40 km/h because the average of 50 and 30 is 40
Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:
Option A
Explanation:
Mechanical waves requires some medium to travel through. They travel faster in the dense medium as compared to a free medium.
The speed of a mechanical wave is fastest in the solid medium and the slowest in the gaseous medium. Hence, as the wave traverses from gaseous medium to the solid medium, its speed increases.
Thus, option A is correct
Answer:
Volume, V = 13564.8 cubic feet
Explanation:
It is given that,
Radius of the cylindrical tank, r = 12 feet
Height of the tank, h = 30 feet
We need to find the water that can be held by a cylindrical tank i.e. we need to find the volume of the tank. It is given by :
V = 13564.8 cubic feet
So, the water held by the tank is 13564.8 cubic feet. Hence, this is the required solution.
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
