A(n) _____ wave is a disturbance in matter that transfers

The Doppler effect is an effect produced by a moving source of sound or electromagnetic waves due to the relative motion of a so

xxMikexx [17]

B. The waves would shorten as the vehicle got closer

4 0

3 years ago

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About three billion years ago, single-celled organisms called cyanobacteria lived in Earth’s oceans. They thrived on the ocean’s

pickupchik [31]

I think its Oxygen.
ancient cyanobacteria produced Earth's first oxygen-rich atmosphere, which allowed the eventual rise of eukaryotes. T<span>he chloroplasts of eukaryotic algae and plants are derived from cyanobacteria</span>

3 0

2 years ago

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

ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

where;

v is speed of sound
L is length of the pipe

f = (3 x 340) / (4 x 0.44)

f = 579.55 Hz

<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f = v/(4L)

251.1 = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

#SPJ1

3 0

10 months ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

bearhunter [10]

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

Total Energy before fall = Total Energy after fall

$E_{i}$ = $E_{f}$

Here, total energy is the sum of kinetic energy and potential energy

$K.E_{i}$ + $P.E_{i}$ = $K.E_{f}$ + $P.E_{f}$ (a)

When banana is at height h, it has

$K.E_{i}$ = 0 and $P.E_{i}$ = mgh

and when it reaches the river, it has

$K.E_{f}$ = 1/2m $v^{2}$ and $P.E_{f}$ = 0

Putting the values in equation (a)

0 + mgh = 1/2m $v^{2}$ + 0

mgh = 1/2m $v^{2}$

<em>cutting 'm' from both sides</em>

<em> </em>gh = 1/2 $v^{2}$

v = $\sqrt{2gh}$

Hence, the velocity of banana before hitting the water is

v = $\sqrt{2gh}$

5 0

2 years ago