Answer:
(c) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.
Explanation:
The formula that gives the angle of the first minimum of the diffraction pattern from a single-slit is
[tex]sin \theta = \frac{\lambda}{w}[/tex]
where
[tex]\lambda[/tex] is the wavelength of the light
w is the width of the slit
We see that the angle is inversely proportional to the width of the slit: therefore, if the width of the slit is reduced (so, w is decreased), the angle that locates the first minimum [tex]\theta[/tex] increases, and so the central bright fringe becomes wider.
Reducing the width of a single slit through which light passes leads to a wider central bright fringe. This occurs because the narrowed slit causes greater divergence of light rays, leading to a larger angle for locating the first dark fringe.
Explanation:
The phenomenon being discussed here pertains to physics, specifically light diffraction and interference. When light passes through a single slit, it exhibits a diffraction pattern with a central bright fringe flanked by smaller, dimmer fringes on either side. This pattern is influenced by the width of the slit and the wavelength of the light source.
For the scenario provided in the question, light passes through a single slit that is then narrowed. The correct answer is (c): The central bright fringe becomes wider because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.
This is explained by the fact that as the slit narrows, the rays of light diverge more upon exiting the slit, due to the wave nature of light. The larger spread of these rays leads to a larger angle for locating the first dark fringe, which subsequently results in a wider central bright fringe.
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Which of he following explains the significance of the observations that some microbes can kill or inhibit the growth of other microbes?
A.) It led to the development of the first antibiotic
B.) It led to the development of the first antiviral
C.) It led to the development of the Theory of Evolution
D.) It led to the development of the Germ Theory of Disease
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.
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.
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.
Explanation: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.
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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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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 106 J/kg . Answer in units of J. 006 (part 2 of 3) 10.0 points Find the amount of heat added to the water to accomplish this process. Answer in units of J.
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]
X-rays with an energy of 265 keV undergo Compton scattering from a target. If the scattered rays are deflected at 41.0° relative to the direction of the incident rays, find each of the following. (a) the Compton shift at this angle _________nm (b) the energy of the scattered x-ray __________keV (c) the kinetic energy of the recoiling electron ___________keV
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)
(b) the energy of the scattered x-rayThe 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
(c) Kinetic energy of the recoiling electron
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
A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle? A) 1.56 × 106 N⋅m2/C B) 6.60 × 105 N⋅m2/C C) 1.42 × 105 N⋅m2/C D) 5.49 × 104 N⋅m2/C E) 4.23 × 104 N⋅m2/C
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]