Answer:
Explanation:
We shall take the help of vector form of displacement . Taking east as i and north as j
4.0m N = 4 j
7.5 m E = 7.5 i
6.8 m S = - 6.8 j
3.7 m E, = 3.7 i
3.6 m S = - 3.6 j
5.3 m W = - 5.3 i
3.7 m N, = 3.7 j
5.6 m W = - 5.6 i
4.4 m S = - 4.4 j
4.9 m W = - 4.9 i
Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i
= -4.6 i -7.1 j
magnitude of displacement = ![\sqrt{(4.6^2+7.1^2)}](https://tex.z-dn.net/?f=%5Csqrt%7B%284.6%5E2%2B7.1%5E2%29%7D)
= 8.46 m
Direction
Tanθ = 7.1/ 4.6
θ = 57⁰ south of west .
distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9
= 49.5 m
Answer: h = 3.34 m
Explanation:
If the hat is thrown straight up, then at its highest point it has no motion and no kinetic energy. All energy is potential energy
PE = mgh
h = PE/mg = 4.92 / (0.150(9.81)) = 3.34352... ≈3.34 m
Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J