(a)
Electronic configuration is given as follows:
![[Kr]4d^{3}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B3%7D)
Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.
Thus, electronic configuration of neutral atom is ![[Kr]4d^{5}5s^{1}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B5%7D5s%5E%7B1%7D)
.
The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.
This corresponds to element: molybdenum. Thus, the tripositive atom will be 
.
(b) The given electronic configuration is ![[Kr]5s^{2}4d^{2}](https://tex.z-dn.net/?f=%5BKr%5D5s%5E%7B2%7D4d%5E%7B2%7D)
.
The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.
This corresponds to element zirconium, represented by symbol Zr.
Answer:
57 N
Explanation:
Were are told that the force
of gravity on Tomas is 57 N.
And it acts at an inclined angle of 65°
Thus;
The vertical component of the velocity is; F_y = 57 sin 65
While the horizontal component is;
F_x = 57 cos 65
Thus;
F_y = 51.66 N
F_x = 24.09 N
The net force will be;
F_net = √((F_y)² + (F_x)²)
F_net = √(51.66² + 24.09²)
F_net = √3249.0837
F_net = 57 N
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N
Umm what are you trying to say
Answer:
Distance from start point is 72.5km
Explanation:
The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:
Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:
Replacing the given values in the equations, the distance is calculated.
