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alexdok [17]
3 years ago
5

In which region of the sun does energy move as waves that transfer from atom to atom

Physics
2 answers:
liq [111]3 years ago
6 0

Answer:photosphere corona core radiative zone chromosphere convective zone. Radiative zone of the Sun does energy move as waves that transfer from atom to atom.

Explanation:photosphere corona core radiative zone chromosphere convective zone. Radiative zone of the Sun does energy move as waves that transfer from atom to atom.

Artist 52 [7]3 years ago
4 0

Answer:

Radiative zone

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A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note:
Sphinxa [80]

F=nmv

where;

n=no. of bullets = 1

m=mass of bullets=2g *10^-3

V=velocity of bullets200m/sec

F=1

loss in Kinetic energy=gain in heat energy

1/2MV^2=MS∆t

let M council M

=1/2V^2=S∆t

M=2g

K.E=MV^2/2

=(2*10^-3)(200)^2/2

2 councils 2

2*10^-3*4*10/2

K.E=40Js

H=mv∆t

(40/4.2)

40Js=40/4.2=mc∆t

40/4.2=2*0.03*∆t

=158.73°C

7 0
3 years ago
Elements from opposite sides of the periodic table tend to form ________.
Ivenika [448]
Ionic compounds is your answer. What happens is one atom donates electron(s) to the other atom, making one positive and the other negative. The opposite atoms attract, forming an ionic bond. 

Hope this helps! :)
4 0
3 years ago
Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift
sammy [17]

Answer:

The doppler effect equation is:

f' = \frac{v +v0}{v - vs}*f

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) -  (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0
3 years ago
An 4-kg ball experiences a force and accelerates at a rate of 1.5 m/s.
ANTONII [103]

Answer:

6.0 N

Explanation:

The strength of a force is expressed as the magnitude of the force in Newton.

The formula to apply here is :

Force= mass * acceleration

F=ma

Mass, m = 4 kg

Acceleration = 1.5 m/s²

Force= 4 *1.5 = 6.0 N

4 0
3 years ago
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