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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Answer:
ΔT = 302 °c
Explanation:
mass (m) = 4.6 g = 0.0046 kg
velocity (v) = 278 m/s
specific heat of lead (c) = 128 J/kg. °c
kinetic energy = 0.5 mx
kinetic energy = 0.5 x 0.0046 x
kinetic energy = 177.8 J
since all the kinetic energy is converted to thermal energy,
kinetic energy = thermal energy (E) = 177.8 J
thermal energy = m x c x ΔT
where ΔT is the temperature change
177.8 = 0.0046 x 128 x ΔT
ΔT = 177.8 / 0.59
ΔT = 302 °c
To solve this exercise it is necessary to apply the equations concerning Work, both by general definition and by conservation of energy.
In other words, the work done by an object due to gravity is the equivalent to that defined by the potential energy equations, that is
Where,
m=mass
g=gravitational acceleration
Change in height
On the other hand we have that the work done by tension is defined by the conservation of kinetic and potential energy, that is to say
Where,
Change in Kinetic Energy
Change in Potential Energy
PART A) As defined by the work done by gravity would be given by,
Therefore the work done by gravity is 25.725kJ
PART B) The work done by the tension applies the energy conservation equation, that is to say
Replacing with our values,
Therefore the work done by tension is 25.9kJ