It's on your exam boards specification or just google it. the foundation paper will have the same equations as higher, but higher just has more.
A
The horizontal force cancels out. The two 4Ns go in opposite directions. So they don't affect the outcome.
The Vertical force is 6N up - 2 N down = 4 N Up
Answer 4 N up
B
The horizontal and vertical forces cancel out. Each gives 3N - 3N =0
The net force is 0
C
You only have horizontal forces on this one
5N - 3N = 2N
The answer is 2N to the right.
Answer:
The. Machine must detect a shift of
1 Hz
Explanation:
Frequency shift is given as
={ ( Vsound +V/ V sound -V) -1}f emitted
So by substitution we have
= { 1540+4E-4/1540-4E)-1) 2*10^6
= 1Hz
Your answer would be : Leyden jar.
Answer:
0.1 rev/s
Explanation:
M = mass of the merry go round = 200 kg
R = radius of merry go round = 6 m
= Moment of inertia of merry go round = (0.5) MR² = (0.5) (200) (6)² = 3600 kgm²
m = mass of the man = 100 kg
= Moment of inertia of merry go round when man sits on it at the edge = (0.5) MR² + mR² = (0.5) (200) (6)² + (100) (6)² = 7200 kgm²
= initial Angular speed of merry-go-round before man sit = 0.2 rev/s
= Angular speed of merry-go-round after man sit = ?
Using conservation of angular momentum
=
(3600) (0.2) = (7200)
= 0.1 rev/s
