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2 years ago

Which of the following is a benefit of globalization?

2 answers:

8 0

Globalization connects foreign powers, increases trade, creates new industries, provides jobs, and modernizes countries.

Romashka-Z-Leto [24]2 years ago

7 0

Answer:

Where the answers choices?

Explanation:

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How did the continental Shelf form? Please help, thank you!! :)

shutvik [7]

The movements of the tectonic plates

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3 years ago

A very long straight current-carrying wire produces a magnetic field of 25 µT at a distance d from the wire. How far will the ma

daser333 [38]

The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.

B ∝ 1/d

B = magnetic field strength, d = distance from wire

Calculate the scaling factor for d required to change B from 25μT to 2.8μT:

2.8μT/25μT = 1/k

k = 8.9

You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT

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3 years ago

During which stage does the birth rate begin to decline?

sveticcg [70]

Answer:

During stage 3 - late expanding (of demogrpahic transition model)

Explanation:

During stage 3, birth rate begins to decline as infant mortality is lower and women have more access to education, family planning, and contraceptives. Children are not needed as "free labor" as they might have been in earlier stages.

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3 years ago

Read 2 more answers

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$