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aksik [14]
4 years ago
10

Which of these properties does not increase as the depth inside earth increases?

Physics
1 answer:
Licemer1 [7]4 years ago
7 0

A.Salinity

The deeper you make your way the hotter it gets because it is getting closer to the inner core which is 4000-4700 Degrees Celsius so that definitely would not be the answer. It also gets more and more dense and the pressure changes and you can see this by the fact that some mantles are liquid but because of the pressure it forces it into a solid. Sorry for the terrible explanation, I am bad at explaining things. Hope this helps!

You might be interested in
Which of the following is the same in all frames and reference?
Otrada [13]

B.Speed of light  

The speed of light is used as a reference and is 3x10^8 m/s no matter what.

Hope this helps :)

7 0
3 years ago
Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de
nirvana33 [79]
Hola!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Vo: 6 m/s 
Altura o distancia recorrida: 40 m
Gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula:

Vf²-Vo²=2×g×h.

Es importante hacer notar que al estar lanzando el ladrilo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo se tiene:

Vf
²= (6 m/s)²+ (2)×(9,81 m/s²)x(40m)
Vf²=820,8 m²/s²
√Vf²=√820,8 m²/s² 
Vf= 28,64 m/s

Que tengas un buen día!


4 0
4 years ago
A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0
3 years ago
What is the label for momentum
andrew11 [14]
P = mv; where m is mass and v is volume
6 0
4 years ago
Read 2 more answers
Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4+??
Sav [38]

Answer:

find answers below

Explanation:

a)

S1 560 e^(j acos 0.707 ) ⋅kVA            S1 = ( ) 395.92 396.04j  kW+ ⋅

S2 := 132 kW⋅                                      S2 = 132 kW⋅

Sd S1 +S2 :=

Sd = 527.92 396.04j  kW

Sd = 660 kVA ⋅

arg Sd = 36.877 deg ⋅

Vd 2.2 e^(− j⋅0⋅deg)⋅kV

current in the line =(Sd/3 Vd) ⋅

⎯

:=

I Line =  79.988 60.006j A

Line = 99.994 A

arg I( ) Line = −36.877⋅deg

r Line := 0.4⋅Ω                          resistance in the line xLine := 2.7⋅Ω

Vsan Vd+ (r Line+ j xLine) ILine

Vsan = ( ) kV 2.394 0.192j + ⋅ Vsan = 2.402 kV⋅ arg V( ) san = 4.584 deg ⋅

Vsab 3 e

j 30 ⋅ ⋅deg ⋅ Vsan := ⋅

Vsab = ( ) kV 3.425 2.361j + ⋅ Vsab = 4.16 kV⋅ arg V( ) sab = 34.584 deg

b)

PLine 3 I ( ) Line

2 ⋅ r

Line := ⋅

PLine = 12 kW⋅

QLine 3 I ( ) Line

2 ⋅ xLine := ⋅ QLine = 80.99 kVAR ⋅

c)

Ss 3 Vsan ⋅ I

Line := ⋅

⎯ Ss = ( ) kVA 539.919 477.03j + ⋅

Ss = 720.5 kVA ⋅ arg S( )s = 41.461 deg ⋅

Ps Re S( )s := Ps = 540 kW⋅

Qs Im S( )s := Qs = 477 kVAR ⋅

S1 S2 + PLine + j QLine + ⋅ = ( ) kVA 539.919 477.03j + ⋅ Check

7 0
3 years ago
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