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satela [25.4K]
3 years ago
15

The diagram below shows different layers of sedimentary rocks.

Physics
2 answers:
Flauer [41]3 years ago
5 0
Bro it is layer f was formed earlier than layer a
Arlecino [84]3 years ago
3 0
Layer F was formed earlier than Layer A
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Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i
leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are \dfrac{\lambda}{2}

Thus; for both speakers; the wavelength of the sound is:

\dfrac{\lambda}{2} = (10+30) cm

\dfrac{\lambda}{2} = (40) cm

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{8}+ \Delta \phi_o

\pi \ rad  = \dfrac{ \pi}{4}+ \Delta \phi_o

\Delta \phi_o  =  \pi -\dfrac{ \pi}{4}

\Delta \phi_o  = \dfrac{ 4\pi - \pi}{4}

\Delta \phi_o  = \dfrac{ 3\pi}{4} \ rad

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad

\Delta \phi = \dfrac{3 \pi}{4} \ rad

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)

A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)

A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)

A = 0.765a

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3 years ago
A gas is compressed in a cylinder from a volume of 20.0 l to 2.0 l by a constant pressure of 10.0 atm. calculate the amount of w
jok3333 [9.3K]
First, we need to convert the pressure in SI units. Keeping in mind that 1 atm = 1.01 \cdot 10^5 Pa:
p=10.0 atm =1.01 \cdot 10^6 Pa

The initial and final volume of the gas are (keeping in mind that 1.0 L = 0.001 m^3):
V_i = 20.0 L=0.020 m^3
V_f = 2.0 L=0.002 m^3

So, the work done on the gas by the surrounding is
W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J
And the final positive sign means that this work corresponds to an increase in internal energy of the gas.
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3 years ago
Water is a fluid, all fluids
Andreyy89
What is the question?
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3 years ago
A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular
ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0

Where,

\theta = Angular Displacement

\alpha =Angular Acceleration

\omega_0 = Angular velocity

\theta_0 =Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

\theta = \frac{1}{2}\alpha t^2

Re-arrange for \alpha,

\alpha = \frac{2\theta}{t^2}

Replacing our values,

\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}

\alpha = 0.139rad/s

Therefore the ANgular acceleration of the mass is 0.139rad/s^2

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Bank A has a leverage ratio of 10, while Bank B has a leverage ratio of 20. Similar losses on bank loans at the two banks cause
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Answer:

<u><em>note:</em></u>

<u><em>find the attached solution:</em></u>

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