A, B, and C are all incorrect because they are talking about springs. I'm assuming you're talking about a 2 dimensional, side to side pendulum. In that case, the answer is D because as the pendulum goes up from one side, gravity accelerates it downward until it swings to the other side. If you're not talking about this type of pendulum, leave a comment and ill come back.
I think it’s by using a magnet because metals are magnetic
1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:
2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:
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
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