Answer:
20m/second
Explanation:
The reason the answer is 20m/second is because to find the speed of the ball in this question you have to divide the distance over the time giving you the result of 20m/second
Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU
Answer:
Against but it really depends on the situation
Answer:
a) 6076 m
b) 43.33 m/s
c) 68 m/s
Explanation:
(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius
s = 2r = 2* 3.38km = 6.76 km or 6760 m
(b) The average velocity would be displacement over unit of time
v = s/t = 6760 / 156 = 43.33 m/s
(c) The length of the chord it's swept in 156s is half of the circle perimeter
c = πr = π3.38 = 10.62 km or 10620 m
The airplane average speed is its chord length over a unit of time
c / t = c / 156 = 68 m/s
