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3 years ago

1. For a reaction to occur, energy must be

Physics

1 answer:

amm18123 years ago

8 0

For a reaction to occur, energy must be absorbed to break chemical bonds

<u>Explanation:</u>

Reactions can be classified as chemical reaction, nuclear reaction, thermal reaction. So in these three reaction types, the nature of energy will only be varying.

But in order to execute a reaction, there should be breaking of existing bonds and then formation of new bonds. So for breaking of the bonds of reactants, energy should be absorbed from the surrounding.

Then the extra energy will be released after forming the products. Thus, the process of absorption of energy will lead to endothermic process and the process of releasing of energy will lead to exothermic reaction. So for a reaction to occur, energy must be absorbed to break the chemical bonds.

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The element in an incandescent light bulb that releases light energy is? A- a phosphor, B- mercury vapor, C- a thin tungsten fil

LenKa [72]

C.) A Thin tungsten filament

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3 years ago

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

$I\alpha \frac{1}{r^{2} }$

Now according to the question the distance is increased by three times than,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }$

Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

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3 years ago

The nucleus of most atoms is composed of which of the following sub-atomic particles?

Blababa [14]

Answer:

protons (+ charge) & neutrons (neutral charge)

however protons has a positive charge so it determined what atom it is.

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3 years ago

Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

3 0

3 years ago

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

$x=v\cos\theta t$

On substituting the known values,

$\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}$

The change in the height, y of the basketball is given by,

$y=-v\sin\theta t+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity.

On substituting the known values,

$\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}$

Final answer:

The parametric equations describing the shot are

$\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}$

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1 year ago