Answer:
a) -4 N
b) +4 N
Explanation:
Draw a free body diagram for each block.
For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.
For the small block, there is 1 force, F pushing to the right.
There are also weight and normal forces in the vertical direction, but we can ignore those.
Sum of forces on the large block in the x direction:
∑F = ma
12 − F = 4a
Sum of forces on the small block in the x direction:
∑F = ma
F = 2a
2F = 4a
Substitute:
12 − F = 2F
12 = 3F
F = 4
The small block pushes on the large block 4 N to the left (-4 N).
The large block pushes on the small block 4 N to the right (+4 N).
The right half will be a new bar magnet of 2cm with north pole on the right side and south pole on teh left.
Roughly 1609 meters in one mile
Answer
Given,
y(x, t) = (3.5 cm) cos(2.7 x − 92 t)
comparing the given equation with general equation
y(x,t) = A cos(k x - ω t)
A = 3.5 cm , k = 2.7 rad/m , ω = 92 rad/s
we know,
a) ω =2πf
f = 92/ 2π
f = 14.64 Hz
b) Wavelength of the wave
we now, k = 2π/λ
2π/λ = 2.7
λ = 2 π/2.7
λ = 2.33 m
c) Speed of wave
v = ν λ
v = 14.64 x 2.33
v = 34.11 m/s