Answer:
The correct option is;
C. 1,715 m
Explanation:
We are given the information from the group of teen at the City edge
Time of arrival of explosion sound = 5 s after sighting
Time of sighting explosion = 5 s before hearing the boom
Speed of sound in air ≈ 343 m/s
Speed of light = 299,792 km/s
Therefore, distance covered by sound in 5 seconds is given by the following equation;
Hence Distance = 343 m/s × 5 s = 1715 m
To check, we compare the time it would take for the light to cover 1715 m
That is 
which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.
Therefore, the distance of the students from the factory is approximately 1,715 m
Contour lines are lines that signifies the elevation on a mountain or hill
Answer:
I = 8.75 kg m
Explanation:
This is a rotational movement exercise, let's start with kinetic energy
K = ½ I w²
They tell us that K = 330 J, let's find the angular velocity with kinematics
w² = w₀² + 2 α θ
as part of rest w₀ = 0
w = √ 2α θ
let's reduce the revolutions to the SI system
θ = 30.0 rev (2π rad / 1 rev) = 60π rad
let's calculate the angular velocity
w = √(2 0.200 60π)
w = 8.683 rad / s
we clear from the first equation
I = 2K / w²
let's calculate
I = 2 330 / 8,683²
I = 8.75 kg m
Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
