Organ pipe a, with both ends open, has a fundamental frequency of 340 hz. the third harmonic of organ pipe b, with one end open,
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Answer:
<h2>a)Explanation:
If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;
S is the altitude of the ballest
u is the initial velocity
a is the acceleration of the body
t is the time taken to strike the ground
Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s
Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g
Substituting this values into the equation of the motion;
a) An expression for the altitude of the ballast after t seconds is therefore
b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);
This means that the ballast strikes the ground after 6secs
c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.
v = 0 + 32(6)
v = 192ft