Answer:
Backdoor
Explanation:
The back door fragment in a program allows user to access backdoor information without necessarily following the common security procedures needed. In this case, once the programmer keys in the username he or she logs in without putting password. Therefore, this is a backdoor fragment.
Answer:
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
Explanation:
Drag force:
Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.
Drag force given as
![F_D=\dfrac{1}{2}\rho\ A\ V^2](https://tex.z-dn.net/?f=F_D%3D%5Cdfrac%7B1%7D%7B2%7D%5Crho%5C%20A%5C%20V%5E2)
So we can say that drag force depends on following properties
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
Answer:
a twisting force that tends to cause rotation
Explanation:
Answer:
The length of bar will be 2.82 m
Explanation:
Given that
d= 15 mm
r= 7.5 mm
Shear stress = 110 MPa
θ = 30° (30° = 30° x π/180° =0.523 rad)
θ = 0.523 rad
G for steel
G= 79.3 GPa
We know that
![\dfrac{\tau}{r}=\dfrac{G\theta }{L}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctau%7D%7Br%7D%3D%5Cdfrac%7BG%5Ctheta%20%7D%7BL%7D)
![\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}](https://tex.z-dn.net/?f=%5Cdfrac%7B110%7D%7B7.5%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cdfrac%7B79.3%5Ctimes%2010%5E3%5Ctimes%200.523%20%7D%7BL%7D)
L= 2. 82 m
The length of bar will be 2.82 m
Answer:
atomic radius R = 0.157 nm
metal atomic weight = 72.27 g/mol
Explanation:
given data
parameters a = 0.413 nm
parameters b = 0.665 nm
parameters c = 0.876 nm
atomic packing factor = 0.536
density = 3.99 g/cm³
to find out
atomic radius and atomic weight
solution
we apply here atomic packing factor (x) that is
atomic packing factor (x) =
..................1
put here value we get
atomic packing factor =
R = ![(\frac{3(x)(abc)}{32\pi })^{1/3}](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%28x%29%28abc%29%7D%7B32%5Cpi%20%7D%29%5E%7B1%2F3%7D)
R = ![(\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%280.536%29%280.413%2A0.665%2A0.876%29%7D%7B32%5Cpi%20%7D%29%5E%7B1%2F3%7D)
atomic radius R = 0.157 nm
and
now we get here metal atomic weight that is
metal atomic weight =
....................2
metal atomic weight =
metal atomic weight = 72.27 g/mol