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elena-s [515]
3 years ago
5

400 N of load can be overcome by an effort of 50 N by using a lever. Calculate the mechanical advantage of the lever.

Physics
1 answer:
mars1129 [50]3 years ago
4 0

Answer:

load (l)=400N

Effort(E)=50N

mechanical advantage (MA)= load ÷Effort

(ma)=400÷50

(ma)=8

Explanation:

I copy pasted from the answer from the same question. Remember to first check if ur question is there

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When considering that in the past human societies developed in greater isolation from one another than today, each of the follow
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Answer:One can easily assume a direct cause and effect relationship between a physical environment and an aspect of culture

Explanation: During the prehistoric and the earlier centuries human societies developed in isolation, there's no interconnection or much of communication between different groups and societies.

Knowledge sharing is not prominent, but in recent times people are more connected to each other,communities and countries interaction takes place through different forums, during the earlier centuries there are no mobile communication equipments,no Television set or the level of sophistication as it is today.

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A boy flying a kite is standing 30 ft from a point directly under the kite. if the string to the kite is 50 ft long, what is the
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Draw a diagram to illustrate the problem as shown in the figure below.
h =  height of the kite above ground.

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Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in
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Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) X_2+3Y_2\rightarrow 2XY_3     \Delta H_1=-370kJ

(2) X_2+2Z_2\rightarrow 2XZ_2    \Delta H_2=-120kJ

(3) 2Y_2+Z_2\rightarrow 2Y_2Z    \Delta H_3=-270kJ

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) 4XY_3\rightarrow 2X_2+6Y_2     \Delta H_1=2\times (+370kJ)=740kJ

(2) 2X_2+4Z_2\rightarrow 4XZ_2    \Delta H_2=2\times (-120kJ)=-240kJ

(3) 6Y_2+3Z_2\rightarrow 6Y_2Z    \Delta H_3=3\times (-270kJ)=-810kJ

The expression for enthalpy of formation of CH_4 will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+740kJ)+(-240kJ)+(-810kJ)

\Delta H=-310kJ

Therefore, the change in enthalpy of the reaction is, -310 kJ

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