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
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The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.
If it accelerates at 20 m/s² during the hit, then
Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .
Answer: 6.9x 107 in standard form is 69,000,000
Answer: The answer is the masses of the objects and the distance between them
Explanation: Gravity is affected by mass and distance between two objects becuase if and object is too far the force of gravity will not be strong. The larger the object, the stronger the force of gravity will be.
