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
Answer:
C 80 m
Explanation:
Given:
v₀ = 30 m/s
a = -10 m/s²
t = 8 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²
Δy = -80 m
The ball lands 80 m below where it started. So the height of the cliff is 80 m.
Answer:
Predicting weather patterns is the answer :-)
Explanation:
1 Bc I just did it and got it right
Particle with more energy move SLOWER than particles with less energy.
