Answer:
35 J
Explanation:
The man is holding the box: this means that he is applying a force vertically upward, to balance the weight of the box (which pushes downward).
Therefore, we can ignore the horizontal displacement of the man, because the force applied (vertically upward) is perpendicular to that displacement (horizontal), so the work done for that is zero.
So, only the vertical motion contributes to the work. The work done by the man is equal to the gain in gravitational potential energy of the box, so:
where
is the weight of the box
is the vertical displacement
Substituting, we find
Momentum question. This is an inelastic collision, so
m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s
0.67m/s
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
3.Es tarde y mi taxi no llega. Estoy ____.
(5 Points)
preocupada
contenta
When air is blown into the open pipe,
L = 
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ=
=3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f= 
⇒f=
≈115 Hz
