# Tutor profile: Rohit R.

## Questions

### Subject: Physics (Newtonian Mechanics)

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/$$s^2$$

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

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 : $$HClO_4(aq)+NH_3(aq)---->NH_4ClO_4$$ $$ΔH_{rxn4}=?$$

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

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.

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: $$x^2+y^2-5x-y+2=0$$

## Contact tutor

needs and Rohit will reply soon.