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Arte-miy333 [17]
1 year ago
12

1. The sedimentary rock known as conglomerate typically forms in _______ environments in which particles can become rounded, suc

h as fast-flowing rivers.
2. An example of _______ is when moving water slows down and particles being transported in the water begin to settle out (sediment) in a new location.

3. _______ is a type of sediment that feels smooth to your fingers but gritty in your mouth.

4. The most common chemical sedimentary rock is _______.
Physics
2 answers:
balu736 [363]1 year ago
7 0

Answer: 1. high-energy

2. deposition

3. Silt

4. limestone

5. soluble

6. tectonic forces

7. before

8. faunal succession

9. nondeposition

10. coarser

Explanation:

ki77a [65]1 year ago
3 0

Answer: (1) Fluvial environment

(2) Deposition

(3) Limestone

(4) Sandstone

Explanation:

  • Conglomerates are those in which the particles are big and rounded. They are formed in the high energy condition, such as fluvial or beach environment having strong winds.
  • Deposition takes place with the rate of transport of fluid or water. If the flow rate is high then the water carry sediments to a far extent, whereas if the flow rate is slow or it stops then the sediments get deposit.
  • Limestone is comprised of calcium carbonate, so when we touch it with our fingers we feel a bit of smoothness and tastes like grit in mouth.
  • Sandstone is the most common one, containing approximately 70-75% of all the sedimentary rocks on earth.
You might be interested in
Which of the following statements cannot be supported by Kepler's laws of planetary motion?
gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio \frac{a^3}{T^2} (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

8 0
2 years ago
To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes a
Gwar [14]

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

5 0
2 years ago
Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark
Nitella [24]

Answer:

Wavelength of incident light, \lambda = 612 nm

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

width of the fringe, \beta = 1.70 cm = 0.017 m

Solution:

To calculate the wavelength of the incident light, \lambda:

\beta = \frac{x\lambda }{d}

\lambda = \frac{\beta d}{x}

\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm

\lambda = 612 nm

4 0
2 years ago
The total mass of the arm shown in is 2.6 kg. Determine the force, F_M required of the "deltoid" muscle to hold up the outstretc
guapka [62]

Answer:

Fm = 51N and Fj = 26N

Summing the moments about the shoulder joint

Sum of anticlockwise moments = sum of clockwise moments

Fm x 12 = mg x 24

Fm = 2.6 x 9.8 × 24/12

Fm = 51N

Summing the forces acting on the arm

Sum of upward forces = sum of downward forces

Fm = Fj + mg

51 = Fj + 2.6 × 9.8

51 = Fj + 25.48

Fj = 51 - 25.48

Fj = 26N

Explanation:

Newtons first law and the principle of moments have been applied in solving this problem.

6 0
2 years ago
Dee is on a swing in the playground. the chains are 2.5 m long, and the tension in each chain is 450 n when dee is 55 cm above t
Leno4ka [110]
Refer to the diagram shown below.

From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
cos θ = 1.95/2.5 = 0.78
θ = cos⁻¹ 0.78 = 38.74°

From the free body diagram, the tension in the chain is 450 N.
F is the centripetal force,
W is Dee's weight.

The components of the tension are
Horizontal component = 450 sin(38.74°) = 281.6 N, acting left.
Vertical component = 450 cos(38.74°) = 351.0 N, acting upward.

Answers:
Horizontal: 281.6, acting left.
Vertical: 351.0 N, acting upward.

8 0
2 years ago
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