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2 years ago

Which techniques can scientists use to determine the characteristics of Earth's layers? Select the two correct answers.

Physics

2 answers:

Rainbow [258]2 years ago

8 0

⭐hello my Friend!

1 and 3 are the right alternatives.

well they can't use x-rays on the ground.

but the 2 could also be but I think it wouldn't be cool for them to do that.

⭐have good studies!

☝️follow the rules of brainly!

Greeley [361]2 years ago

5 0

Answer:

1 and 3

Explanation:

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Can someone pls help me answer this I’ll give brainliest to whoever actually answers it

sergejj [24]

Answer:

Limestone classifacation: sedimentary rock

Sandstone: sedimentary rock

7 0

2 years ago

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago

Cual es la importancia de movimiento en la sociedad?

Marta_Voda [28]

Answer:

It allows for transportation and other things

Explanation:

Why are you asking this on the physics section

5 0

2 years ago

An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach

Gelneren [198K]

Answer: $5 m/s^{2}$

Explanation:

This problem is related to vertical motion, and the equation that models it is:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

5 0

2 years ago

if the charge of each two particles is doubled and the seperation between them is also doubled. the force between the two partic

jok3333 [9.3K]

Explanation is in this link

https://doubtnut.app/physics-question/when-the-distance-between-two-charged-particles-is-doubled-the-force-between-them-becomes-422318076

4 0

3 years ago