Answer: ionic compound
Explanation:
An ionic compound is formed when one element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.
For formation of sodium chloride:
Electronic configuration of sodium:
![[Na]=1s^22s^22p^63s^1](https://tex.z-dn.net/?f=%5BNa%5D%3D1s%5E22s%5E22p%5E63s%5E1)
Sodium atom will loose one electron to gain noble gas configuration and form sodium cation with +1 charge.
![[Na^+]=1s^22s^22p^63s^0](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D1s%5E22s%5E22p%5E63s%5E0)
Electronic configuration of chlorine:
![[Cl]=1s^22s^22p^63s^23p^5](https://tex.z-dn.net/?f=%5BCl%5D%3D1s%5E22s%5E22p%5E63s%5E23p%5E5)
Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.
![[Cl^-]=1s^22s^22p^63s^23p^6](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D1s%5E22s%5E22p%5E63s%5E23p%5E6)
The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic compound.
Incomplete question. Full text is:
"<span>Give an example of a situation in which you would describe an object's position in (a) one-dimension coordinates (b) two-dimension coordinates (c) three-dimension coordinates"
Solution
(a) One dimension example: a man walking along a metal plank. We just need to specify one coordinate, the distance from the beginning of the plank.
(b) Two-dimension example: a ball moving on a circle. In this case, we need two coordinates: (x,y) to specify the position of the ball at every instant, since it is moving on a 2-D plane.
(c) The position of an airplane in the air: in this case we need 3 coordinates, the height, the latitude and the longitude of the airplane.</span>
Answer:
a) Maximum height reached above ground = 2.8 m
b) When he reaches maximum height he is 2 m far from end of the ramp.
Explanation:
a) We have equation of motion v²=u²+2as
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0²=5.6² + 2 x -9.81 x s
s = 1.60 m
Height above ground = 1.2 + 1.6 = 2.8 m
b) We have equation of motion v= u+at
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0= 5.6 - 9.81 x t
t = 0.57s
Now considering horizontal motion of skateboarder.
We have equation of motion s =ut + 0.5 at²
u = 6.6cos58 = 3.50 m/s
a = 0 m/s²
t = 0.57
Substituting
s =3.5 x 0.57 + 0.5 x 0 x 0.57²
s = 2 m
When he reaches maximum height he is 2 m far from end of the ramp.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
<u>Determine the Thrust developed</u>
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : <em>calculate the area of the duct </em>
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
<em>next : calculate the velocity of propeller</em>
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
<em>Finally determine the thrust developed </em>
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
