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3 years ago

The acceleration of an object increases by the force applied to it and decreases based on it's mass.

Physics

1 answer:

xenn [34]3 years ago

5 0

Explanation:

Force is defined as mass times acceleration.

Mathematically, F = m × a

where F = force

m = mass

a = acceleration

This shows that force is directly proportional to acceleration. So, when greater force is applied then acceleration will also increase.

Whereas acceleration is inversely proportional to mass. So, when there is increase in mass then there will be decrease in acceleration.

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What is called the process of finding exact quantity of a substance?

Tema [17]

Answer:

Measurement is called the process of finding exact quantity of a substances.

7 0

2 years ago

A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift th

Alina [70]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

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$\large \bold {ANSWER}$

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

$\large \bold {EXPLANATION}$

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, because generally when it is mobile, this information is highlighted in the question.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.

3 0

2 years ago

Which of the following is the largest unit? Hectogram Dekagram Decigram Microgram

MAXImum [283]

Answer:

So far it seems that hectogram us the largest unit.

Dekagram is a metric unit of mass or weight, equal to 10 grams.

Hectogram is a metric unit of mass equal to 100 grams.

Decigram is a metric unit of mass and weight equal to ¹/₁₀ gram.

Microgram is a metric unit of mass equal to one millionth of a gram.

8 0

3 years ago

What force is needed to propel a 421-kg car with an acceleration of 3.77 m/s2?

nekit [7.7K]

brainly.com/question/11542618?answering=true&answeringSource=greatJob%2FquestionPage

7 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

4 years ago