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3 years ago

This illustration represents the compoundA)carbon oxide.B)carbon dioxide.C)carbon monoxide.EliminateD)monocarbon oxide.

Physics

2 answers:

pentagon [3]3 years ago

3 0

B.) Carbon Dioxide because the carbon is surrounded by oxygen

maxonik [38]3 years ago

3 0

<u>Answer:</u> The name of the given compound is carbon dioxide.

<u>Explanation:</u>

We are given a compound having 1 carbon atoms and 2 oxygen atoms.

Covalent compound is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound. These are usually formed when two non-metals react.

$CO_2$ is a covalent compound because sharing of electrons takes place between carbon and oxygen atoms. Both the elements are non-metals and hence, will form covalent bond.

The nomenclature of covalent compound is given by:

The less electronegative element is written first, which is carbon.
The more electronegative element is written then (which is oxygen), and a suffix is added with it. The suffix added is '-ide'.
If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on..

So, the name of the given compound is carbon dioxide.

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

AVprozaik [17]

Answer:

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Explanation:

Given data

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To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

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$mg=1/2*C*p_{air}*v^{2}*A_{area}\\ v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s$

For speed in km/h(kilometers per hour)

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3 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

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3 years ago