### Answer 1: c: 80s

**Solution:**

Apply Equations of motion to calculate the maximum acceleration:

V = U + A x T

70 = 55 + 120 x a

Maximum acceleration, a = 0.125 m/s^{2}

Maximum deceleration = 4 x 0.125 = 0.5 m/s^{2}

Now apply equations of motion to train starting at 70 m/s and coming to rest at this rate of deceleration:

V = U + A x T

0 = 40 + 0.5 x T

Which gives the time taken, T = 80 s

### Answer 2: b. 528 m

**Solution:**

Apply equations of motion to calculate the distance the plane travels, while accelerating from rest to 65 m/s

V^{2} = U^{2} + 2 x A x

65^{2} = 0 + 2 x 4 x S

Which gives the distance, S = 528 m

### Answer 3: b. 40 kW

**Solution:**

Apply the pump power equation to solve this problem

P = Q x Y x H / (E / 100)

The flow rate, Q must be converted to standard units:

Q = 2,500 m^{3}/hour = 2,500 / (60 x 60) m^{3}/s = 0.694 m^{3}/s

The specific weight of fluid (water), Y = density x g = 1000 kg/m^{3} x 9.81 m/s^{2} = 9,800 N/m^{3}

Therefore

P = 0.694 x 5.9 x 9,800 / (100/100)

P = 40,000 W = 40 kW

### Answer 4: a. The fatigue strength.

### Answer 5: d. 900 m/s

**Solution:**

If we calculate the time it takes the axle to rotate 20, then we can calculate the time it takes the bullet to travel the distance between the two discs.

The discs rotate at 3,00 rpm which is equal to 3,000 / 60 = 50 revolutions per second

20º is 20º / 360º = 0.0556 of a complete revolution

Therfore the disc rotates 20º in 0.0556 / 50 = 0.0011 s

Which means the bullet travels 1 m in the same amount of time

The linear velocity of the bullet is therefore 1 / 0.0011 m/s = 900 m/s

### Answer 6: c. 7.5%

**Solution:**

The $900 per year represents the interest alone and doesn’t include payment of the principal balance

Therefore the question is simply a matter of calculating the percentage of $12,000 that is equal to $900

900 / 12,000 = 0.075 = 7.5%

### Answer 7: b. A scalar quantity is not completely described by its magnitude.

### Answer 8: a. 985 kg

**Solution:**

Use the formula for centripetal force:

F = M*R*(W^2)

965 = M x 200 x W^{2}

Calculate the angular velocity, W from the angular velocity equation W = V/R

W = 14 / 200 = 0.07

Therefore 965 = M x 200 x 0.07^{2}

So the mass, M = 985 kg

### Answer 9: d. 11.5º

**Solution:**

Use the formula for banking of curves:

A= ATAN((V^2)/(G*R))

A = ATAN ((10^{2)} / (9.8 x 50))

A = 11.5

So the desired banking angle is 11.5º

### Answer 10: c. 90 m

**Solution:**

First, calculate the X and Y components of the initial velocity

V_{x} = V x cos53º V_{y} = V x sin 53º

V_{x} = 25 x 0.6 V_{y} = 25 x 0.8

V_{x} = 15 m/s V_{y} = 20 m/s

Vertical motion:

Calculate the time taken for the projectile to reach its maximum height:

V_{final} = V_{initial} – g x T

0 = 20 – 9.8 x T

T = 2 s

Calculate the maximum height:

Y = V_{y} x T x SIN(A) – g x T^{2} / 2

H = 20– 9.8 x 2^{2 }/ 2

H = 20 m

H_{max} – 20 + 60 = 80 m

Free fall from the maximum height:

H = ½ x g x T^{2}

80 = ½ x 9.8 x T^{2}

T = 4 s

Therefore the total time = 4 + 2 = 6 s

Horizontal motion:

X = V x T = 15 x 6

X = 90 m

### Answer 11: a. 0

**Solution:**

The Gibbs Phase Rule states that:

F = C – P + 2

Where:

F = Degrees of Freedom

C = Number of Components

P = Number of phases (solid, liquid or gas)

At the triple point of water, C = 1 (water) and P = 3 (solid, liquid, gas)

Therefore, F = 1 – 3 + 2 = 0