Answer appears here.
c. 80 s
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
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 S
65^{2} = 0 + 2 x 4 x S
Which gives the distance, S = 528 m
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
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
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%
b. A scalar quantity is not completely described by its magnitude.
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
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
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
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)
Therfore, F = 1 - 3 + 2 = 0
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