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Vladimir [108]
3 years ago
15

What is the formula to calculate the thickness of a test tube?​

Physics
1 answer:
Maslowich3 years ago
4 0

Answer:

Having the inside dimensions (ID) and the outside dimensions (OD) will allow you to figure out the wall thickness on tubing. You would need to subtract the ID from the OD and then divide by two. This number is the wall thickness.

Explanation:

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What are the conditions of the thermonuclear fusion reaction in the sun?​
Verizon [17]

High temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons. Fusion requires temperatures of about 100 million Kelvin (approximately six times hotter than the sun's core).

study \: well

8 0
2 years ago
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?​
myrzilka [38]
(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)
5 0
3 years ago
What is the equation for potential energy
meriva
PE = mgh
where
m = mass
g = acceleration due to gravity
h = height
3 0
3 years ago
If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
Oxana [17]

Answer:

The answer to your question is:      V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

V1 = 1.5 l

V2 = ?

Formula

                          P1V1 = P2V2

                          V2 = (P1V1) / P2

                          V2 = (200 x 1.5) / 300

                          V2 = 1 l

6 0
4 years ago
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
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