To show your engaged in the learning topic and to show your undivided attention
Cuts, sneeze, vomit, spit, cry
Answer:
velocity. height. weight. possition. place. energy. force.
Explanation: 50/50 % chance they are wrong and write.
Answer:
a) 35.75 ft/s
b) 45 ft
Explanation:
<u>Given </u>
Weight W = 100 lbf
mass(m) = 100*32.174/32.2=99.92 lb
decrease in kinetic energy ΔKE = -500 ft.lbf
increase in kinetic energy ΔPE= 1500 ft.lbf
initial velocity V_1 = 40 ft/s
initial height h_1 = 30 ft/s
The gravitational acceleration g = 32.2 ft/s2 Required
(a) Final velocity V_2 (a) Final elevation h_2
<u>Solution </u>
Change in kinetic energy is defined by
ΔKE = .5*m *( V_2 ^2-V_1^2)
Change in potential energy is defined by
ΔKE = W *( h_2 -h_1 )
Then,
-500=.5*99.92*1/32.174*(V_2 ^2-40^2)
V_2=35.75 ft/s
1500 = 100 x (h_2 — 30)
h_2= 45 ft
Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
