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3 years ago

The force of 75 N must overcome in order to get the truck to start moving. What type of friction must be overcome?

Chemistry

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

Static Friction.

Explanation:

Friction is the force that resists the relative motion between the surfaces sliding against each other.

Static friction is friction between objects that are not in relative motion with each other.

The coefficient of static friction, typically denoted as μs,

Static friction arises due to surface roughness( relative term)

The static friction force can be overcome by an applied maximum force

F max = μs x N

N= normal force

Any force smaller than F max attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction.

Any force larger than F max overcomes the force of static friction and causes sliding to occur.

This maximum force is sometimes called the limiting value also. Here that value is 75 N.

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maksim [4K]

Answer:

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Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

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