Analog information is the correct answer.
Answer:
(i) -556 rad/s²
(ii) 17900 revolutions
(iii) 11250 meters
(iv) -55.6 m/s²
(v) 18 seconds
Explanation:
(i) Angular acceleration is change in angular velocity over time.
α = (ω − ω₀) / t
α = (10000 − 15000) / 9
α ≈ -556 rad/s²
(ii) Constant acceleration equation:
θ = θ₀ + ω₀ t + ½ αt²
θ = 0 + (15000) (9) + ½ (-556) (9)²
θ = 112500 radians
θ ≈ 17900 revolutions
(iii) Linear displacement equals radius times angular displacement:
s = rθ
s = (0.100 m) (112500 radians)
s = 11250 meters
(iv) Linear acceleration equals radius times angular acceleration:
a = rα
a = (0.100 m) (-556 rad/s²)
a = -55.6 m/s²
(v) Angular acceleration is change in angular velocity over time.
α = (ω − ω₀) / t
-556 = (0 − 15000) / t
t = 27
t − 9 = 18 seconds
Change in speed = (acceleration) x (time)
4 minutes = 240 seconds
Change in speed = (40 m/s²) x (240 seconds)
Change in speed = <em>9,600 m/s</em>
What you're actually describing here is a car pulling 4 G's for 4 minutes, and ending up going 21,475 miles per hour.
The driver would definitely NOT get a speeding ticket, because nobody could catch him.
Also, his car would heat up and shoot flames from atmospheric friction.
(He could avoid this with some fancy steering, leave the atmosphere, and end up in low-Earth-orbit.)
Actually, I hope there's nobody in the car. His experience wouldn't be pretty.
Answer:
1.23 m/s²
Explanation:
Given:
v₀ = 0 m/s
v = 11.1 m/s
t = 9 s
Find: a
Equation:
v = at + v₀
Plug in:
11.1 m/s = a (9 s) + 0 m/s
a = 1.23 m/s²
The runner's acceleration is 1.23 m/s².
Answer:
fr = 65.46 N
, a = 8.74 m / s² and vf = 19.25 m / s
Explanation:
We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight
sin 21.2 = Wx / W
cos21.2 = Wy / W
Wx = W sin21.2
Wy = W cos 21.2
We form Newton's equations
X axis
Wx -fr = m a
Y Axis
N- Wy = 0
N = Wy
fr = μ N
fr = μ (W cos 21.2)
fr = 0.113 63.4 9.8 cos 21.2
fr = 65.46 N
We replace and calculate the acceleration
W sin 21.2 - μ W cos 21.2 = m a
a = g (sin21.2 - μ cos 21.2)
a = 9.8 (without 21.2 - 0.113 cos 21.2)
a = 8.74 m / s²
This acceleration is along the slope of the slide, so we can calculate the distance
d = 21.2 m
vf² = v₀² + 2 a d
vf² = 0 + 2 a d
vf = √(2 8.74 21.2)
vf = √ (370,576)
vf = 19.25 m / s
