Because metallic bonds involve all of the metal atoms in a piece of metal sharing all of their valence electrons with "delocalized" bonds.
Answer:
x = 5.79 m
Explanation:
given,
mass of the car = 39000 Kg
spring constant = 5.7 x 10⁵ N/m
acceleration due to gravity = 9.8 m/s²
height of the track = 25 m
length of spring compressed = ?
using conservation of energy
potential energy is converted into spring energy




x = 5.79 m
the spring is compressed to x = 5.79 m to stop the car.
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
The only balanced equation is B. If you look at the equation and break it down you can see that in:

→

Starting from the left side of the equation there are 2 Nitrogen atoms, and 2 oxygen atoms as indicated by the subscript.
To balance the equation, the number of atoms of each element in the right side equation should be equal to left. By putting the numerical coefficient of 2, you will distribute that to each element. So you will end up with 2 nitrogen atoms and 2 oxygen atoms on the left side of the equation. Thus, the equation is balanced.
The answer again, is B.
Answer:
The maximum volume is 1417.87 
Explanation:
<u>Optimization Using Derivatives</u>
We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches
When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

Operating

To find the maximum value of V, we compute the first derivative and equate it to zero

Simplifying by 12

Completing squares


We have two values for x


The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16
We'll keep only the solution

The width is

The length is

And the height

The maximum volume is
