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Inga [223]
4 years ago
15

What is heat of combustion?

Physics
1 answer:
BlackZzzverrR [31]4 years ago
4 0
The heat of combustion<span> is the total energy released as </span>heat<span> when a substance undergoes complete </span>combustion<span> with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon or other organic molecule reacting with oxygen to form carbon dioxide and water and release </span>heat<span>.

According to </span><span>Wikipedia :)</span>
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Which of the following would likely be the best reflector of heat energy?
expeople1 [14]
Answer: A, a polished pale yellow surface would likely be the best reflector of heat energy. 
6 0
4 years ago
Read 2 more answers
a tungsten and coil has a resistance of 12 ohm at 15 degree celsius is the temperature coefficient of resistance of tungsten is
Dennis_Churaev [7]

Answer:

The resistance of the tungsten coil at 80 degrees Celsius is 15.12 ohm

Explanation:

The given parameters are;

The resistance of the tungsten coil at 15 degrees Celsius = 12 ohm

The temperature coefficient of resistance of tungsten = 0.004/°C

The resistance of the tungsten coil at 80 degrees Celsius is found using the following relation;

R₂ = R₁·[1 + α·(t₂ - t₁)]

Where;

R₁ = The resistance at the initial  temperature = 12 ohm

R₂ = The resistance of tungsten at the final temperature

t₁ = The initial temperature = 15 degrees Celsius

t₂ = The final temperature = 80 degrees Celsius

α = temperature coefficient of resistance of tungsten = 0.004/°C

Therefore, we have;

R₂ = 12×[1 + 0.004×(80 - 15)] = 15.12 ohm

The resistance of the tungsten coil at 80 degrees Celsius = 15.12 ohm.

6 0
3 years ago
Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elem
Aleksandr-060686 [28]

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus,  atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

7 0
3 years ago
Each of 100 identical blocks siting on a frictionless surface is connected to the next bloc by a massless string. The first bloc
Anit [1.1K]

Answer:

A) 1 N

B) 50 N

Explanation:

Let us consider that the string does not deform.

To solve this problem lets consider the whole system as two parts. In the initial case, the first part will be de 100N being exterted to the whole system and in the second the 100 blocks system.

In this case we can imagine as the whole system being pulled by 100 N, and therefore its acceleration will be:

a = 100 N /(100 m)

where m stantds for the mass of one block

Now, the whole system and its individual parts must move with the same acceleration otherwise the string would stretch.

Now lets consider the first part of the system as the first block, and the second part as the other 99 blocks.

The Tension of the string pulling the 99 blocks must be so that it exterts the enough force to move that 99blocks-system at an acceleration a, since that sub-system has a mass of 99m

T1 = 99 m * a = (99 m) * (100 N/ 100 m) = 99 N

Now lets consider an intermidiate sub-system, where the first part is made of n blocks and the second susbsystem is made of (100 -n) blocks

Following the same logic, the tension of the corresponding string must be the acceleration of the whole systems times the mass of the second subsystem:

Tn = (100 -n)m * ( 100 N / 100 m ) = (100 -n) N

a)

Therefore the tension in the string connecting block 100 to block 99 must be

<u>T99 = 1 N</u>

<u />

b)

And

<u>T50 = 50 N</u>

6 0
3 years ago
A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial
Sveta_85 [38]
B- the student changed too many variables
7 0
3 years ago
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