Answer:
t₁ - t₂ = 0.0011 s
Explanation:
given,
y(x, t) = (6.0 mm) sin( kx + (600 rad/s)t + Φ)
now,
y m = 6 mm ω = 600 rad/s
y₁ = + 2.0 mm y₂ = -2 .0 mm
now,
2 = (6.0 mm) sin( kx + (600 rad/s)t + Φ)
-2 = (6.0 mm) sin( kx + (600 rad/s)t + Φ)
so,
kx + (600 rad/s)t₁ + Φ = 
......(1)
we have multiplied with π/180 to convert angle into radians
kx + (600 rad/s)t₂ + Φ = 
......(2)
subtracting both the equation (1)-(2)
600(t₁-t₂) = 
now,
t₁ - t₂ = 0.0011 s
time does any given point on the string take to move between displacements is equal to 0.0011 s
Answer:
15.1 N
Explanation:
mass of block (m) = 4 kg
angle of inclination = 36 degrees
applied force (P) = 31 N
acceleration due to gravity (g) = 9.8 m/s^{2}
since the block is moving at a constant speed, it means the acceleration is 0 and therefore the summation of all the forces acting on the body is 0
therefore
P - f - mgsinθ = 0
where
- P = applied force
- f = frictional force
- m = mass
- g = acceleration due to gravity
when P = 31 N and the block is pushed upward
31 - f - (4 x 9.8 x sin 36) = 0
f = 7.96 N
now that we have the value of the frictional force we can find P required to lower the block, our equation becomes p + f - mgsinθ = 0 since the block is to be lowered
P + f - mgsinθ = 0
P = mgsinθ - f
P = (4 x 9.8 x sin 36) - 7.96 = 15.1 N
