The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul
2 answers:
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Answer:
the coin does not slide off
Explanation:
mass (m) = 5 g = 0.005 kg
distance (r) = 15 cm = 0.15 m
static coefficient of friction (μs) = 0.8
kinetic coefficient of friction (μk) = 0.5
speed (f) = 60 rpm
acceleration due to gravity (g) = 9.8 m/s^{2}
lets first find the angular speed of the table
ω = 2πf
ω = 2 x π x 60 x
ω = 6.3 s^{-1]
Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding
static force (Fs) = ma
static force (Fs) = μs x Fn = μs x m x g
Fs = 0.8 x 0.005 x 9.8 = 0.0392 N
Fs = ma
0.0392 = 0.005 x a
a = 7.84 m/s^{2}
= a x r
= 7.84 x 0.15
Vmax = 1.08 m/s
ωmax =
ωmax = = 7.2 s^{-1}
now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm
Fmax =
Fmax = = 68.7 rpm
since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.
Answer:
a) Stone v_{y} = - 7.25 ft / s , vₓ = 0.362 ft / s
b) tennis ball v_{y} = -3.16 ft / s , vₓ = 0.634 ft / s
c) golf ball v_{y} = - 1,536 ft / s, vx = 0.634 ft / s
2) golf ball
Explanation:
1) The average speed is defined with the displacement interval in the given time interval
v =( -x₀) / Δt
let's use this expression for each object
a) Stone
It tells us that it is released from y₀ = 10 ft and reaches the floor at
t = 0.788 s, but in the problem they tell us that the calculation is for t = 1.38 s
= (0-10) / 1.38
v_{y} = - 7.25 ft / s
in this interval a distance of x_{f} = 0.500 ft was moved away from the building (x₀ = 0 ft)
vₓ = (0.500- 0) / 1.38
vₓ = 0.362 ft / s
In my opinion it makes no sense to keep measuring the time after the stone has stopped.
b) tennis ball
It leaves the building at a height of y₀= 10ft and at the end of the period it is at a height of y_{f} = 5.63 ft, all this in a time of t = 0.788 + 0.591 = 1.38 s
the average vertical speed is
= (5.63 - 10) / 1.38
v_{y} = -3.16 ft / s
for horizontal velocity the ball leaves the building xo = 0 reaches the floor
x₁ = 0.500 foot and when bouncing it travels x₂ = 0.375 foot, therefore the distance traveled
x_{f} = x₁ + x₂
x_{f} = 0.500 + 0.375
x_{f} = 0.875 ft
we calculate
vₓ = (0.875 - 0) / 1.38
vₓ = 0.634 ft / s
c) The golf ball
the vertical displacement y₀ = 10 ft, and y_{f} = 7.88 ft
v_{y} = (7.88 - 10) / 1.38
v_{y} = - 1,536 ft / s
the horizontal displacement x₀ = 0 ft to the point xf = 0.875 ft
vₓ = (0.875 -0) / 1.38
vₓ = 0.634 ft / s
2) in this part we are asked for the instantaneous speed at the end of the time interval
a) the stone is stopped so its speed is zero
v_{y} = vₓ = 0
b) the tennis ball
It is at its maximum height so its vertical speed is zero
v_{y} = 0
horizontal speed does not change
vₓ = 0.634 ft / s
c) The golf ball
they do not indicate that it is still rising. Therefore its vertical speed is greater than zero
v_{y} > 0
horizontal speed is constant
vₓx = 0.634 ft / s
the total velocity of the object can be found with the Pythagorean theorem
v = √ (vₓ² + v_{y}²)
When reviewing the results, the golf ball is the one with the highest instantaneous speed at the end of the period