Ultraviolet (UV) radiation is a part of the electromagnetic spectrum that reaches the Earth from the Sun. It has wavelengths shorter than those of visible light, making it invisible to the naked eye. These wavelengths are classified as UVA, UVB, or UVC, with UVA the longest of the three at 317 nm to 400 nm. Both the U.S. Department of Health and Human Services and the World Health Organization have identified UV as a proven human carcinogen. Many experts believe that, especially for fair-skinned people, UV radiation frequently plays a key role in melanoma, the deadliest form of skin cancer, which kills more than 8000 Americans each year. UVB has a wavelength between 280 nm and 317 nm. Determine the frequency ranges of UVA and UVB. UVA Hz (smaller value) Hz (larger value) UVB Hz (smaller value) Hz (larger value)

Final answer:

The change in kinetic energy of the bullet-block system can be calculated using the principle of conservation of kinetic energy. The initial kinetic energy of the bullet is given by 0.5 * mass of the bullet * (initial velocity of the bullet)^2, and the final kinetic energy is given by 0.5 * mass of the bullet * (final velocity of the bullet)^2. The change in kinetic energy is the difference between the final and initial kinetic energies.

Explanation:

The change in kinetic energy of the bullet-block system can be calculated using the principle of conservation of kinetic energy. The initial kinetic energy of the bullet is given by 0.5 * mass of the bullet * (initial velocity of the bullet)^2, and the final kinetic energy is given by 0.5 * mass of the bullet * (final velocity of the bullet)^2. The change in kinetic energy is the difference between the final and initial kinetic energies.

Using the values given in the question:

Mass of the bullet = 22.3 g = 0.0223 kg

Initial velocity of the bullet = 1000 m/s
Final velocity of the bullet = 100 m/s

Initial kinetic energy of the bullet = 0.5 * 0.0223 kg * (1000 m/s)^2 = 111.5 J
Final kinetic energy of the bullet = 0.5 * 0.0223 kg * (100 m/s)^2 = 11.15 J

Change in kinetic energy of the bullet-block system = Final kinetic energy - Initial kinetic energy = 11.15 J - 111.5 J = -100.35 J

Answer:

C) 21 m/s

Explanation:

The general formula of the Doppler effect is:

[tex]f'=(\frac{v+v_r}{v+v_s})f[/tex]

where

f' is the apparent frequency

f is the original frequency

v is the speed of the wave in the medium

[tex]v_r[/tex] is the velocity of the receiver, positive if the receiver is moving towards the source

[tex]v_s[/tex] is the velocity of the source, positive if the source is moving away from the receiver

Here we have

f = 440 Hz

f' = 415 Hz

v = 343 m/s

[tex]v_r = 0[/tex] (the observer is stationary)

[tex]v_s[/tex] is positive since we are considering when the train has passed the observer, so it is moving away from him

So we can rewrite the formula as

[tex]f'=(\frac{v}{v+v_s})f[/tex]

And solving for [tex]v_s[/tex], we find the speed of the train

[tex]v_s = v(\frac{f}{f'}-1)=(343 m/s)(\frac{440 Hz}{415 Hz}-1)=20.7 m/s \sim 21 m/s[/tex]

By applying Doppler's effect of a wave, the speed of this train is equal to 21 m/s.

Given the following data:

Maximum frequency = 440 Hz.

Apparent frequency = 415 Hz.

Speed of sound in air = 343 m/s.

Observer speed = 0 m/s (since his stationary).

Doppler effect can be defined as the change in frequency of a wave with respect to an observer that is in motion and moving relative to the source of the wave.

Mathematically, Doppler's effect of a wave is given by this formula:

[tex]F_o = \frac{V+V_r}{V+V_s}F[/tex]

Substituting the given parameters into the formula, we have;

[tex]415 = \frac{343+0}{343+V_s} \times 440\\\\142345+415V_s=150920\\\\415V_s=150920-142345\\\\415V_s=8575\\\\V_s=\frac{8575}{415} \\\\V_s=20.66[/tex]

Speed = 20.66 ≈ 21 m/s.

