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Marrrta [24]
2 years ago
7

If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water

knocks two rocks together, you'll barely hear the sound.
Match the words.

The air-water interface is an example of boundary. The( )portion of the initial wave energy is way smaller than the( )portion. This makes the( ) wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can( ) .

1. reflect more efficiently
2. transmitted
3. travel directly to your ears
4. boundary
5. reflected
6. discontinuity
Physics
1 answer:
Svetradugi [14.3K]2 years ago
4 0

Answer:

The air-water interface is an example of<em> </em>boundary. The <u><em>transmitted</em></u><em> </em> portion of the initial wave energy is way smaller than the <u><em>reflected</em></u><em> </em> portion. This makes the <u><em>boundary</em></u>  wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can <u><em>travel directly to your ear</em></u>.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at
ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The  expression is   F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

b

F_{No}= 7771.125 \ N

c

 F_p = 2.2*10^{6} N

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

          The altitude of container in Denver is  d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m

           The surface area of the container lid is A = 0.0155m^2

           The altitude of container in New Orleans  is sea-level

           The air pressure in Denver is  P_D = 79000 \ Pa

            The air pressure in new Orleans is P_{ro} = 100250 \ Pa

Generally force is mathematically represented as

            F_{No} = \Delta P A

  So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

   The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa

So the \Delta P which is the difference in pressure within and outside the container is  

           \Delta P = P_{No} - \frac{P_{sea}}{2}

Therefore

                F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

Now substituting values

                F_{No} =   0.0155 [100250 - \frac{101000}{2}]

                       F_{No}= 7771.125 \ N

The force to remove the lid in Denver is  

           F_p = \Delta P_d A

So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

 The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa    

 At  sea level the air pressure in Denver is mathematically represented as

              P_D = \rho g h

     =>     g = \frac{P_D}{\rho h}      

Let height at sea level is h = 1

  The air pressure at height d_D

             P_d__{D}} = \rho gd_D

    =>     g = \frac{P_d_D}{\rho d_D}

  Equating the both

                 \frac{P_D}{\rho h}  = \frac{P_d_D}{\rho d_D}

                 P_d_D =  P_D * d_D

Substituting value  

                   P_d__{D}} = 1828.2 * 79000

                    P_d__{D}} = 1.445*10^{8} Pa

    So

              \Delta P_d  = P_{d} _D - \frac{P_{sea}}{2}

=>          \Delta P_d  = 1.445 *10^{8} - \frac{101000}{2}    

                        \Delta P_d = 1.44*10^{8}Pa

  So

               F_p = \Delta P_d A

                  = 1.44*10^8 * 0.0155

              F_p = 2.2*10^{6} N

               

                 

             

             

6 0
2 years ago
A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going
tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2

For the sound

Distance = Speed × Time

\text{Distance}=340\times (2.5-t)

Here, distance traveled by the stone and sound is equal

\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0

t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds

The time taken to fall down is 2.4158 seconds

h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m

Height of the buidling is 28.6260196842 m

7 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
1 year ago
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
1 year ago
A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
blsea [12.9K]

Answer:

a) Q = C*emf

b)  Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let V_{R} = Voltage dropped across the Resistor

V_{c} = Voltage dropped across the capacitor

Applying KVL;

emf - V_{R}  - V_{c} = 0\\.........................(1)

Since the connection is in series, the same current flow through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Putting V_{c} and V_{R} into equation (1)

emf - IR - Q/C = 0

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Putting the values of t and I₀ into the formula for I written above

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

Q = K* Q_{initial} \\Q_{initial}  = C *emf\\Q_{final}  = KCemf

4 0
2 years ago
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