) Exact formula:

$B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }$

$B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }$

B = 0.00000199 T

B = 0.000002 T

Hence, this is the required solution.

4 0

3 years ago