Explanation:
work input is the work done by the machine in lifting the drum = Force × distance
100N × 10m
= 1000 J
I would say A or B if there’s a text can u post that as well but I’m just going by what it sound like
You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same
Answer:
Explanation:
v = u + at
v₃ = 100 +30.0(3.00) = 190 m/s
s = vt + ½at²
y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m
x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m
a) v² = u² + uas
s = (v² - u²) / 2a
ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m
b) t₁ = 3.00 s
t₂ = (190sin53) / 9.80 = 15.5 s
t₃ = √(2(1522) / 9.80) = 17.6 s
t = 3.00 + 15.5 + 17.6 = 36.1 s
c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m