All categories

2 years ago

1. Rocks change as a result of what?

Physics

2 answers:

lisov135 [29]2 years ago

3 0

1.Erosion 2.by color,texture etc

aliina [53]2 years ago

3 0

Answer by YourHope:

Hi! :)

1. Rocks change as a result of Erosion!

2. Rocks are sorted by color and texture!

3. Melted rock deep in the earth is called magma!

4. The process that occurs when melted rock cools is called igneous rock!

5. How does the rate of Cooling affect crystal formation in rocks? It affects the size of the crystals that form within each of the solid rocks. Rapid cooling produces small crystals, while slower cooling producing larger ones!

You might be interested in

Hair color is a trait that’s controlled by many genes working together. Each gene contributes to this trait only in a small prop

Katena32 [7]

Codominence i think, its both parents genes are used

3 0

2 years ago

The __________ pedal in a stick shift vehicle enables a driver to shift gears.

mamaluj [8]

The answer is C. Clutch

8 0

3 years ago

Read 2 more answers

Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a

mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where, taking upward as positive direction:

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the pebble is launched horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

$v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s$ (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

$v_x = 20.0 m/s$

So, the final speed of the pebble as it strikes the ground is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s$

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0

2 years ago

We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t

Arisa [49]

Answer:

$31.6\:\mathrm{m/s}$

Explanation:

The elastic potential energy of a spring is given by $Us=\frac{1}{2}kx^2$ , where $k$ is the spring constant of the spring and $x$ is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by $KE=\frac{1}{2}mv^2$ , where $m$ is the mass of the object and $v$ is the object's velocity.

Thus, we have:

$Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2$

Substituting given values, we get:

$\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}$

4 0

2 years ago

Why do we call the microscope a compound light microscope

Ede4ka [16]

Explanation:

Types of light microscope

1. Compound , and 2. Stereo Microscope

Compound microscope has two lens system also called compound lens system. The objective lens and the eyepiece lens. The magnification provided by the objective lens is compounded by the eyepiece lens, the a higher magnification is observed.

6 0

2 years ago