Two cannon balls weighing 13 kg and 19 kg are chained together and fired horizontally with a velocity of 165 m/s from the top of
1 answer:
Answer:
7 m to the left of the line of fire.
213.55 m from the foot of the wall.
7.25 m above the ground.
Explanation:
Let's assume the height is the y-direction, the line of fire is in x-direction. Since there is no external force acting in z-direction, z-coordinate of the center of mass of he ball should be zero. In order to make z-coordinate zero other ball should fall symmetrically with respect to z-axis. So, z-coordinate of 19 kg ball = -7 m.
Hence, 7 m to the left of the line of fire.
The balls do not have any external force in x-direction. Thus, in x-direction, the center of mass should move with constant velocity. x-coordinate of center of mass at t = 1.5 s is:
Hence, 213.55 m from the foot of the wall.
Height fallen by the center of mass at t = 1.5 s.
Hence, y-coordinate of the center of mass is:
Hence, 7.25 m above the ground.
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Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
153.2 J
Explanation:
Let's first list our given parameters;
mass (m) of the block = 10 kg
which slides down ( i.e displacement) = 2 m
kinetic coefficient of friction (μk) = 0.2
In the diagram shown below; if we take an integral look at the component of force in the direction of the displacement; we have
Fcos 40°
100 (cos 40°)
76.60 N
Workdone by the friction force can now be determined as:
W = × displacement
W = 76.60 × 2
W = 153.2 J
∴ the work done by the friction force = 153.2 J
Answer:
Your question is lacking some information attached is the missing part and the solution
A) AB = AD = BD = 0, BC = LC
AC =
B) AB = AD = BC = BD = 0
AC =
Explanation:
A) Forces in all members due to the load L in position A
assuming that BD goes slack from an inspection of Joint B
AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C
B) steps to arrive to the answer is attached below
AB = AD = BC = BD = 0
AC =
Answer:
a. 4.279 MPa
b. 3.198 MPa to 4.279 MPa
c. 0.939 MPa
d. Below 3.198 MPa
Explanation:
From the given parameters
M = 1.75 MPa
M at 1.6 MPa gives A/A* = 1.2502
M at 1.8 MPa gives A/A* = 1.4390
Therefore, by interpolation, we have M = 1.75 MPa gives A
However, we shall use M = 1.75 MPa and A
Similarly,
P/P₀ = 0.1878
a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have
A/A* = 1.387. by interpolation M
Therefore P = P₀ × P
Which shows that the nozzle is choked for back pressures lower than 4.279 MPa
b) Where there is a normal shock at the exit of the nozzle, we have;
M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa
Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406
Where the normal shock occurs at the nozzle exit, we have
Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as = 4.279 MPa
Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa
c) At M = 1.75 MPa and P
d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane