The x-coordinates of two objects moving along the x-axis are given as a function of time t. x1 = (4 m/s)t and x2 = −(159 m) + (2
4 m/s)t − (1 m/s2)t2 . Calculate the magnitude of the distance of closest approach of the two objects. x1 and x2 never have the same value.
1 answer:
The x-coordinate of the first object is
x₁(t) = (4 m/s)t
The x-coordinate of the second object is
x₂(t) = -(159 m) + (24 m/s)t - (1 m/s²)t²
The distance between the two objects is
x(t) = x₂ - x₁
= - 159 + 24t - t² - 4t
= -t² + 20t - 159
Write this equation in the standard form for a parabola.
x = -[t² - 20t] - 159
= -[(t - 10)² - 100] - 159
= -(t-10)² - 59
This parabola has a vertex at (-10, -59), and it is downward.
Because the maximum value of x is negative, the two objects never touch
The closest distance between the objects is 59 m.
The two graphs confirm that the analysis is correct.
Answer: The closest approach is 59 m.
x₁(t) = (4 m/s)t
The x-coordinate of the second object is
x₂(t) = -(159 m) + (24 m/s)t - (1 m/s²)t²
The distance between the two objects is
x(t) = x₂ - x₁
= - 159 + 24t - t² - 4t
= -t² + 20t - 159
Write this equation in the standard form for a parabola.
x = -[t² - 20t] - 159
= -[(t - 10)² - 100] - 159
= -(t-10)² - 59
This parabola has a vertex at (-10, -59), and it is downward.
Because the maximum value of x is negative, the two objects never touch
The closest distance between the objects is 59 m.
The two graphs confirm that the analysis is correct.
Answer: The closest approach is 59 m.
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The question is incomplete. Here is the complete question.
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
,
where is a reference intensity. for sound waves,
is taken to be
. Note that log refers to the logarithm to the base 10.
Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. ? Express the sound intensity numerically to the nearest integer.
Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. ? Express the sound intensity numerically to the nearest integer.
Part C: Calculate the change in decibels ( and
) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.
Answer and Explanation: Using the formula for sound intensity level:
A)
β = 10
<u>The sound Intensity level with intensity 10x is </u><u>10dB</u>.
B)
β = 20
<u>With intensity 100x, level is </u><u>20dB</u>.
C) To calculate the change, take the f to be the factor of increase:
For :
β = 3
For :
β = 6
For :
β = 9
Change is
,
= 3,6,9 dB
If you look below, we are using the centripetal acceleration formula. You just need to plug your numbers in and it should work!