Does the image above correctly illustrate how the coriolis effect impacts global winds?

Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the

masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

$\mu = \alpha *N$

Where,

$\alpha =$ Dipole momento associated with an Atom

N = Number of atoms

$\alpha$ y previously given in the problem and its value is 2.8*10^{-23}J/T

$L = 5.8cm = 5.8*10^{-2}m$

$A = 1.5cm^2 = 1.5*10^{-4}m^2$

The number of the atoms N, can be calculated as,

$N = \frac{\rho AL}{M_{mass}}*A_n$

Where

$\rho =$ Density

$M_{mass} =$ Molar Mass

A = Area

L = Length

$A_n =$ Avogadro number

$N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)$

$N = 7.4041*10^{23}atoms$

Then applying the equation about the dipole moment associated with an iron bar we have,

$\mu = \alpha *N$

$\mu = (2.8*10^{-23})*(7.4041*10^{23})$

$\mu = 20.72Am^2$

PART B) With the dipole moment we can now calculate the Torque in the system, which is

$\tau = \mu B sin(90)$

$\tau = (20.72)(2.2)$

$\tau = 45.584N.m$

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

3 0

3 years ago

magine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and rad

DerKrebs [107]

Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.

For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.

If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.

6 0

3 years ago

With which part of the brain is awareness typically associated?

weeeeeb [17]

Answer:cerebral cortex

Explanation:

4 0

3 years ago

Read 2 more answers

In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for

Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0

3 years ago

A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo

kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

⇒ (300 N) d = 1100 N•m

⇒ d ≈ 3.7 m

6 0

2 years ago