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A) [tex]1.36\cdot 10^{-4}T[/tex]

The magnetic field at the center of a coil of N turns is given by

[tex]B=\frac{\mu_0 N I}{2R}[/tex]

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

[tex]R=\frac{6.0 m}{2}=3.0 m[/tex] is the radius of the coil

Substituting,

[tex]B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T[/tex]

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) [tex]3.26\cdot 10^{-23}N[/tex]

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

[tex]F=qvB[/tex]

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

[tex]q=+e=1.6\cdot 10^{-19} C[/tex] is the charge

[tex]v=1.5 m/s[/tex] is the velocity

[tex]B=1.36\cdot 10^{-4}T[/tex] is the magnetic field strength

Substituting,

[tex]F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N[/tex]

The magnetic field at the center of the coil is 0.34 T. The force on a positive ion in the water flowing above the coil points North. Without the value of the charge, the magnitude of the force on an ion cannot be determined.

Part A) The magnitude of the magnetic field at the center of the coil, given that the field strength, B is calculated by the equation B = μI/(2r), where μ is the permeability of free space (4*10^-7 T.m/A), I is the current (6.5A), and r is the radius of the loop, which is the diameter divided by 2 (3m), is approximated as 0.34 T (Tesla).

Part B) The direction of the force on a positive ion in the water above the center of the coil is determined by the Right-Hand Rule-2. Given that the water is flowing east and the field is directed downward, the force on a positive ion would point to the north.

Part C) The magnitude of the force on an ion with a charge +e in this electromagnetic field can be calculated with the equation F = qvB, where q is the charge (+e), v is the velocity of the ion (1.5 m/s), and B is the magnetic field. However, without the actual value of the charge, this can't be determined from the current information.

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Answer:

B

Explanation:

In a DC current, the current supplied is steady motion (a straight horizontal line in a graph) and this is what makes it had to transform. The alternating current takes a sine wave motion in a graph. This means is voltage varies from zero to peak and reverses polarity. The rate at which it achieves this is its frequency in Hertz.

Final answer:

DC current is a unidirectional flow of electric charge, unlike AC current which periodically reverses direction. The true statement about DC current is that there can only be one voltage supplied in a DC circuit (option B).

Explanation:

The question concerns the characteristics of DC current. Direct current, or DC, is the flow of electric charge in only one direction, which is characteristic of systems with a constant voltage source, such as a battery. Unlike alternating current, or AC, which periodically reverses direction, DC's flow is unidirectional. The given statements regarding DC current must be assessed for their correctness. To clarify:

Given this information, we can conclude that the true statement regarding DC current is B. There can only be one voltage supplied in a DC circuit.

A) [tex]3.6\cdot 10^5 J[/tex]

The power used by an object is defined as

[tex]P=\frac{E}{t}[/tex]

where

E is the energy used

t is the time elapsed

In this problem, we have

P = 100 W is the power of the light bulb

t = 1 h = 3600 s is the time elapsed

Solving for E, we find the amount of energy used by the light bulb:

[tex]E=Pt = (100 W)(3600 s)=3.6\cdot 10^5 J[/tex]

B) [tex]1.1 \cdot 10^2 m/s[/tex]

The kinetic energy of an object is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is the mass

v is the speed of the object

In this problem, we have a man of mass

m = 65 kg

we want its kinetic energy to be

[tex]E=3.6\cdot 10^5 J[/tex]

Therefore, we can calculate its speed from the previous formula:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.6\cdot 10^5 J)}{65 kg}}=105.2 m/s = 1.1 \cdot 10^2 m/s[/tex]

A 100-watt light bulb uses 360,000 joules of energy per hour. A 65 kg person would need to run at approximately 105 m/s to have the same amount of kinetic energy.

Energy = Power × Time = 100 W × 3600 s = 360000 J

Therefore, a 100-watt light bulb uses 360,000 joules of energy per hour.

