Series,effective resistance =R₁+R₂+R₃...
parallel,effective resistance 1/R=1/R₁ +1/R₂ +1/R₃...
Here,effective resistance 1/R =1/7 +1/5+ 1/4+1/2
=1.092
R = 1/1.092 =0.915Ω
voltage V=9 V
current I=V/R
I=9 / 0.915
=9.83 A
Answer:
Approximately 
. (Assuming that 
, and that the tabletop is level.)
Explanation:
Weight of the book:
.
If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, 
.
As a side note, the 
and 
on this book are not equal- these two forces are equal in size but point in the opposite directions.
When the book is moving, the friction 
on it will be equal to
, the coefficient of kinetic friction, times
, the normal force that's acting on it.
That is:
.
Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 
applied force. The net force on the book shall be:
.
Apply Newton's Second Law to find the acceleration of this book:
.
Answer:
(1) A sound wave a mechanical wave because mechanical waves rely on particle interaction to transport their energy, they cannot travel through regions of space that are void of particles. Sound is a mechanical wave and cannot travel through a vacuum. These particle-to-particle, mechanical vibrations of sound conductance qualify sound waves as mechanical waves. Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave. The answer is(B) it travels in the medium.
(2) An ocean wave is an example of a mechanical transverse wave
The compression is the part of the compressional wave where the particles are crowded together. The rarefaction is the part of the compressional wave where the particles are spread apart. The answer is (C) Compression.
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,
Ep=m*g*h=0.75*9.8*1.5=11.025 J
Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,
E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J
Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek
Ek=(1/2)*m*v²=12.5 J.
Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,
Ep=m*g*h=0.7*9.8*7=48.02 J
The order of examples starting with the lowest energy:
1. book, 2. ball, 3. stone, 4. brick
