Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer: A liquid to gas
Explanation: I just got it wrong :(
Answer:
67.9 kg*m/s
Explanation:
Pi = 38 kgm/s
F = 88.3N and ∆t = 0.338s
Final momentum Pf = Pi + F∆t = 38 + (88.3)(0.338) = 38 + 29.8454
=) Pf = 67.8454 kgm/s = 67.85kg*m/s
Your answer is 67.9kg*m/s with three significant figures
hope this helps your troubles!
A large elliptical galaxy. Hope this helped