Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2
)
Answer:
108 N
Explanation:
From the question,
Applying
F' = mgμ................ Equation 1
Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.
From the question,
Given: m = 18 kg, μ = 0.6, g = 10 m/s²
Substitute these values into equation 1
F' = 18×0.6×10
F' = 108 N
Answer:
4 seconds
Explanation:
There are links between cars, they are in between
1-2, 2-3, 3-4, 4-5, 5-6, 6-7, 7-8, 8-9, 9-10, 10-11, 11-12, 12-13, 13-14, 14-15, 15-16, 16-17.
Here 1 represents the first car which is firmly attached to the engine. So, there are 16 links and each link has a slack of 9 m.
So, total slack
9×16 = 144 cm
Speed of train = 36 cm/s
Time taken by the pulse to travel the length of train is 4 seconds.
The question is incomplete. The complete question is :
A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:
J(t) = Jo (1 - exp (-t/to))
Where,
Jo= 3m^2/ GPA
to= 200s
Determine
a) the strain after 100's (before stress is reversed)
b) the residual strain when stress falls to zero.
Answer:
a)-60GPA
b) 0
Explanation:
Given t= 0,
σ = 20Mpa
Change in σ= 0.2Mpas^-1
For creep compliance material,
J(t) = Jo (1 - exp (-t/to))
J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa
a) t= 100s
E(t)= ΔσJ (t - Jo)
= 0.2 × 3 ( 100 - 200 )
= 0.6 (-100)
= - 60 GPA
Residual strain, σ= 0
E(t)= Jσ (Jo) ∫t (t - Jo) dt
3 × 0 × 200 ∫t (t - Jo) dt
E(t) = 0
Answer:
cells
Explanation:
they are the smallest living organisms
