On a large scale, no. Sound travels through space by the compression and relaxation of molecules. When you speak to someone on Earth, each inflection of voice compresses local molecules, and relaxes the molecules around the compressed ones. In space, however, the large lack of molecules means that the compression waves can’t be transmitted from the sound to the listener.
Answer:
Explanation:
Given
initially steam is at 
and converted to 
ice
Let m be the mass of steam
latent heat of fusion and vaporization for water is
Heat required to convert steam in to water at
Heat required to lower water temperature to 
Heat required to convert water to ice at is
So this energy is equal to kinetic energy of bullet of mass m moving with velocity v
most of the earth is covered with oceans
Answer:
L = 8694 Kg.m²/s
Explanation:
r = 270 ĵ m
v = 14 î m/s
m = 2.3 kg
θ = 90º
L = ?
We can apply the equation
L = m*v*r*Sin θ
L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
