Answer:
Explanation:
In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.
The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.
Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.
To find the weight of the book we simply multiply the mass of the book by gravity.
W = m*g
W = 1.3[kg] * 9.81[m/s^2]
W = 12.75 [N]
Answer:
To identify the momentum of object 1, you must multiply mass (m) and velocity(v) to find momentum.
Object 1 has momentum of 8 kg. m/s before collision.
Object 1 has momentum of 0 kg. m/s before collision.
The combined mass after the collision had a total momentum of 8 kg. m/s.
Explanation:
Momentum of the object is given by,
Momentum = mass × velocity
For object 1:
Momentum = mass × velocity
Momentum = 2 × 4
Momentum = 8 kg. m/s
For object 2:
Momentum = mass × velocity
Momentum = 6 × 0
Momentum = 0 kg. m/s
For object 1 + object 2:
Momentum = mass × velocity
Momentum = 8 × 1
Momentum = 8 kg. m/s
To identify the momentum of object 1, you must multiply mass (m) and velocity(v) to find momentum.
Object 1 has momentum of 8 kg. m/s before collision.
Object 1 has momentum of 0 kg. m/s before collision.
The combined mass after the collision had a total momentum of 8 kg. m/s.
Answer:
a) I = 0.0198 kg m²
, b) I = 21.85 kg m²
Explanation:
For this exercise we will use the definition of moment of inertia
I = ∫ r² dm
For body with high symmetry they are tabulated
sphere I = 2/5 m r²
bar with respect to center of mass I = 1/12 m L²
let's calculate the mass of each body
bar
ρ = m / V
m = ρ V
m = ρ l w h
where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm
m = 32820 1 0.008 0.04
m = 10.5 kg
Sphere
M = ρ V
V = 4/3 pi r³
M = rgo 4/3 π r³
give us the density 37800 kg / m³ and the radius of 5 cm
M = 37800 4/3 π 0.05³
M = 19.8 kg
a) asks us for the moment of inertia of the sphere with respect to its center of mass
I = 2/5 M r²
I = 2/5 19.8 0.05²
I = 0.0198 kg m²
b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes
I = I_cm + M d2
where d is the distance from the body to the point of interest
I_cm = 0.0198 kg m²
the distance to the pivot point is
l = length of the bar + radius of the sphere
l = 1 + 0.05 = 1.005 m
I = 0.0198 + 19.8 1.05²
I = 21.85 kg m²
Answer:
Momentum, p = 34.937 kg-m/s
Explanation:
It is given that,
Force acting on the bowling ball, F = 48.2 N
Velocity of bowling ball, v = 7.13 m/s
We have to find the momentum of the ball. Momentum is given by :
p = mv........(1)
Firstly, calculating the mass of bowling ball using second law of motion. The force acting on the ball is gravitational force and it is given by :
F = m g (a = g)
m = 4.9 kg
Now putting the value of m in equation (1) as :
p = 34.937 kg-m/s
Hence, this is the required solution.
