Acceleration x time = velocity
Since you're given acceleration and time, just plug the values into the equation.
3
x 1.1 s = ?
Solve that equation, and remember your velocity should be in m/s.
Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan
Answer:
8.8 m and 52.5 m
Explanation:
The vertical component and horizontal component of water velocity leaving the hose are
Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level
t = 3.49 or t = 0.58
We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down
t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building
Answer:
Explanation:
This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.
