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Tutor profile: Rohit R.

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Rohit R.
Science enthusiast with a passion for teaching.
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

How much force must one apply perpendicular to the wall on the box to keep it from slipping, given mass of box is 10kg and $$μ _s=0.2$$, g= 9.8m/$$s^2$$

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Rohit R.
Answer:

When we apply a force(F) to the box the wall gives a normal reaction N, as the block is in equilibrium in this direction. We can equate : $$F=N$$ $$(1)$$ In the vertical direction: Mg is acting vertically downwards, to resist this friction acts vertically upwards. For the box to not slip and fall: Frictional force must be equal to Mg We know the frictional force is given by: $$μ _sN$$ Therefore, $$μ _sN=Mg$$ using $$(1)$$ We can write, $$μ _sF=Mg$$ $$F=\frac{Mg}{μ _s}$$ Now substituing, $$μ _s=0.2 , M=10kg , g=9.8m/s^2$$ $$F=\frac{98N}{0.2}$$ $$F=490N$$ Therefore, the force we need to apply on the box to keep it from slipping is 490N.

Subject: Physical Chemistry

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Question:

Given the heat of reactions of the given for the following: $$HClO_4(aq)+Li(OH)(aq)---->LiClO_4(aq)+H_2O$$ ---1 $$ΔH_{rxn1}=-57.3kJ/mole$$ $$CH_3COOH(aq)+NH_3(aq)--->NH_4CH_3OO(aq)$$---2 $$ΔH_{rxn2}=-49.184kJ/mole$$ $$CH_3COOH(aq)+Li(OH)(aq)---->LiCH_3COO(aq)$$-3 $$ΔH_{rxn3}=-51.46kJ/mole$$ find the heat of reaction of : $$HClO_4(aq)+NH_3(aq)---->NH_4ClO_4$$ $$ΔH_{rxn4}=?$$

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Rohit R.
Answer:

Using Hess law, this can be solved. Hess law: The total change in enthalpy in a reaction is equal to the sum of all the enthalpy changes needed to reach that reaction. Adding 1 +2 we get: $$HClO_4(aq)+Li(OH)(aq)+CH_3COOH(aq)+NH_3(aq)---->LiClO_4(aq)+NH_4CH_3OO(aq)$$ Now we substract 3 from this, $$HClO_4(aq)+Li(OH)(aq)-Li(OH)+CH_3COOH(aq)-CH_3COOH(aq)+NH_3(aq)---->LiClO_4(aq)+NH_4CH_3OO(aq)-LiCH_3COO(aq)$$ after cancelling we get: $$HClO_4(aq)+NH_3(aq)-->LiClO_4(aq)+NH_4CH_3OO(aq)-LiCH_3COO(aq)$$ $$HClO_4(aq)+NH_3(aq)-->Li^++ClO_4^-(aq)+NH_4^++CH_3OO^-(aq)-Li^+-CH_3COO^-(aq)$$ Cancelling the common terms we get: $$HClO_4(aq)+NH_3(aq)---->NH_4ClO_4$$ Using Hess law: $$ΔH_{rxn4}=ΔH_{rxn1}+ΔH_{rxn2}-ΔH_{rxn3}$$ $$ΔH_{rxn4}=-57.3kJ/mole+-49.184kJ/mole+-51.46kJ/mole$$ $$ΔH_{rxn4}=-55.02kJ/mole$$

Subject: Algebra

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Question:

Find the point of intersections of the given two graphs $$f(x)=x^2-4x+2 $$ $$g(x)=x-2$$ and using those points as the endpoints of a diameter of the circle, find the equation of the circle.

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Rohit R.
Answer:

Point of intersection of two graphs basically means that we need to find (x,y) values where graphs intersect. First, we find the x values of the intersection points: $$f(x)=g(x)$$ $$x^2-4x+2=x-2$$ $$x^2-5x-x+2+2=0$$ $$x^2-5x+4=0$$ $$x^2-x-4x+4=0$$ Taking x and 4 common, we get : $$x(x-1)-4(x-1)=0$$ Now, we take (x-1) common, $$(x-1)(x-4)=0$$ This term becomes 0 at $$x=1 , x=4$$ To get the corresponding values of y, We substitute the value of x in either g(x) or f(x). $$g(x)=x-2$$ $$g(1)=(1)-2=-1$$ $$g(4)=(4)-2=2$$ Therefore, the point of intersections are: $$(1,-1),(4,2)$$ Now, using this as endpoints of a diameter we are required to find the equation of a circle. The equation of a circle, whose endpoints of a diameter are $$(x_1,y_1) and (x_2,y_2)$$ is given by: $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$ In our problem, $$(x_1,y_1)=(1,-1) , (x_2,y_2)=(4,2)$$ Therefore, the equation of circle is: $$(x-1)(x-4)+(y-(-1))(y-2)=0$$ $$(x-1)(x-4)+(y+1))(y-2)=0$$ Multiplying the terms: $$x^2-4x-1x+4+y^2-2y+y-2=0$$ Rearranging them: $$x^2+y^2-5x-y+2=0$$ The equation of the circle is: $$x^2+y^2-5x-y+2=0$$

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