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3 years ago

What happens when sediment builds up over time

2 answers:

6 0

When sediment has built up over time layers of rock start to form, starting with sedimentry rocks, then metamorphic rocks

Arlecino [84]3 years ago

6 0

When sediments build up layers form, like the Grand canyon. over time sediments built up to form the grand canyon

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What properties of the wave define why it is found within this area of the spectrum?

MatroZZZ [7]

Answer:

Explanation:

Speed and medium are properties of a wave which is in common within an area of the spectrum of visible light. i.e;

Their medium of propagation

They both travel at the same speed ( speed of light )

The properties mentioned above are properties that define that wave is found within an area spectrum for visible light.

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7 0

2 years ago

Hang two sheets of paper vertically from adjacent corners. The sheets should be parallel and close to each other with a small ga

Mrrafil [7]

Answer:

a. The sheets move toward each other and the gap narrows.

Explanation:

This exercise is related to fluid mechanics, when blowing between the two sheets, we can apply Bernoulli's equation, where the index 2 is the space between the two sheets

P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y²

if the two leaves are at the same height

y₁ = y₂

whereby

P₁ + ½ ρ g v₁² = P₂ + ½ ρ v₂²

for the air velocity between the leaves let us use the continuity equation

A₁ v₁ = A₂ v₂

the area between the leaves is less than the external area, so the air speed must increase. If we use this in Bernoulli's equation, increasing the speed 2 (between the leaves) to maintain equality the pressure must decrease.

If the pressure decreases, the blades should move closer

When resisting the answers, the correct one is a

4 0

3 years ago

The force supplied by a spring depends on what?

Morgarella [4.7K]

The force and motion

8 0

3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

2 years ago

A 6.0-μF air-filled capacitor is connected across a 100-V potential source (a battery). After the battery fully charges the capa

Dovator [93]

Answer:

Change in Q = 2.1x 10^-3 C

Explanation:

We are given that

The Initialcapacitance C1 = 6.0μF

Initial charge oncapacitor

Q1 = C1 V

= 6.00 x 10^-6 x 100

= 6.00 x 10^-4 C

So the Final capacitance C2 will be

= K x C1 = 4.5 x 6.00 x 10^-6

= 2.7 x 10^ -5 F

So to get Finalcharge

We use Q2 = C2 x V

= 2.7 x 10^ - 5 x 100

= 27 x 10^ -4 C

So Charge flown in thecapacitor is change in Q

Which is = Q2 - Q1

= 27 x 10^-4 - 6.0 x 10^ -4

Change in Q = 2.1x 10^-3 C

5 0

3 years ago