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4 years ago

What is an input force?

2 answers:

nataly862011 [7]4 years ago

6 0

Force is a pull or a push acting on a body at rest or in motion resulting from its interaction with another body. Input force is the force that you put on a machine while Output force is the force the machine exerts on an object. The output distance is when the output force moves the machine a certain distance while the input distance is when the input distance is when the input force moves the machine a certain distance.

Alchen [17]4 years ago

6 0

Answer: Option (D) is the correct answer.

Explanation:

When a force is applied by a person to a system then this force is known as input force.

It is also known as initial force that needs to be applied in order to start a machine.

For example, when a piston in a system is pressed inwards by a person then an input force is applied to the system.

Thus, we can conclude that an input force the force a person applies to a simple machine.

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The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago

For some transformation having kinetics that obey the Avrami equation (Equation 11.17), the parameter n is known to have a value

Alex787 [66]

Answer:

t ≈ 235.47 secs

Explanation:

<u>Calculate how long it will take the transformation to go to 88% completion</u>

Given that Avrami equation ( <em>y </em>)= 1 - exp( - kt^n )

n = 2.3

t = 145 secs

reaction = 50% after t = 145 secs

Reaction = 88% after t = ?

attached below is a detailed solution

3 0

3 years ago

6. The graph below shows the heating curve for ethanol (from –200C to 150C). Calculate the amount of heat (kJ) required for each

Kazeer [188]

This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

$Q_T=Q_1+Q_2+Q_3+Q_4+Q_5$

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

$Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\
\\
Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\
\\
Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\
\\
Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\
\\
Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ$

Finally, we add them up to get the result:

$Q_T=0.208kJ+0.106kJ+0.470kJ+0.838kJ+0.136kJ\\
\\
Q_T=1.758kJ$

Learn more about heating curves: brainly.com/question/10481356

7 0

2 years ago

The density of nitric acid (HNO3) is 1.41 g/ml. how many ml are in 20.0g of the nitric acid?

soldi70 [24.7K]

Answer:

14.2

Explanation:

to find mL we must cancel grams we do this by the equation:

20/1.41 = 14.2 mL

7 0

3 years ago

For many purposes we can treat ammonia (NH) as an ideal gas at temperatures above its boiling point of -33.C. Suppose the temper

levacccp [35]

Answer:

In homeothermic (“warm-blooded”) animals, body temperature is carefully

regulated. The hypothalamus, located in the brain, acts as the master ther-

mostat to keep body temperature constant to within a fraction of a degree

Celsius in a healthy animal. If the body temperature starts to deviate much

from the desired constant level, the hypothalamus causes changes in blood

flow and initiates other processes, such as shivering or perspiration, to bring

the temperature back to normal. What evolutionary advantage does a con-

stant body temperature give the homeotherms (e.g., birds and mammals)

over the poikilotherms (e.g., reptiles and insects), whose body temperatures

are not kept constant? What are the disadvantages?

Explanation: Basic chemical understanding as revealed upwards

6 0

3 years ago