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tia_tia [17]
3 years ago
10

Which of the following is true about a hypothesis?

Physics
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

i think you need to show a picture

Explanation:

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I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph
arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

I_{CM} = \dfrac{2}{5}mR^2.....(1)

now, moment of inertia about tangent

I= \dfrac{2}{5}mR^2 + mR^2

I= \dfrac{7}{5}mR^2...........(2)

dividing equation (1)/(2)

\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}

\dfrac{I_{CM}}{I}=\dfrac{2}{7}

I_{CM}=\dfrac{2}{7}I

the correct answer is option E

4 0
3 years ago
Will an electromagnet be obtained if steel is placed inside the solenoid?​
iogann1982 [59]

Answer:A solenoid is a simple electromagnetic device consisting of a coiled electric wire, wrapped in a 3D circular pattern. When electric current is passed through the wire, the solenoid acts like a magnet with N and S poles at the ends of the coil.

When a ferromagnetic material rod is permanently placed inside the solenoid, the metal greatly increases the magnetic effect and becomes a permanent electromagnet. Moreover, it can also be used as an electrical switch by drawing in or pushing out a ferromagnetic material like an iron rod. Depending on the directions of the rod and the electrical current the switching action takes place.

Given figure  represents the solenoid as electromagnet and the switching action.

Explanation:

7 0
3 years ago
A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
lora16 [44]

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

5 0
3 years ago
This experiment deals with the relationship between volume and temperature for a fixed amount of gas at constant pressure. Desig
creativ13 [48]

Your experiment should keep one thing constant and measure the other. So vary the temp and measure the pressure. You will get a set of data that relates pressure with temp.
<span>PV = nRT
So
P and T are directly proportional.
</span>These experiments are one of either Boyle-Mariottte's, Gay-Lussac'a or Charles' law.

8 0
3 years ago
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