Some amount of ideal gas with internal energy U and initial temperature 1000C was compressed to half of volume meanwhile absolute pressure inside of a container increased twice. We can say that internal energy of this gas after compression in terms of U is (20.2, 20.1, 19.4, 19.5) Group of answer choices


Answer 1


U. With no variation.


Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.

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Which of the following is a description of the Remote Associates Test (RAT)?



The description is outlined throughout the clarification section following, and according to the given word.


  • Throughout the 1960s, Sarnoff Mednick created the RAT as a tool used for testing imaginative convergent thought. Through each RAT test query lists a set of terms, which demands that we have a single additional term that will tie any of the others around.  
  • Those other words may also be related in something like a variety of ways, such as through creating a compound word or perhaps a semantic connexon.

What was the main idea of Malthus theory of population



The idea that population growth is potentially exponential while the growth of the food supply or other resources is linear.


The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?


The particle reach its minimum velocity at time 1.06 sec.

The function is given as


Differentiating the above equation with respect to time, to obtain the velocity


For maximum and minimum values, put dx/dt=0


On solving the equation, t=0, 1.06

Therefore at time t=1.06 sec, the particle has the minimum value of velocity.

The particle reaches its minimum velocity at t = 0 s or t = 16/15 s

Further explanation

Acceleration is rate of change of velocity.

\large {\boxed {a = (v - u)/(t) } }

\large {\boxed {d = (v + u)/(2)~t } }

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!


x = ( 5t^3 - 8t^2 + 12) ~ m

To find the velocity function, we will derive the position function above.

v = (dx)/(dt)

v = 5(3)t^(3-1) - 8(2)t^(2-1)

v = ( 15t^2 - 16t ) ~ m/s

Next to calculate the time to reach its minimum speed, then v = 0 m/s

0 = ( 15t^2 - 16t )

0 = t( 15t - 16)

\large {\boxed {t = 0 ~s ~ or ~ t = 16/15 ~ s} }

Learn more

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change?a. 4
b. 2/pi
c. 2
d. (2)^1/2



the answers the correct one is a  4


The centripetal acceleration is by

           a = v² / R

angular and linear velocities are related

           v = w R

let's substitute

           a = w² R

for initial condition

           a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

           a = (2w₀)² R

           a = 4 (w₀² R)

           a = 4 a₀

when reviewing the answers the correct one is a

If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)





Radius of Pulley r=12 cm

mass of block m=60 gm

mass of Pulley M=430 gm

Block descend h=50 cm

Applying Conservation of Energy

Potential Energy of block convert to rotational Energy of pulley and kinetic energy of block


mgh=(1)/(2)I\omega ^2+(1)/(2)mv^2

where I=moment of inertia


and for rolling \omega =(v)/(r)




v=\sqrt{(2* 60* 9.8* 0.5)/(430+60)}

v=\sqrt{(60* 9.8)/(490)}


v=1.095 m/s

Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.




  \phi = 1.78 *10^(-7) \  Weber


 L  = 1.183 *10^(-7) \  H


From the question we are told that

   The radius is  r = 6 \ cm =  (6)/(100) =  0.06 \ m

   The current it carries is  I  = 1.50 \ A


The  magnetic flux of the coil is mathematically represented as

       \phi = B  * A

Where  B is the  magnetic field which is mathematically represented as

         B  =  (\mu_o  * I)/(2 *  r)

Where  \mu_o is the magnetic field with a constant value  \mu_o  =  4\pi * 10^(-7) N/A^2

substituting  value

          B  =  (4\pi * 10^(-7)   * 1.50 )/(2 *  0.06)

          B  =  1.571 *10^(-5) \ T

The area A is mathematically evaluated as

       A  = \pi r ^2

substituting values

       A  = 3.142 *  (0.06)^2

       A  = 0.0113 m^2

the magnetic flux is mathematically evaluated as    

        \phi = 1.571 *10^(-5) * 0.0113

         \phi = 1.78 *10^(-7) \  Weber

The self-inductance is evaluated as

       L  =  (\phi )/(I)

substituting values

        L  =  (1.78 *10^(-7) )/(1.50 )

         L  = 1.183 *10^(-7) \  H