D protons
k12 student here
The constant in Newton's law of gravitation relating gravity to the masses and separation of particles, equal to 6.67 × 10-11N m2 kg-2.
Answer:
3 significant figures are in 1.02m
Explanation:
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:
I. 0 m/s
II. 20 m/s
III. Part BC
Explanation:
I. Determination of the initial velocity.
From the diagram given above,
The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s
II. Determination of the maximum velocity attained.
From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.
III. Determination of the part of the graph that represents zero acceleration.
It important that we know the meaning of zero acceleration.
Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.
From the graph given above, the car has a constant velocity between B and C.
Therefore, part BC illustrates zero acceleration.