Answer:
v_f = 3 m/s
Explanation:
From work energy theorem;
W = K_f - K_i
Where;
K_f is final kinetic energy
K_i is initial kinetic energy
W is work done
K_f = ½mv_f²
K_i = ½mv_i²
Where v_f and v_i are final and initial velocities respectively
Thus;
W = ½mv_f² - ½mv_i²
We are given;
W = 150 J
m = 60 kg
v_i = 2 m/s
Thus;
150 = ½×60(v_f² - 2²)
150 = 30(v_f² - 4)
(v_f² - 4) = 150/30
(v_f² - 4) = 5
v_f² = 5 + 4
v_f² = 9
v_f = √9
v_f = 3 m/s
12 MPH
I DIDNT do the math my brother did hes in colledge so good lucks guys
Answer:
405 m
Explanation:
Given:
v₀ = 0 m/s
a = 2.5 m/s²
v = 45 m/s
Find: Δx
v² = v₀² + 2aΔx
(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx
Δx = 405 m
C) Seesaw is the best answer in my opinion
