How much heat is absorbed by a 45g iron skillet when its temperature rises from 6oC to 21oC?

The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex]     (1)

Where:

[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]h=4.136(10)^{-15}eV.s[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c=3(10)^{8}m/s[/tex] the speed of light in vacuum.

[tex]\theta=41\°[/tex] the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are deflected at [tex]41\°[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(41\°))[/tex]   (2)

[tex]\Delta \lambda=\lambda' - \lambda_{o}=5.950(10)^{-13}m[/tex]   (3)

But we are asked to express this in [tex]nm[/tex], so:

[tex]\Delta \lambda=5.950(10)^{-13}m.\frac{1nm}{(10)^{-9}m}[/tex]  

[tex]\Delta \lambda=0.000595nm[/tex]  (4)

The initial energy [tex]E_{o}=265keV=265(10)^{3}eV[/tex] of the photon is given by:

 [tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex]    (5)

From this equation (5) we can find the value of [tex]\lambda_{o}[/tex]:

[tex]\lambda_{o}=\frac{h.c}{E_{o}}[/tex]    (6)

[tex]\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{265(10)^{3}eV}[/tex]    

[tex]\lambda_{o}=4.682(10)^{-12}m[/tex]    (7)

Knowing the value of [tex]\Delta \lambda[/tex] and [tex]\lambda_{o}[/tex], let's find [tex]\lambda'[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}[/tex]

Then:

[tex]\lambda'=\Delta \lambda+\lambda_{o}[/tex]  (8)

[tex]\lambda'=5.950(10)^{-13}m+4.682(10)^{-12}m[/tex]  

[tex]\lambda'=5.277(10)^{-12}m[/tex]  (9)

Knowing the wavelength of the scattered photon [tex]\lambda'[/tex]  , we can find its energy [tex]E'[/tex] :

[tex]E'=\frac{h.c}{\lambda'}[/tex]    (10)

[tex]E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{5.277(10)^{-12}m}[/tex]    

[tex]E'=235.121keV[/tex]    (11) This is the energy of the scattered photon

If we want to know the kinetic energy of the recoiling electron [tex]E_{e}[/tex], we have to calculate all the energy lost by the photon in the wavelength shift, which is:

[tex]K_{e}=E_{o}-E'[/tex]  (12)

[tex]K_{e}=265keV-235.121keV[/tex]  

Finally we obtain the kinetic energy of the recoiling electron:

[tex]E_{e}=29.878keV[/tex]  

Answer:

The first one:

the energy of the scattered x-ray

The answer for last on:

Kinetic energy of the recoiling electron

The federal government and industry are the main sources of funding for research, providing funding to Principal Investigators (PIs) through their organizations. The U.S. economy has increasingly relied on industry-funded research, but the government still plays a significant role in funding research.

The federal government is one of the main sources of funding for research, along with industry. The government provides funding to Principal Investigators (PIs) through the organizations they work for. Over time, the U.S. economy has relied more heavily on industry-funded research and development (R&D). However, the government still plays a significant role in funding research, especially in areas where private firms are not as active.

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Most funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through their organizations.

Explanation:

Most of the funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through the organizations for which they work. The federal government, through agencies such as the National Institutes of Health (NIH) and the National Science Foundation (NSF), provides grants for research in various fields. Industry-funded research, on the other hand, is supported by companies and private entities who invest in research and development projects.

Learn more about Research here:

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The probable question can be: Complete the following sentence. Most of the funding for research comes from the federal government or ______ and is provided to Principal Investigators (PIs) through the organizations for which they work.

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1. 408.4 J

The work done by a gas is given by:

[tex]W=p\Delta V[/tex]

where

p is the gas pressure

[tex]\Delta V[/tex] is the change in volume of the gas

In this problem,

[tex]p=1.01\cdot 10^5 Pa[/tex] (atmospheric pressure)

[tex]\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3[/tex] is the change in volume

So, the work done is

[tex]W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J[/tex]

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

[tex]Q = m \lambda_v[/tex]

where

m is the mass of the water

[tex]\lambda_v = 2.26\cdot 10^6 J/kg[/tex] is the specific latent heat of vaporization

The initial volume of water is

[tex]V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3[/tex]

and the water density is

[tex]\rho = 1000 kg/m^3[/tex]

So the water mass is

[tex]m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg[/tex]

So, the amount of heat added to the water is

[tex]Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J[/tex]

Some microbes can kill or inhibit the growth of other microbes led to the development of the first antibiotic.

Answer: Option A

Explanation:

Antibiotics are the substance that restricts the bacteria growth and replication. These are designed against microbes and it targets bacterial infections in or on a human body. Bacteria's comes under the microbe category that cause harm to the human body.  

The substance that targets, restricts and kills microbial cells includes antibiotics, antiseptics and antiviral. The antibiotics that we use today can be produced in labs and they can be found in nature also. Hence, antibiotics are the one that kills or restricts the microbe's growth.

Answer:

[tex]6.60\cdot 10^5 Nm^2/C[/tex]

Explanation:

The electric flux through the rectangle is given by

[tex]\Phi = E A cos \theta[/tex]

where

E is the electric field strength

A is the area of the rectange

[tex]\theta[/tex] is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

[tex]A=2.50 m \cdot 5.00 m=12.5 m^2[/tex]

and the angle is

[tex]\theta=65.0^{\circ}[/tex]

so, the electric flux is

[tex]\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C[/tex]