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
By Boyle's law:
P₁V₁ = P₂V₂
70*8 = P<span>₂*4
</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
</span>
P<span>₂ = 140 kiloPascals.</span>
Answer:
<em>W=700 Joule</em>
Explanation:
<u>Physics Work
</u>
Is the dot product of the force vector by the displacement vector
When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version
The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is
