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grin007 [14]
3 years ago
6

What temperature has the same value in both the fahrenheit and kelvin scales?

Physics
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

574.25 °

Explanation:

To know this, we need to get the expressions to calculate °F and °K.

In the case of Kelvin:

°K = °C + 273

in fahrenheit:

°F = 9/5(°C + 32)

Now, in order to know the temperature which both Kelvin and Fahrenheit are the same, we need to use both above equations, and solve for °C.

°C = °K - 273 (1)

°C = 5/9(°F - 32) (2)

Using 1 and 2 into a same expression:

°K - 273 = 5/9(°F - 32)

With this, we need to know now the moment which K = F, so, all we need to do is replace the F for the K in the above expression. Doing this, we have:

°K - 273 = 5/9(°K - 32)

°K = 0.555(°K - 32) + 273

°K = 5/9K - 17.78 + 273

°K - 5/9K = 255.22

4/9°K = 255.22

°K = 255.22 * 9 / 4

°K = 574.25 °

And this is the temperature in which both Kelvin and Fahrenheit are the same.

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At what distance on the axis of a current loop is the magneticfield half the strength of the field at the center of the loop? Gi
olchik [2.2K]

Answer:

x = 1.26 R

Explanation:

For this exercise let's find the magnetic field using the Biot-Savart law

            B = μ₀ I/4π ∫ ds x r^ / r²

In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop

suppose that the spiral eta in the yz plane, therefore the axis is in the x axis

         B = μ₀ I/4π ∫ ds / (R² + x²)

     

The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is

            B = Bₓ i^

using trigonometry

            Bₓ = B cos θ

we substitute

            Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)

the cosine function is

            cos θ = R /√(x² + R²)

The differential is

            ds = R dθ

we substitute

             Bₓ = μ₀ I/4π ∫ (R dθ)  R /√( (x² + R²)³ )

we integrate from 0 to 2π

              Bₓ =μ₀ I/4π R² / √(x² + R²)³   2pi

therefore the final expression is

            B = μ₀ I R²/ 2√(x² + R²)³   i^

In our case the distance is requested where B is half of B in the center of the bone loop x = 0

Spire center field   x=0

              B₀ = μ₀ I/2R

Field at the desired point (x)

              B = B₀ / 2

               

we substitute

              R² /√(x² + R²)³ = ½  1 /R

              2R³ =√(x² + R²)³

              (x² + R²)³ = 4 (R²)³

              (x²/R² + 1)³ = 4

               

The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)

           B = μ₀ I/2x³ 

we substitute

            R² / x³ = 1/2   1 / R

            2R³ = x³

          x = ∛2  R

            x = 1.2599 R

3 0
3 years ago
What type of charge do individual hair strands have when standing on end due to static electricity?
aalyn [17]

Answer:

C. I think

Explanation:

C. permanent positive charges

8 0
3 years ago
A car accelerates from rest at 3m/s2 along a straight road how far has the car traveled after 4s?
murzikaleks [220]

Answer:

24 m

Explanation:

x = 1/2at^{2} + v_{0}t + x_{0}\\a = 3 m/s^{2}\\v_{0} = 0

7 0
3 years ago
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the thro
koban [17]

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

           

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

4 0
3 years ago
The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radi
guapka [62]

Answer:

(A) E = 2.633 * 10^19 N/C

(B) E = 1.096 * 10^13 N/C

(C) E = -1.096 * 10^13 N/C

Explanation:

Parameters given:

Number of protons, N = 92

Radius of Uranium nucleus = 7.4 * 10^-15 m

Electronic charge, ẹ = 1.6023 * 10^-19 C

Electric field at a point R due to a charge Q is given as

E = (k*Q) / (R^2)

Where k =Coulombs constant

(A) Since there are 92protons,the total Electric field due to the protons will be:

E = (k*e*N) / (R * R)

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (7.4 * 10^-15)^2

E = 2.633 * 10^19 N/C

(B) At the position of the electrons, R = 1.1 * 10^-10m. Therefore, Electric field will be:

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = 1.096 * 10^13 N/C

(C) There are 92 electrons in the Uranium atom and electrons have a charge - e, hence, the Electric field due to the electrons at the nucleus will be:

E = -(k*e*N) / (R * R)

E = -(9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = -1.096 * 10^13 N/C

7 0
3 years ago
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