No.
The acceleration of gravity on or near Earth's surface is 9.8 m/s² ,
not 20 m/s² .
If it were 20 m/s², then you would weigh almost exactly double
what you really weigh now.
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
![v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]](https://tex.z-dn.net/?f=v_%7Bx%7D%20%3D%2040%2Acos%2828%29%5C%5CV_%7Bx%7D%20%3D%2035.31%20%5Bkm%2Fh%5D)
In order to find the vertical component, we must use the sine function of the angle.
![V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]](https://tex.z-dn.net/?f=V_%7By%7D%3D40%2Asin%2828%29%5C%5CV_%7By%7D%20%3D%2018.77%20%5Bkm%2Fh%5D)
Natural sources of light include our sun and other stars, where the source of energy is nuclear energy (recall that the moon does not produce light but merely reflects sunlight), lightning, where the source is electrical, and fire, where the energy source is chemical.
Answer:
Explanation:
Total weight
My weight+weight of belongings
660+1100=1760N.
a. Work done by the elevator to travel a total height of 15.2m
Using newton law of motion
ΣF = ma
There are only two forces acting upward, the weight and the reaction by the elevator
Also note it is moving at constant velocity then, a=0
N - W=0
Then, N=W
N=1760N
So, workdone is given as
Wordone, =force × distance
Work done=1760×15.2
W=26,752J
W=26.752KJ
b. Work done on me alone is still need to go through the same process but will remove the weight of the belonging
Therefore,
Weight now = 660N
And using the same equation of motion
ΣF = ma
Comstant velocity, a=0
N - W=0
N=W
N=660N
Then, workdone
W=F×d
W=660×15.2
W=10,032J
W=10.032KJ
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
