Navigation is the art of measuring distances in order to be able to get from one place to another
Answer:
During a typical school day all forms of eneergy is being utilised and also transfer of energy takes place from one form to another.
Explanation:
Chemical energy- A bunsen burner burning a beaker filled with water.
Heat energy- The water in the beaker absorbing the heat from the burner.
Electrical energy- Running Fans and lights in a classroom by switches.
Solar energy- Solar energy harnessed by solar panels to run the fans and lights by converting it into electrical energy.
Potential energy- A ball being held by a student at a certain height possesses energy due to gravity.
Kinetic energy- The same ball being left by the boy from a certain height produces kinetic energy
<span>Force = Work done / distance = 4Nm / 2m = 2N</span>
Answer:
2.68 hours
Explanation:
A.) Suppose the wind blows out from the west (with the air moving east). The pilot should then head her plane to northwest direction to move directly north.
B.) Given that plane flies at a speed of 102 km/h in still air. And the wind blows out from the west (with the air moving east) at a speed of 46 km/h.
The plan resultant speed can be calculated by using pythagorean theorem.
Resultant Speed = Sqrt( 102^2 + 46^2 )
Resultant Speed = Sqrt( 12520)
Resultant speed = 111.89 km/h
From the definition of speed,
Speed = distance/time
Where distance = 300 km
Substitute the resultant speed and the distance into the formula.
111.89 = 300/time
Time = 300/111.89
Time = 2.68 hours
Therefore, it take her 2.68 hours to reach a point 300 km directly north of her srarting point
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
