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
Regions in the milky way where density waves have caused gas clouds to crash into each other are called clumps.Clumps are molecular clouds (interstellar clouds) with higher density,where lots of dust and gs cores resides. These clouds are the beginning of stars.
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell.
Therefore, we can say KE = PE
We know that PE = m·g·h
Which means KE = m·g·h
We can solve for h:
h = KE / m·g
= 20 / (0.15 · 9.8)
= 13.6m
The correct answer is: the ball has fallen from a height of 13.6m.
Answer:
protons
atomic number
Note: The same atomic number can be associated with several different values of atomic mass, but an element can have only 1 atomic number,
The centripetal acceleration is given by
where v is the tangential speed and r the radius of the circular orbit.
For the car in this problem,
and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
