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guapka [62]
2 years ago
9

Kim has a metal casting company which makes commemorative coins. She has 0.12 cubic meters of silver which she needs to make int

o 1.07e5 coins of thickness 2 mm. If she uses all of the silver, what is the diameter in centimeters of each coin
Physics
1 answer:
a_sh-v [17]2 years ago
5 0

Answer:

The coin has a diameter of 2.67 cm

Explanation:

First, we need to find the volume of each coin by dividing the total volume of silver by the number of coins. We have also to do a conversion of units in terms of centimeters as follows:

V=0.12 m^3\times (\frac{100cm}{1m})^3=12000\ cm^3\\V_c=\frac{120000\ cm^3}{1.07\times10^5}= 1.121 cm^3

Then, we define the coin as a tiny cilinder to determine its diameter. In that order we use the cilinder's volumen equation as follows:

V=\pi r^2h\\r = \sqrt \frac{V}{\pi h}= \sqrt\frac{1.121 cm^3}{\pi \times 0.2cm}=1.336 cm

Finally, we know that the diameter is twice the radius, therefore the diameter of each coin is 2.67 cm.

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You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be
Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

7 0
3 years ago
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Evgen [1.6K]

Answer:

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Explanation:

8 0
3 years ago
Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is
goblinko [34]

Answer:

d \frac{x}{l} = m×  λ⇒ d = λ ×m×l / x

= 630×10^{-9} m × 3×3m/ 45×10^{-3} m

= 1.26×10^{-4}m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0
2 years ago
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
Where coastal rock is softer, waves erode the land faster ​
prohojiy [21]

Answer:

Headlands and bays are created where there are bands of hard and soft rock which meet the coastline at right angles. Softer rock is eroded more quickly and erodes backwards to form bays (which may have beaches). The harder rocks are more resistant to erosion and jut out into the sea to form exposed headlands

5 0
2 years ago
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