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2 years ago

6

Hundreds of millions of years ago, all the continents on Earth were connected to form a giant “supercontinent” called Pangaea. H

ow might ocean currents have differed during the time of Pangaea compared to today? Explain your answer.

1 answer:

Snezhnost [94]2 years ago

5 0

The ocean currents differed because Pangaea moved.

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Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0

2 years ago

Consider two waves X and Y traveling in the same medium. The two carry the same amount of energy per unit time, but X has one-se

RideAnS [48]

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy = 1 / 7

λy / λx = 7 / 1

7 0

2 years ago

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

=43m -16m

=27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V = (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12 ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0

3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

2 years ago

At high elevations or high latitudes, some of the water that falls on land does not immediately soak in , run off, evaporate or

LUCKY_DIMON [66]

Answer:

The water is stored in ice sheets and as snow

Explanation:

Temperature reduces with an increase in altitudes. The standard laps rate is 6.5°C per 1,000 m gained in elevation

At very high elevations, therefore, the air is usually very cold such that when an elevation of 4,500 meters is reached at the equator, it is possible to observe snowfall and the water remain temporarily stored on the surface of the mountain as ice and snow

7 0

2 years ago