Yes, it is true to a certain that the number of outer shell electrons determines the chemical properties of an element, only because this determines how the element interacts with other elements.

3 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

2 years ago

An electromagnet produces a magnetic eld of 0.95 T in a cylindrical region of radius 3.0 cm between its poles. A straight wire c

Daniel [21]

Answer:

0.855 N

Explanation:

Using

F = BILsinФ.................... Equation 1

Where F = magnitude of the force exerted on the wire, B = magnetic Field, I = current, L = Length of the wire in the magnetic Field, Ф = angle between the wire and the magnetic field.

Given: B = 0.95 T, I = 15.0 A, L = 2r, r = radius = 3.0 cm = 0.03 m L = 2×0.03 = 0.06 m, Ф = 90° (perpendicular)

Substitute into equation 1

F = 0.95(15)(0.06)sin90°

F = 0.855 N.

Hence the magnitude of force that is exerted on the wire = 0.855 N

6 0

3 years ago