Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F
Answer:
stress = 16.9 MPa
Explanation:
The stress in the cable can be calculated as:
Where F is the force and A is the area. So, the area can be calculated as:
Where r is the radius. Since the radius is half the diameter, the radius is 4.0 mm and the area will be equal to:
Then, replacing the force F by 850 N, and A by 50.24 mm², we get that the stress is equal to:
Therefore, the answer is 16.9 MPa
(a) At its maximum height, the ball's vertical velocity is 0. Recall that
Then at the maximum height 
, we have
(b) The time the ball spends in the air is twice the time it takes for the ball to reach its maximum height. The ball's vertical velocity is
and at its maximum height, 
so that
which would mean the ball spends a total of about 5.6 seconds in the air.
(c) The ball's horizontal position in the air is given by
so that after 5.6 seconds, it will have traversed a displacement of
When firing straight up:
v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained.
Then,
0 = u^2 - 2gh
u = Sqrt (2gh) ---- (1)
When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45
Maximum height, H = U^2*(Sin Ф)^2/2g
substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
H = [h*(Sin 45)^4] = h/4
Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.