KE = ½ mv²

Where m is mass in kilograms and v is velocity in meters per second. Plugging in the values, we can solve for v:

360,000 J = ½ × 65 kg × v²

v² = (2 × 360,000 J) / 65 kg = 11076.92 m²/s²

v = √11076.92 m²/s²

v ≈ 105 m/s

Hence, a 65 kg person would need to run at approximately 105 meters per second to have 360,000 joules of kinetic energy, which is equivalent to the energy used by a 100-watt light bulb in one hour.

Answer:

(c) The current in each resistor is the same.

Explanation:

When two resistors are connected in series, we have the following:

- The resistors are connected such that the current passing through the two resistors is the same

- The voltage of the battery is equal to the sum of the voltage drops across each resistor

- the equivalent resistance of the circuit is equal to the sum of the individual resistances:

R = R1 + R2

So, let's analyze each statement:

(a) The resistor with the smaller resistance carries more current than the other resistor. --> FALSE. The current through the two resistors is the same.

(b) The resistor with the larger resistance carries less current than the other resistor. --> FALSE. The current through the two resistors is the same.

(c) The current in each resistor is the same. --> TRUE.

(d) The potential difference across each resistor is the same. --> FALSE: the potential difference across each resistor is given by

V=RI

where I (the current) is the same for both resistors, while R (the resistance) is not, so V is also different for the two resistors.

(e) The potential difference is greatest across the resistor closest to the positive terminal --> FALSE. According to

V=RI

the potential difference depends only on the value of the resistance, so it doesn't matter which resistor is connected to the positive terminal.

In a series circuit, the resistors have the same current flowing through them but can have different potential differences. The resistor with smaller resistance carries more current, while the one with larger resistance carries less current. The potential difference across each resistor is not the same.

In a series circuit, the resistors are connected one after the other, so the current passing through each resistor is the same. However, the potential difference (voltage) across each resistor can be different depending on their resistances.

Therefore, the correct statements are:

(a) The resistor with the smaller resistance carries more current than the other resistor. Since the current is the same in both resistors, the one with smaller resistance will have a larger potential difference across it. This means it carries more current.

(b) The resistor with the larger resistance carries less current than the other resistor. Again, since the current is the same in both resistors, the one with larger resistance will have a smaller potential difference across it. This means it carries less current.

(d) The potential difference across each resistor is the same. This statement is incorrect. As mentioned before, the potential difference across each resistor can be different depending on their resistances.

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Answer:

[tex]3.4\cdot 10^4 N/m[/tex]

Explanation:

The spring system in the taptap obey's Hooke's law, which states that:

[tex]F=kx[/tex]

where

F is the magnitude of the force applied

k is the spring constant

x is the compression/stretching of the spring

In this problem:

- The force applied is the weight of the driver of mass m = 69 kg, so

[tex]F=mg=(69 kg)(9.8 m/s^2)=676.2 N[/tex]

- The compression of the spring is

[tex]x=2\cdot 10^{-2} m=0.02 m[/tex]

So, the spring constant is

[tex]k=\frac{F}{x}=\frac{676.2 N}{0.02 m}=3.4\cdot 10^4 N/m[/tex]

The spring constant of the spring compressed 2x10⁻² m by a 69 kg driver is 3.4x10⁴ N/m.    

We can use the Hooke's law equation to find the spring constant:

[tex] F = -kx [/tex]   (1)  

Where:

F: is the Hooke's force

k: is the spring constant =?

x: is the distance of compression of the spring = -2x10⁻² m. The negative sign is because it is compressing (negative direction).

The force of equation (1) is equal to the weight force (W) of the driver, so:

[tex] W = F [/tex]  

[tex] mg = -kx [/tex]   (2)

Where:

m: is the mass of the driver = 69 kg

g: is the acceleration due to gravity = 9.81 m/s²

Solving equation (2) for k, we have:

[tex] k = -\frac{mg}{x} = -\frac{69 kg*9.81 m/s^{2}}{-2\cdot 10^{-2} m} = 3.4 \cdot 10^{4} N/m [/tex]  

Therefore, the spring constant is 3.4x10⁴ N/m.

You can learn more about Hooke's law here:

I hope it helps you!                  

