Answer:
Maximum acceleration will be
Explanation:
We have given mass of the object m = 2 kg
Spring constant k = 55.6 N/m
Amplitude is given as A = 0.045 m
We know that maximum acceleration in SHM is given by
Maximum acceleration
We know that
So maximum acceleration =
Answer:
The speed with which the man flies forward is 5.5 m/s
Explanation:
The mass of the man = 100 kg
The mass of the scooter = 10 kg
The speed with which the man was traveling on the scooter = 5 m/s
The speed of the scooter after it hits the rock = 0 m/s
Let v represent the speed with which the man flies forward
The formula for momentum, P, is P = Mass × Velocity
The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;
The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s
The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v
When the momentum is conserved, we have;
550 kg·m/s = 100 kg × v
∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.
The speed with which the man flies forward = v = 5.5 m/s
Answer:
57 %
Explanation:
input power = 16.4 kW = 16.4 x 10^3 W = 16400 W
Water pumped per second = 67 L/s
Mass of water pumped per second, m = Volume of water pumped epr second x density of water
m = 67 x 10^-3 x 1000 = 67 kg/s
height raised, h = 14 m
Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W
efficiency = output power / input power = 9380 / 16400 = 0.57
% efficiency = 57 %
thus, the efficiency of the pump is 57 %.
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