A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm​

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Answer 1
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What is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s

Answers

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer:

a

 (dE)/(dt) =-  2.72 *10^(15) \  N/C \cdot s

b

The  direction of the electric field is opposite that of the current              

Explanation:

From the question we are told that

   The current is  I  =  17\ A

   The diameter of the ring is  d =  3.0 \ cm  = 0.03 \ m

   

Generally the  radius is mathematically represented as

       r =  (d)/(2)

       r =  (0.03)/(2)

       r =  0.015 \  m

The  cross-sectional area is mathematically represented as

       A =  \pi r^2

=>     A = 3.142 *  (0.015^2)

=>    A = 7.07 *10^(-4 ) \  m^ 2

Generally  according to ampere -Maxwell equation we have that

      \oint \= B \cdot  \= ds =  \mu_o  I + \epsilon_o  \mu _o( d \phi )/(dt )

Now given that \= B  =  0 it implies that

     \oint \= B \cdot  \= ds = 0

So

    \mu_o  I + \epsilon_o  \mu _o( d \phi )/(dt ) = 0

Where  \epsilon _o is the permittivity of free space with value \epsilon_o  =  8.85*10^(-12 ) \   m^(-3) \cdot kg^(-1)\cdot  s^4 \cdot A^2

            \mu_o is the permeability of free space with value  

\mu_o  =   4\pi * 10^(-7) N/A^2

      \phi is magnetic flux which is mathematically represented as

       \phi  =  E * A

Where E is the electric field strength

  So  

       \mu_o  I + \epsilon_o  \mu _o   ( d [EA] )/(dt )  = 0

=>   (dE)/(dt) =-  (I)/(\epsilon_o * A )

=>   (dE)/(dt) =-  (17)/(8.85*10^(-12) * 7.07*10^(-4) )

=>   (dE)/(dt) =-  2.72 *10^(15) \  N/C \cdot s

The  negative  sign shows that the  direction  of  the electric field is opposite that of the current

           

       

What is it called when a Rock forms due to heat and pressure in the earth?

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It is called Metamorphic rocks!!

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which it will resonate. Express your answers using two significant figures separated by commas.

Answers

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=(320)/(4)\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

Final answer:

The resonant frequencies of a string depend on its length, tension, and linear mass density. For a string resonating in four loops at 320 Hz, three possible smaller frequencies could be 160 Hz, 106.7 Hz, and 80 Hz.

Explanation:

When a string resonates, it vibrates at certain frequencies called its resonant frequencies. The resonant frequencies of a string depend on factors such as its length, tension, and linear mass density. In this case, the string resonates in four loops at a frequency of 320 Hz.

Three other possible resonant frequencies at which the string could vibrate with smaller loops include:

  1. 160 Hz: This is half the frequency of the given resonant frequency, which means the string vibrates with twice the number of loops.
  2. 106.7 Hz: This is one third of the given resonant frequency, which means the string vibrates with three times the number of loops.
  3. 80 Hz: This is one fourth of the given resonant frequency, which means the string vibrates with four times the number of loops.

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A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then travels for 30 minutes at 80 km/h. (a) Sketch a plot of her displacement versus time and speed versus time. (b) Calculate her average speed.

Answers

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

t_(hours)=t_(min)*(1 h)/(60 min)\n30min*(1 h)/(60 min)=0,5h\n45min*(1 h)/(60 min)=0,75h

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

d=v*t\n80km/h*0.5h=40km\n100km/h*0.75h=75km

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

v_(average)=(155km)/(2.25h)=68.9km/h

Energy is the capacity to do work, but not to produce heat

Answers

Energy and Work Energy is the capacity to do work or to produce heat. Internal energy is the sum of kinetic energy and the potential energy. ... The KE would increase because heating something causes an increase in temperature.

Final answer:

Energy is the capacity to do work but not to produce heat. In physics, energy can exist in various forms, including mechanical and thermal energy.

Explanation:

Energy is the capacity to do work and is an important concept in physics. In the context of this question, it is stated that energy is the capacity to do work but not to produce heat. This highlights the distinction between the two forms of energy. For example, mechanical energy can be used to perform work on an object and cause it to move, while thermal energy is associated with heat and not directly related to work. However, it's important to note that energy can be converted from one form to another, such as converting mechanical energy to thermal energy in a friction process.

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