Answer:

[tex]7.96 \Omega[/tex]

Explanation:

First of all, we can calculate the power dissipated by the resistor. We have:

E = 181 J is the energy produced

t = 9.99 s is the time interval

So, the power dissipated is

[tex]P=\frac{E}{t}=\frac{181 J}{9.99 s}=18.1 W[/tex]

But the power dissipated can also be written as

[tex]P=\frac{V^2}{R}[/tex]

where

V = 12.0 V is the potential difference across the resistor

R is the resistance

Solving for R, we find

[tex]R=\frac{V^2}{P}=\frac{(12.0 V)^2}{18.1 W}=7.96 \Omega[/tex]

Each automobile adds 0.24 kg of chlorine to the atmosphere each year due to CFC-12 leakage. The total amount of chlorine added each year is 24,000,000 kg.

To calculate the amount of chlorine added to the atmosphere each year due to auto air conditioners, we need to consider the amount of CFC-12 leaked by each automobile.

Given that each automobile contains 1.2 kg of CFC-12 and leaks 20% per year, we can calculate the amount of chlorine added per automobile per year:

Chlorine added per automobile per year = 1.2 kg of CFC-12 x 20% = 0.24 kg.

Since there are 100 million automobiles, the total amount of chlorine added to the atmosphere each year due to auto air conditioners is:

Total chlorine added per year = Chlorine added per automobile per year * Number of automobiles = 0.24 kg * 100,000,000 = 24,000,000 kg.

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Hello! :)

A study of motion is basically like physics.

Hence, using the telescope, Galileo discovered the mountains on the moon, the spots on the sun, and four moons of Jupiter. His discoveries provided the evidence to support the theory that the earth and other planets revolved around the sun.

Hope this helped and I hope I answered in time!

Good luck!

~ Destiny ^_^

Answer:

[tex]v=14.5(m/s)[/tex].

Explanation:

This is a projectile motion problem, so for solving it, we will require some equations of uniform motion and free-fall.

First, we need to recognize the data that we have so we can start solving the problem.

For Free -Fall:

[tex]g= -9.8(m/s^{2})[/tex]

[tex]\Delta y=2.4(m)[/tex]

For uniform motion:

[tex]\Delta x=17(m)[/tex]

The time that takes to the ball to travel 17m horizontally and to hit the crossbar at 2.4m of height is the same, so time is a common variable for free-fall and uniform motion.

Using the equations:

[tex]\Delta y=v_{y0}t+0.5gt^{2}[/tex] and

[tex]\Delta x=v_{x}t[/tex]

and noticing that

[tex]v_{y0}=v*sin(38)[/tex] and

[tex]v_{x}=v*cos(38)[/tex],

we obtain

[tex]\Delta y=v*sin(38)t+0.5gt^{2}[/tex] and

[tex]\Delta x=v*cos(38)t[/tex]

wich is a system of two equations and to variables that can be easily solve.

Making

[tex]v=\frac{\Delta x}{t*cos(38)}[/tex]

we get

[tex]\Delta y=(\frac{\Delta x}{t*cos(38)})sin(38)t+0.5gt^{2}[/tex],

[tex]\Delta y=\Delta x*tan(38)+0.5gt^{2}[/tex],

[tex]\Delta y-\Delta x*tan(38)=0.5gt^{2}[/tex],

[tex]t=\sqrt{\frac{\Delta y-\Delta x*tan(38)}{0.5g}}[/tex],

[tex][tex]t=\sqrt{\frac{2.4-17*tan(38)}{0.5*(-9.8)}}[/tex]

so

[tex]t=1.50s[/tex],

now the speed can be easily compute from one of our equations. Using

[tex]v=\frac{\Delta x}{t*cos(38)}[/tex],

[tex]v=\frac{17}{1.5*cos(38)}[/tex],

[tex]v=14.5(m/s)[/tex].

To find the initial speed of the ball, we can analyze the projectile motion of the ball. We can break down the initial velocity of the ball into horizontal and vertical components, and then use kinematic equations to solve for the initial speed. We can calculate the time it takes for the ball to reach the goal crossbar and then use that time and the horizontal distance to find the initial speed.

