Answer:
The time of oil tanker will be 80 sec.
Explanation:
Given that,
He walks at 3 km/hour.
If the container ship is 100 m long, and travelling at 12 km/hour.
The distance will be 200 m.
We need to calculate the relative velocity
Using formula of relative velocity
Put the value into the formula
The distance is 200 m.
We need to calculate the time
Using formula of time
Put the value into the formula
Hence, The time of oil tanker will be 80 sec.
I will definitely feel bad about what happen, try to help if i can by doing fundraising's, going to the family of the included people and donate things for their mental stage. <span />
Answer:
It reaches its peek at 1, then begins to decrease until it reaches zero, but then rises back up to one before decreasing again
Known Variables:
Δy = 1.5 m
v0 = 30m/s at 60° above the horizontal.
t = 4.0 s
a. vfy = v0y + at
0 = 30sin(60) + (-9.8)t
-25.98 = -9.8t
t = 2.65.
4-2.65 = 1.35s.
Δy = v0yt + at^2/2
Δy = 25.98(2.65) + (-9.8)(2.65)^2/2
Δy = 34.44 m
1.5 + 34.44 = 35.94 m.
Δy = v0t + at^2/2
Δy = (-9.8)(1.35)^2/2
Δy = -8.93.
35.94 - 8.93 = 27.01 m.
b. vfy = v0y + at
0 = 30sin(60) + (-9.8)t
-25.98 = -9.8t
t = 2.65.
Δy = v0yt + at^2/2
Δy = 25.98(2.65) + (-9.8)(2.65)^2/2
Δy = 34.44 m
1.5 + 34.44 = 35.94 m.
vfy = v0y + at
vfy = (-9.8)(1.35)
vfy = -13.23 m/s.
I might have gone wrong somewhere so please check my answer! Cheers!
