Only velocity uses direction of travel in its calculations.
Answer:
a) 0.462 m/s^2
b) 31.5 rad/s
c) 381 rad
d) 135m
Explanation:
the linear acceleration is given by:

the angular speed is given by:

to calculate how many radians have the wheel turned we need the apply the following formula:

the distance is given by:


A system that repeats to and from its mean or rest point. that executes harmonic motion. a few examples I've heard of are since the springtime a mass-spring system,a swing, simple pendulum, one more example is a steel ball rolling in a curved is this what you need or do you need three more sentences dish. to get S.H.M a body just displaced away from the resting position and of course then is released. the human body oscillates due to the reinforce that pulls it back do you need anything else answered on this and I'll answer it
The answer is 165.3 cm³.
P1 * V1 / T1 = P2 * V2 / T2
The initial sample:
P1 = 84.6 kPa
V1 = 215 cm³
T1 = 23.5°C = 23.5 + 273 K = 296.5 K
At STP:
P2 = 101.3 kPa
V2 = ?
T2 = 273 K
Therefore:
84.6 * 215 / 296.5 = 101.3 * V2 / 273
61.34 = 101.3 * V2 / 273
V2 = 61.34 * 273 / 101.3
V2 = 165.3 cm³
Answer:
The railroad tracks are 13 m above the windshield (12 m without intermediate rounding).
Explanation:
First, let´s calculate the time it took the driver to travel the 27 m to the point of impact.
The equation for the position of the car is:
x = v · t
Where
x = position at time t
v = velocity
t = time
x = v · t
27 m = 17 m/s · t
27 m / 17 m/s = t
t = 1.6 s
Now let´s calculate the distance traveled by the bolt in that time. Let´s place the origin of the frame of reference at the height of the windshield:
The position of the bolt will be:
y = y0 + 1/2 · g · t²
Where
y = height of the bolt at time t
y0 = initial height of the bolt
g = acceleration due to gravity
t = time
Since the origin of the frame of reference is located at the windshield, at time 1.6 s the height of the bolt will be 0 m (impact on the windshield). Then, we can calculate the initial height of the bolt which is the height of the railroad tracks above the windshield:
y = y0 + 1/2 · g · t²
0 = y0 -1/2 · 9.8 m/s² · (1.6 s)²
y0 = 13 m