Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s
Answer:
No, the pendulum's period of oscillation does not depend on initial angular displacement.
Explanation:
Given that,
For small angle, the pendulum's period of oscillation depend on initial angular displacement from equilibrium.
We know that,
The time period of pendulum is defined as
Where, l = length of pendulum
g = acceleration due to gravity
So, The time period of pendulum depends on the length of pendulum and acceleration due to gravity.
It does not depend on the initial angular displacement.
Hence, No, the pendulum's period of oscillation does not depend on initial angular displacement.
Answer:
The total normal force acting on the system is approximately 58.8 N
Explanation:
The masses arranged in the stack are;
3 kg, 2 kg, and 1 kg
The mass of the stack system, m = 3 kg + 2 kg + 1 kg = 6 kg
Weight = The force of gravity on an object = m·g
Where;
m = The mass of the object
g = The acceleration due to gravity ≈ 9.8 m/s²
∴ The weight of the stack system, W ≈ 6 kg × 9.8 m/s² ≈ 58.8 N
The direction of the weight force = Perpendicular to the surface (acting downwards)
From Newton's third law of motion, the normal force acts perpendicular to the plane and it is equal in magnitude to the force acting perpendicular to the plane
∴ The magnitude of the total normal force acting on the system = The magnitude of the weight of the system ≈ 58.8 N
The (magnitude of the) total normal force acting on the system ≈ 58.8 N
Answer : The atomic weight of the element is, 121.75 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
As we are given that,
Mass of isotope 1 = 120.9038 amu
Percentage abundance of isotope 1 = 57.25 %
Fractional abundance of isotope 1 = 0.5725
Mass of isotope 2 = 122.8831 amu
Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %
Fractional abundance of isotope 2 = 0.4275
Now put all the given values in above formula, we get:
![\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20element%7D%3D%5Csum%5B%28120.9038%5Ctimes%200.5725%29%2B%28122.8831%5Ctimes%200.4275%29%5D)
Therefore, the atomic weight of the element is, 121.75 amu
