Answer and Explanation:
The answer is attached below
Answer:
80.16 m/s^2
at t=2 s
x=42.3 m
y=16 m
z=14 m
Explanation:
solution
The x,y,z components of the velocity are donated by the i,j,k vectors.
![v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2](https://tex.z-dn.net/?f=v_%7Bx%7D%3D16t%5E%7B2%7D%20%20%5C%5Cv_%7By%7D%3D4t%5E%7B3%7D%5C%5Cv_%7Bz%7D%3D5t%2B2)
acceleration is a derivative of velocity with respect to time.
![a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5](https://tex.z-dn.net/?f=a_%7Bx%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%20v_%7Bx%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B16t%5E%7B2%7D%5D%3D32t%5C%5Ca_%7By%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%20v_%7By%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B4t%5E%7B3%7D%5D%3D12t%5E%7B2%7D%20%5C%5Ca_%7Bz%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%20v_%7Bz%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B5t%2B2%5D%3D5)
evaluate acceleration at 2 seconds
![a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}](https://tex.z-dn.net/?f=a_%7Bx%7D%20%3D32%2A2%3D64m%2Fs%5E%7B2%7D%5C%5C%20a_%7By%7D%20%3D12%2A2%5E%7B2%7D%20%3D48m%2Fs%5E%7B2%7D%5C%5Ca_%7Bz%7D%20%3D5m%2Fs%5E%7B2%7D)
the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.
![=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2](https://tex.z-dn.net/?f=%3D%5Csqrt%7Ba_%7Bx%7D%5E2%20%2Ba_%7By%7D%5E2%2Ba_%7Bz%7D%20%5E2%20%7D%20%5C%5C%3D%5Csqrt%7B64%5E2%20%2B48%5E2%2B5%20%5E2%20%7D%5C%5C%3D80.16m%2Fs%5E2)
position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.
![x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\](https://tex.z-dn.net/?f=x%3D%5Cint%5Climits%20%7Bv_%7Bx%7D%20%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E2_0%20%7B%7B16t%5E2%7D%20%5C%2C%20dt%3D42.7m%5C%5C%5C%5Cy%3D%5Cint%5Climits%20%7Bv_%7By%7D%20%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E2_0%20%7B%7B4t%5E3%7D%20%5C%2C%20dt%3D16m%5C%5C%5C%5C%5C%5C%5C%5C%5C%5Cz%3D%5Cint%5Climits%20%7Bv_%7Bz%7D%20%7D%20%5C%2C%20dx%3D%5Cint%5Climits%5E2_0%20%7B%7B5t%2B2%7D%20%5C%2C%20dt%3D14m%5C%5C%5C%5C)
Answer:
skskkdkdkfkgkgkgkkgkgkgigooigigi lol
Explanation:
Oof
Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.