Is the following nuclear reaction balanced? Need answered ASAP!
2 answers:
You might be interested in
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: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
