# ENGAA Prep Thread 2019 (for 2020 Admissions)

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#41

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Can you explain question 12 of Section 2 please?

**TravisBickle02**)Can you explain question 12 of Section 2 please?

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#42

(Original post by

Look at the attached image

**nidza100**)Look at the attached image

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#43

(Original post by

We are given that when the speed is v0, acceleration is half the acceleration of freefall, so kv0^n = 1/2 mg, so mg = 2kv0^n. We also know that when the object reaches terminal velocity, the air resistance is equal to the weight (ignoring any other forces). If we call terminal velocity V, then kV^n = mg, meaning kV^n = 2kv0^n. The k's cancel, meaning V^n = 2v0^n, and if we increase all the terms by the power of (1/n), we get V = 2^(1/n)v0, which is B.

**Gfhfnxrj**)We are given that when the speed is v0, acceleration is half the acceleration of freefall, so kv0^n = 1/2 mg, so mg = 2kv0^n. We also know that when the object reaches terminal velocity, the air resistance is equal to the weight (ignoring any other forces). If we call terminal velocity V, then kV^n = mg, meaning kV^n = 2kv0^n. The k's cancel, meaning V^n = 2v0^n, and if we increase all the terms by the power of (1/n), we get V = 2^(1/n)v0, which is B.

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#44

Hi,

I’m currently studying the OIB spécialité S (which is a mix of A level Literature, History and Geography combined with the French Scientific Baccalaureate). This makes my course rather broad and severely limits the amount of Physics which we are taught (we haven’t studied any real mechanics yet not have we studied radioactivity). I have been working hard for the past 8 months on catching up the gap, although I do find the ENGAA paper really difficult. I’m currently getting 60% on section 1 and 45% in section 2. I’ve tried to go though my mistakes and understand them, but there is one question I still don’t understand.

Could anyone please explain to me question 18 in section 2? I am really confused as to how I should solve it.

Thanks

I’m currently studying the OIB spécialité S (which is a mix of A level Literature, History and Geography combined with the French Scientific Baccalaureate). This makes my course rather broad and severely limits the amount of Physics which we are taught (we haven’t studied any real mechanics yet not have we studied radioactivity). I have been working hard for the past 8 months on catching up the gap, although I do find the ENGAA paper really difficult. I’m currently getting 60% on section 1 and 45% in section 2. I’ve tried to go though my mistakes and understand them, but there is one question I still don’t understand.

Could anyone please explain to me question 18 in section 2? I am really confused as to how I should solve it.

Thanks

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#46

(Original post by

And 15, 18, 19, 20 please?

**TravisBickle02**)And 15, 18, 19, 20 please?

18) Let I1 equal the current from the 20V power supply, I2 equal the current in the line with the ammeter, and I3 equal the current around the 10V power supply. If we look at the loop made with the first and middle line and use the formula e.m.f = sum of Current x resistance, we get:

20 =40I1 +40I2

If we do the same with the loop in the middle and the end, we get:

10 = 40I3 + 40I2

Finally, if we do it with the loop made from the first and last lines, we get:

10 = 40I1 + 40I3 (10 because the 20V cell and the 10V cell are in the opposite direction, meaning we subtract 10 off from 20).

Using the second and third formula, we get I1=I2. The ammeter reads I2, so we replace I1 with I2, and we get I2 = 0.25A. The voltmeter reading is for the resistor that has a current I3. Substituting I2 into the third equation, we get I3 = 0. If there is no current flowing through the resistor, there is no voltage across it, meaning the Voltmeter reading has to be 0. A

19) If we look at the pressure of water acting on top of the rock:

we know u = 0 (starts from rest), s = 45 and a = 10 (air resistance is negligible and assumption through basically every ENGAA Q is that g = 10). From this, we can work out t using s = ut + 1/2at^2. From this, we get t = 3. We know 40kg of water hits the rock surface every second, so in 3 seconds 120 kg of water hits the rock surface, meaning 1,200 N of force acting on it, and as the rock surface has a surface area of 2, the pressure acting on top is 600 Pa.

Now if we look at the pressure acting at the bottom:

We know the pressure of a fluid acting on a submerged object = height x density of the fluid x acceleration of freefall. We are given the average depth of water on the rock surface is 0.05m, and that the density of water is 1000 kg m^-3, so using the formula, we get the pressure as 500 Pa.

The total pressure acting on the rock is 600 + 500 = 1100 Pa. E

20) We are given that X emits light when the P.D across it exceeds 2V. The 200 ohm resistor has a voltage of 2V and the e.m.f is 12V. By using simple ratios, we can figure out that the resistance of the thermistor is 1000 ohms if the P.D of X is 2V. If the resistance becomes greater than 1000 ohms, the P.D of X falls below 2V, meaning no light is emitted, meaning the resistance of the thermistor cannot exceed 1000 ohms. If we then substitute 1000 as R in the given equation, and the fact we know 1000 ohms is the maximum resistance, we get 1000 > Rob^(-uT) (I know it's mu and not u, but I don't know how to put the greek letters into here). using logs, we get logb(1000) > -uT + logb(R0), rearranging further to get uT > logb(R0) - logb(1000), and finally T > (logb(R0) - logb(1000))/u, or T > (1/u) (logb(R0) - logb(1000)) A

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#47

(Original post by

15) Admittedly this is the one question I have yet to solve, will probably edit tomorrow to show the answer

18) Let I1 equal the current from the 20V power supply, I2 equal the current in the line with the ammeter, and I3 equal the current around the 10V power supply. If we look at the loop made with the first and middle line and use the formula e.m.f = sum of Current x resistance, we get:

20 =40I1 +40I2

If we do the same with the loop in the middle and the end, we get:

10 = 40I3 + 40I2

Finally, if we do it with the loop made from the first and last lines, we get:

10 = 40I1 + 40I3 (10 because the 20V cell and the 10V cell are in the opposite direction, meaning we subtract 10 off from 20).

