2. What is the feather's initial velocity (before it is dropped) in m/s?

Use examples to explain how the geosphere interacts with two other of Earth's spheres. Explain the interaction for each using co

Orlov [11]

The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere.

7 0

3 years ago

Read 2 more answers

Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H

djyliett [7]

Answer:

The wavelength of the infrared wave is <u>0.0001 m</u>.

Explanation:

Given:

Frequency of an infrared wave is, $f=3.0\times 10^{12}\ Hz$

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

$v=c=3.0\times 10^8\ m/s$

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

$v=f\lambda$

Where, $\lambda \to \textrm{Wavelength of infrared wave}$

Now, rewriting the above formula in terms of wavelength, $\lambda$ , we get:

$\lambda=\dfrac{v}{f}$

Now, plug in $3.0\times 10^8$ for 'v', $3.0\times 10^{12}$ for 'f' and solve for $\lambda$ . This gives,

$\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m$

Therefore, the wavelength of the infrared wave is 0.0001 m.

5 0

3 years ago

The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below. The vertical axis is the maximum

Rudiy27

Answer:

Check the explanation

Explanation:

A) 7th Harmonic. (Of an open ended pipe, odd harmonics are allowed (3rd overtone))

b) f = n v / 4 L

n = 7

f = 7 x 350 / 4 x 0.41 = 1493.9 Hz

c) Let level of water H, If reduces the effective length of pipe

Using, f = n v / 4 Leff

n = 1

251.8 = 1 x 350 / 4 ( 0.41 - H)

H = 0.0625m

H = 6.25 cm

5 0

3 years ago

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

or

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

or

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

or

I = $0.0109[1-2.717^{-1}]$

or

I = 0.00688 A

or

I = 6.88 mA

5 0

3 years ago

A certain plucked string produces a fundamental frequency of 150 hz. Which frequency is not one of the harmonics produced by tha

Irina-Kira [14]

Answer:

Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...

Explanation:

Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.

For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc

If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc

Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...

6 0

3 years ago