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3 years ago

6

Which property remains unaffected during the rock cycle?

2 answers:

atroni [7]3 years ago

7 0

Which property remains unaffected during the rock cycle?

A. Texture of the rock.

B. Mass of the material.

C. Structure of the rock.

D. Compositions of the minerals

The answer is ( B ) Mass of material

aivan3 [116]3 years ago

6 0

the answer is B.) Mass of material.

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1- ¿Cuál sería la energía de un objeto de 50 newton de peso que se encuentra sobre una estantería de 3 metros de altura? ¿Qué ti

Liula [17]

Responder:

<h3>150 Nm </h3><h3>Energía potencial </h3>

Explicación:

El tipo de energía que posee el objeto se conoce como energía potencial. <u>La energía potencial es la energía que posee un objeto, mi virtud de su posición. </u>

Energía potencial = masa * aceleración debido a la gravedad * altura

Dado que Force = masa * aceleración debido a la gravedad

Energía potencial = Fuerza * altura

Fuerza dada = 50N y altura = 3 m

Energía potencial = 50 * 3

Energía potencial = 150 Nm

6 0

3 years ago

On the asteroid Cere, the acceleration due to gravity is about 0.27 m/s^2. How long would it take a ball to fall to the ground i

KiRa [710]

Answer:

F. 25.82 s

Explanation:

Given:

Δy = 90 m

v₀ = 0 m/s

a = 0.27 m/s²

Find: t

Δy = v₀ t + ½ at²

90 m = (0 m/s) t + ½ (0.27 m/s²) t²

t = 25.82 s

7 0

3 years ago

What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg

scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

$F_c = F_g$

$\frac{mv^2}{r} = \frac{GmM}{r^2}$

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

$v^2 = \frac{GM}{r}$

At the same time we know that the period is equivalent in terms of the linear velocity to,

$T = \frac{2\pi}{\frac{v}{r}}$

$v = \frac{2\pi r}{T}$

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

$v^2 = \frac{GM}{r}$

$( \frac{2\pi r}{T})^2 = \frac{GM}{r}$

$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$

Replacing we have,

$T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}$

$T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})$

$T= 1.74hour$

Therefore the correct answer is C.

7 0

3 years ago

Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a

MariettaO [177]

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

5 0

3 years ago

Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).

JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

$M_1$ = Mass of Earth

$M_2$ = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

$F=\frac{GM_1m}{r^2}$

The force of gravity on an object on the other planet is

$F=\frac{GM_2m}{(2r)^2}$

As the forces are equal

$\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1$

So, the other planet would have 4 times the mass of Earth

6 0

4 years ago