Using the second and third formula, we get I1=I2. The ammeter reads I2, so we replace I1 with I2, and we get I2 = 0.25A. The voltmeter reading is for the resistor that has a current I3. Substituting I2 into the third equation, we get I3 = 0. If there is no current flowing through the resistor, there is no voltage across it, meaning the Voltmeter reading has to be 0. A

19) If we look at the pressure of water acting on top of the rock:

we know u = 0 (starts from rest), s = 45 and a = 10 (air resistance is negligible and assumption through basically every ENGAA Q is that g = 10). From this, we can work out t using s = ut + 1/2at^2. From this, we get t = 3. We know 40kg of water hits the rock surface every second, so in 3 seconds 120 kg of water hits the rock surface, meaning 1,200 N of force acting on it, and as the rock surface has a surface area of 2, the pressure acting on top is 600 Pa.

Now if we look at the pressure acting at the bottom:

We know the pressure of a fluid acting on a submerged object = height x density of the fluid x acceleration of freefall. We are given the average depth of water on the rock surface is 0.05m, and that the density of water is 1000 kg m^-3, so using the formula, we get the pressure as 500 Pa.

The total pressure acting on the rock is 600 + 500 = 1100 Pa. E

20) We are given that X emits light when the P.D across it exceeds 2V. The 200 ohm resistor has a voltage of 2V and the e.m.f is 12V. By using simple ratios, we can figure out that the resistance of the thermistor is 1000 ohms if the P.D of X is 2V. If the resistance becomes greater than 1000 ohms, the P.D of X falls below 2V, meaning no light is emitted, meaning the resistance of the thermistor cannot exceed 1000 ohms. If we then substitute 1000 as R in the given equation, and the fact we know 1000 ohms is the maximum resistance, we get 1000 > Rob^(-uT) (I know it's mu and not u, but I don't know how to put the greek letters into here). using logs, we get logb(1000) > -uT + logb(R0), rearranging further to get uT > logb(R0) - logb(1000), and finally T > (logb(R0) - logb(1000))/u, or T > (1/u) (logb(R0) - logb(1000)) A

**Gfhfnxrj**)15) Admittedly this is the one question I have yet to solve, will probably edit tomorrow to show the answer

18) Let I1 equal the current from the 20V power supply, I2 equal the current in the line with the ammeter, and I3 equal the current around the 10V power supply. If we look at the loop made with the first and middle line and use the formula e.m.f = sum of Current x resistance, we get:

20 =40I1 +40I2

If we do the same with the loop in the middle and the end, we get:

10 = 40I3 + 40I2

Finally, if we do it with the loop made from the first and last lines, we get:

10 = 40I1 + 40I3 (10 because the 20V cell and the 10V cell are in the opposite direction, meaning we subtract 10 off from 20).

Using the second and third formula, we get I1=I2. The ammeter reads I2, so we replace I1 with I2, and we get I2 = 0.25A. The voltmeter reading is for the resistor that has a current I3. Substituting I2 into the third equation, we get I3 = 0. If there is no current flowing through the resistor, there is no voltage across it, meaning the Voltmeter reading has to be 0. A

19) If we look at the pressure of water acting on top of the rock:

we know u = 0 (starts from rest), s = 45 and a = 10 (air resistance is negligible and assumption through basically every ENGAA Q is that g = 10). From this, we can work out t using s = ut + 1/2at^2. From this, we get t = 3. We know 40kg of water hits the rock surface every second, so in 3 seconds 120 kg of water hits the rock surface, meaning 1,200 N of force acting on it, and as the rock surface has a surface area of 2, the pressure acting on top is 600 Pa.

Now if we look at the pressure acting at the bottom:

We know the pressure of a fluid acting on a submerged object = height x density of the fluid x acceleration of freefall. We are given the average depth of water on the rock surface is 0.05m, and that the density of water is 1000 kg m^-3, so using the formula, we get the pressure as 500 Pa.

The total pressure acting on the rock is 600 + 500 = 1100 Pa. E

20) We are given that X emits light when the P.D across it exceeds 2V. The 200 ohm resistor has a voltage of 2V and the e.m.f is 12V. By using simple ratios, we can figure out that the resistance of the thermistor is 1000 ohms if the P.D of X is 2V. If the resistance becomes greater than 1000 ohms, the P.D of X falls below 2V, meaning no light is emitted, meaning the resistance of the thermistor cannot exceed 1000 ohms. If we then substitute 1000 as R in the given equation, and the fact we know 1000 ohms is the maximum resistance, we get 1000 > Rob^(-uT) (I know it's mu and not u, but I don't know how to put the greek letters into here). using logs, we get logb(1000) > -uT + logb(R0), rearranging further to get uT > logb(R0) - logb(1000), and finally T > (logb(R0) - logb(1000))/u, or T > (1/u) (logb(R0) - logb(1000)) A

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#48

(Original post by

Look at the attached image

**nidza100**)Look at the attached image

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#50

The worked solutions were very helpful as well as the thoughtful explanations, thanks.

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#54

(Original post by

Anyone have worked solutions for 17?

**TravisBickle02**)Anyone have worked solutions for 17?

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#55

(Original post by

whose scared for tomorrow?

**2013zislam**)whose scared for tomorrow?

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#57

(Original post by

How did you get cos(a)=1/2

**TravisBickle02**)How did you get cos(a)=1/2

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#59

(Original post by

Does anybody know if there is a break between Section 1 and 2?

**granted1111**)Does anybody know if there is a break between Section 1 and 2?

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