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4 years ago

6

O lugar mais fundo do mar é a Trincheira de Mariana, localizada a 11 km de profundidade. Essa estrutura imensa está localizada n

o norte ocidental do Oceano Pacífico.
Qual alternativa apresenta corretamente a profundidade na Trincheira de Mariana, em ordem de grandeza?

1 answer:

ipn [44]4 years ago

3 0

Sorry i dont speak spanish

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In the lab, You have a highly reactive metal like sodium. It is so reactive that water sets it off. How would you store it?

kenny6666 [7]

Answer:

Sodium is stored under paraffin oil

Explanation:

Group 1 metals are known to be very highly reactive. The chemical reactivity of the members of the group decreases down the group.

Sodium reacts violently with water and upon exposure to the atmosphere. As a result of this, it must be stored under paraffin oil.

8 0

3 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

5 0

3 years ago

Deimos's Orbit. Deimos orbits Mars at a distance of 23,460 km from the center of the planet and has a period of 1.263 days. Assu

mihalych1998 [28]

Answer:

M = 5.882 10²³ kg

Explanation:

Let's use Newton's second law to analyze the satellite orbit around Mars.

F = m a

force is universal attraction and acceleration is centripetal

a = v²/ R

the modulus of velocity in a circular orbit is constant

v= d/T

the distance of the cicule is

d =2pi R

a = 2pi R/T

we substitute

- G m M / R² = m ( $- \frac{4\pi^2 R^2 }{T^2 R}$ )

G M = $\frac{ 4\pi ^2 R^3 }{T^2 }$

M = $\frac{4 \pi ^2 R^3 }{ G T^2 }$

the distance R is the distance from the center of the planet Mars to the center of the satellite Deimos

R = 23460 km = 2.3460 10⁷ m

the period of the orbit is

T = 1,263 days = 1,263 day (24 h / 1 day) (3600s / h)

T = 1.0912 10⁵ s

let's calculate

M = $\frac{4 \pi ^2 ( 2.3460 \ 10^7)^3 }{5.67 10^{-11} \ (1.0912 \ 10^5)^2 }$

M = 509.73418 10²¹ /8.66640 10⁻¹

M = 58.817 10²² kg

M = 5.882 10²³ kg

4 0

3 years ago

When an oxygen atom forms an ion, it gains two electrons. What is the electrical charge of the oxygen ion? A. +2 B. -1 C. -2 D.

Alecsey [184]

C. 2-

electrons have a negative charge, thus adding two electrons to an atom will give it a charge of 2 times -1 or -2

6 0

3 years ago

Read 2 more answers

Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some

erica [24]

If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.

Calculation:

Step-1:

It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.

The gauge pressure at a particular depth of ocean water is calculated as:

$P=\rho g h$

Here $\rho$ is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.

Step-2:

Now we are substituting the values to calculate the pressure at the depth of 2452-m.

$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$

Learn more about gauge pressure here,

brainly.com/question/14012416

#SPJ4

5 0

2 years ago