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:
<h2>
2.8×10^-3</h2>
<u>0.0028125N</u><u> </u>
Answer:
(1) V = 0.2 J (2) 0.05J
Explanation:
Solution
Given that:
K = 160 N/m
x = 0.05 m
Now,
(1) we solve for the initial potential energy stored
Thus,
V = 1/2 kx² = 0.5 * 160 * (0.05)²
Therefore V = 0.2 J
(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.
By using the energy conversion, we have the following
ΔV = mgh
=(0.1/9.8) * 9.8 * 1.5 = 0.15J
The internal energy = 0.2 -0.15
=0.05J
Answer:
a)
b)
Explanation:
The gravitational force on the satellite is calculated with Newton's Gravitation Law:
Where 
is Earth's mass, 
is the satellite mass, 
is the distance between their centers, where 
is the height of the satellite (from Earth's surface) and 
is Earth's radius, and 
is the gravitational constant.
a) With these values we then have:
b) And the fraction this force is of the satellite’s weight <em>W=mg</em> is:
