GS S3

14.

The acceleration of an object which is moving in a circle (centripetal acceleration)acts toward the center of the circle when the object is at constant speed. However, since the speed of the plane is decreasing, it experiences a deceleration (= negative acceleration) which acts along the tangent to the circle opposite to the direction of its velocity. The resultant acceleration vector can be obtained by finding the vector sum of A. and C., and this gives vector B. 

(Had the speed been increasing, we would add C. plus a vector to be drawn in the direction of the velocity at point 'P')

15.

Pipe with the largest area. Continuity eq states that Av= constant ∴ fluid will have its lowest velocity when Area is at its greatest. Then extrapolating from Bernoulli's, where P + (1/2)ρv² + ρgh = const; region with the highest pressure is also the one with the smallest velocity.
23. Question is asking how changing the [catalyst] would affect the rate of reaction or the amount of product. Catalysts do not affect rates. Therefore, there would be no change in the partial pressure of any of the reactants.
24.First, ask yourself if the rxn is exothermic or endothermic. How do you do that? Is ∆H positive or negative? Negative= exothermic. Then ask yourself if you add or subtract heat from either side, how this would affect the rate of the rxn (more products vs reactants)? 
For an exothermic reaction, if heat is reduced than the rxn would shift to the right, more products produced.

38.nmedium = c / vmediumfor n air to equal 1.00, c and v medium must be the same.
41.sin θcritical= n2/n1

This is all you should recall when thinking about critical angle if n2 and n1 are given, plug them in. 

= 1/1.3= 1/(4/3)= 3/4=0.75
42.sin θcritical= n2/n1

Now: sin θc2 = (nair + x)/nwater

Thus we increased the numerator by some value "x" which means sin θc2 > sin θc, which also means θc2 > θc.
50.Since MgCO3 ↔ Mg2+ + CO32-, Ksp = [Mg2+][CO32-] = s x s = s2 where s is the solubility. From the table provided, s = 1.30 x 10-3 mol L-1. 
Thus Ksp = [Mg2+][CO32-] = s2 = (1.30 x 10-3)2 = 1.69 x 10-6 ≈ 1.7 x 10-6. Mathfax: (13)2= 169 ∴ (1.3)2 = 1.69

52.lowest solubility= highest stability. When the size of the anion matches that of the cation, crystal structure is more uniform.
CO32- anion is approximately the same size as Sr2+ because the passage says in the 4th paragraph that if the cation and anion are about the same size then when they are together (SrCO3), it is more stable.
56.
a constant can be any calculated value.

In this particular case, we have:

PV = nRT

But the y axis is PV/nRT so if we have PV = nRT, we must divide both sides by nRT to acquire what the graph is showing on the y axis:

PV = nRT

so:

PV/nRT = nRT/nRT

which means:

PV/nRT = 1

This is not something to memorise. This is strictly an application of basic math rules (see GS GAMSAT book 6 GAMSAT Math chapters).75.A simple torque force problem: let the fulcrum (= the center of gravity of the bar) be the pivot point :

∑L; = CCW - CW = (100 kg x g x 0.5 m) - (75 kg x g x x) = 0

Thus (100 kg x g x 0.5 m) = (75 kg x g x x)

g cancels, manipulate to get: x = 50/75 = 2/3 m = 0.67 m (CW is to the right of the fulcrum).
76.momentum is conserved by KE isn't. Total energy doesnt change
77.Using the principle of Conservation of Momentum:

Momentum before collision = Momentum after collision

(1200 kg x 7.5 m s-1) + (8000 kg x 3.0 m s-1) = (1200 kg x 3.0 m s-1) + (8000 kg x x m s-1)

Divide through by 400: (3)(7.5) + (20)(3) = (3)(3) + 20x

Isolate x: (7.5 + 20 - 3)3/20 = x

Thus x = (24.5)(3/20) = 73.5/20 = 3 13.5/20 ≈ 3 7/10 = 3.7 m s-1.

{Recall: momentum is a vector; in this problem the sign for the velocities are always positive because we are told that the vehicles are always moving in the same (i.e. northerly) direction}
78.
1. Determine if the car after its 300N forward thrust is now in motion (i.e. overcomes frictional force)
2. If it is not in motion, then the sum of the forces must be zero. Therefore, frictional force is 300N in the opposite direction.
79.The normal force on an object always acts perpendicular to the surface, in this case, the road. Friction depends on the normal force (see Q33). Since only a component of the weight of the car acts perpendicular to the hill, the value of the normal force decreases.Since the coefficient of friction remains the same, the value of the maximum frictional force decreases. 

**Once an object is on an incline: N < weight of the object as determined by the cosine of the angle of the incline
90.Layers in graphite can move relative to one another/ slide. Hence, it can be used as a lubricant.
91.ASK yourself what heats are we given? Combustion, formation, fusion.

From Table I we are not given ΔHformation; rather, we are provided with a new parameter which the table describes as the enthalpy of combustion ΔHc of carbon. The end product of combustion (the oxide) of carbon is carbon dioxide. Now we can use Hess's Law knowing that the ΔHc for Cgraphite= -393.3 kJ mol-1 and for Cdiamond= -395.1 kJ mol-1. We can summarize the process as follows:
Cgraphite → CO2 ΔHc = -393.3 kJ mol-1

CO2 → Cdiamond ΔHc = 395.1 kJ mol-1

Cgraphite → Cdiamond ΔHc = 1.8 kJ mol-1

{Notice the change in direction of the equation for Cdiamond was necessary in order to cancel the CO2; thus the sign forΔHc was changed from negative to positive}
98.

On the Surface: Don't convert mL to litres because it will cancel out.

x = molarity of Mg3(PO4)2 --> Solve for x

(.005M Mg2+)(5mL) + (3x)(15mL) = (20mL)(.05M Mg2+) 

Note the 3 in front of x because there are 3 potential Mg2+ generated from each Mg3(PO4)2 in aqueous solution.

.025 + 45x = 1 
45x = 0.975 thus x = 0.022

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