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3 years ago

What can be inferred about the cave in the photo above?

Physics

2 answers:

lana66690 [7]3 years ago

8 0

This cave is located below the water table because of the way the rock has been formed. The way the rock has been formed has been done because of the water forming that overtime.

Therefore your answer is,

It is located below the water table.

Alekssandra [29.7K]3 years ago

4 0

The answer is c because of how the rocks are formed, and because the cave is filled with water

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On Planet Y, which has no air, a dropped object falls 9 m in 3 seconds. What is g, the acceleration due to gravity, on that plan

slavikrds [6]

Answer:

a = 2 m/s^2

which agrees with the third answer option provided.

Explanation:

Recall the kinematic formula for displacement under the action of a constant acceleration "a":

yf - yi = 1/2 a t^2

using the information provided this equation becomes:

9 = 1/2 a (3)^2

solve for a:

9 * 2 / 9 = a

then a = 2 m/s^2

which agrees with the third answer option provided.

7 0

3 years ago

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

Nastasia [14]

Answer:

750 nm

Explanation:

$d$ = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

$D$ = screen distance = 4.8 m

$y$ = position of first bright fringe = $\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m$

$n$ = order = 1

Position of first bright fringe is given as

$y = \frac{nD\lambda }{d}$

$0.002 = \frac{(1)(4.8)\lambda }{0.0018}$

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

3 0

3 years ago

A car changes speed from 25 m/s to 10 m/s in 240 seconds. Describe its acceleration.

Dmitry_Shevchenko [17]

Explanation:

Acceleration is change in velocity over change in time:

a = Δv / Δt

a = (10 m/s - 25 m/s) / (240 s - 0 s)

a = -0.0625 m/s²

So the car decelerates at 0.0625 m/s².

7 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

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lorasvet [3.4K]

Answer:

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4 0

2 years ago