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3 years ago

PLEASE HELP!! What type of bond does this molecule have?

Physics

2 answers:

ehidna [41]3 years ago

8 0

Looks like a double covalent bond.

ladessa [460]3 years ago

7 0

Answer : The correct option is, double covalent bond.

Explanation:

Ionic bond : It is defined as the bond that is formed by complete transfer of electrons from one atom to another atom.

That means, the atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom.

The ionic bond is usually formed between a metal and a non-metal.

Covalent bond : It is defined as the bond which is formed when sharing of electrons takes place between the atoms.

This is usually formed between two non-metals.

In the given image, two oxygen atoms combine with 1 carbon atom and form $CO_2$ as a molecule.

In $CO_2$ molecule, carbon is a non-metal and oxygen is also non-metal. So, carbon and oxygen will form a covalent bond.

That means, two oxygen atoms combine with 1 carbon atom by two covalent bond or double covalent bond.

Hence, the correct option is, double covalent bond.

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A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t

iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

$T = 2\pi \sqrt{\frac{l}{g} }$

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

$T' = 2\pi \sqrt{\frac{2l}{g} }$

$T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }$

$T' = \sqrt{2} * T$

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

#SPJ4

8 0

1 year ago

Identify the law, write the equation and calculate the answer to the problem below.

lyudmila [28]

Find refractive index first

$\\ \rm\Rrightarrow \mu=\dfrac{c}{v}$

$\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}$

$\\ \rm\Rrightarrow \mu =0.75$

Now

$\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu$

$\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75$

$\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr$

$\\ \rm\Rrightarrow sinr=0.94$

$\\ \rm\Rrightarrow r=sin^{-1}(0.94)$

$\\ \rm\Rrightarrow r=70^{\circ}$

5 0

2 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$