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:
(a) Ff = 0.128 N
(b μk = 0.1135
Explanation:
kinematic analysis
Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:
vf=v₀+a*t Formula (1)
d= v₀t+ (1/2)*a*t² Formula (2)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s
Calculation of the acceleration of the hockey puck
We apply the Formula (1)
vf=v₀+a*t v₀=5.8 m/s , vf=0
0=5.8+a*t
-5.8 = a*t
a= -5.8/t Equation (1)
We replace a= -5.8/t in the Formula (2)
d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s
15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²
15.1= 5.8*t-2.9*t
15.1= 2.9*t
t = 15.1 / 2.9
t= 5.2 s
We replace t= 5.2 s in the equation (1)
a= -5.8/5.2
a= -1.115 m/s²
(a) Calculation of the frictional force (Ff)
We apply Newton's second law
∑F = m*a Formula (3)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Look at the free body diagram of the hockey puck in the attached graphic
∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²
-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation
Ff = 0.128 N
(b) Calculation of the coefficient of friction (μk)
N: Normal Force (N)
W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight
g: acceleration due to gravity =9.8 m/s²
∑Fy = 0
N-W=0
N = W
N = 1.127 N
μk = Ff/N
μk = 0.128/1.127
μk = 0.1135
Answer:
122.84 J
Explanation:
Since plate is square, area, A is given by
The distance between plates, d, is given in the question as 2mm=0.002m
Charge on plate, Q, as given in the question is 240
Assuming mica dielectric constant, k of 7
Capacitance, C is given by
C=
Stored energy, E is given by
E=
Therefore, the stored energy is 122.84 J