It 43 degrees
You take 94+43= 137
180- 137= 43
Answer: V = 5.22m/s
Step-by-step explanation:
Given that the mass of the particles is proportional to the fifth power of the speed. That is
M = k V^5
Where M = mass and V = speed
K = constant of proportionality
A certain river normally flows at a speed of 3 miles per hour
V = 3 mph
M = unknown
M = k × 3^5
M = 243K
K = M/243 ........(1)
What must its speed be in order to transport particles that are 16 times as massive as usual
M = 16M
Using same formula
I.e M = KV^5
16M = M/243 × V^5
M will cancel out
16 = V^5/243
V^5 = 3888
V = 5.22m/s
ME=(mv^2)/2+mgh in this case:
ME=(5*4^2)/2+5*9.8*3
ME=40+147
ME=187 J