To find the initial speed of the ball, we need to analyze the projectile motion of the ball. Given that the ball passes over the goal, 2.4 m above the ground, we can use the kinematic equations to solve for the initial speed. In this case, we know the horizontal distance (17 m) and the vertical distance (2.4 m above the ground). We can break down the initial velocity of the ball into horizontal and vertical components.

The horizontal component of the initial velocity (Vx) can be found using the equation:

Vx = V * cos(theta)

where V is the initial speed and theta is the angle of the kick. The vertical component of the initial velocity (Vy) can be found using the equation:

Vy = V * sin(theta)

Using these values, we can calculate the time it takes for the ball to reach the goal crossbar, which is the same as the time it takes for the ball to hit the ground. The time can be found using the equation:

t = 2 * Vy / g

where g is the acceleration due to gravity. Finally, we can use the time and the horizontal distance to find the initial speed using the equation:

Vx = distance / t

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Answer:

5 mg, [tex]5\cdot 10^{-3}g[/tex]

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

[tex]m=5\cdot 10^{-3}g[/tex]

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

[tex]1 g : 1000 mg = 0.005 g : x[/tex]

and we find

[tex]x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg[/tex]

To express the mass of the pill in grams, you can write it in scientific notation as 5.0 x 10^-3 g or in milligrams as 5 mg.

To express the mass of the pain-relieving pill in grams, we can either use scientific notation or a metric prefix. The mass of the pill is given as 0.005 g. To express it in scientific notation, we write it as 5.0 x 10-3 g. This means the mass is 5.0 times 10 raised to the power of -3 grams. Alternatively, we can express the mass in milligrams. Since 1 gram is equal to 1000 milligrams, the mass of the pill can be written as 5 mg.

Explanation:

Photons have momentum, this was proved by he American physicist Arthur H. Compton after his experiments related to the scattering of photons from electrons (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, 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=30\°[/tex] the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at [tex]30\°[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))[/tex]   (2)

[tex]\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m[/tex]   (3)

Now, the initial energy [tex]E_{o}=400keV=400(10)^{3}eV[/tex] of the photon is given by:

 [tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex]    (4)

From this equation (4) we can find the value of [tex]\lambda_{o}[/tex]:

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

[tex]\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}[/tex]    

[tex]\lambda_{o}=3.102(10)^{-12}m[/tex]    (6)

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]  (7)

[tex]\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m[/tex]  

[tex]\lambda'=3.427(10)^{-12}m[/tex]  (8)

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]    (9)

[tex]E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}[/tex]    

[tex]E'=362.063keV[/tex]    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron [tex]E_{e}[/tex], we have to calculate all the energy lost by the photon, which is:

[tex]E_{e}=E_{o}-E'[/tex]  (11)

[tex]E_{e}=400keV-362.063keV[/tex]  

Finally we obtain the energy of the recoiling electron:

[tex]E_{e}=37.937keV[/tex]  

The energy of the recoiling electron after the Compton scattering is 38 KeV.

The wavelength of the photons after the Compton scattering is calculated as follows;

λ = λ₀ + λc(1 - cosθ)

[tex]\lambda' = \frac{hc}{E_0} \ + \ 2.43 \times 10^{-12}(1 - cos\theta)\\\\\lambda' = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400,000 \times 1.6 \times 10^{-19}} \ + \ 2.43 \times 10^{-12}(1 - cos30)\\\\\lambda ' = 3.108 \times 10^{-12} \ + \ 3.26 \times 10^{-13}\\\\\lambda ' = 3.434 \times 10^{-12} \ m[/tex]

The energy of photons after the Compton scattering is calculated as follows;

[tex]E ' = \frac{hc}{\lambda '} \\\\E ' = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.434 \times 10^{-12} } \\\\E ' = 5.79\times 10^{-14} \ J\\\\E' = \frac{5.79\times 10^{-14} \ J}{1.6\times 10^{-19} J/eV} = 362,000 \ eV = 362 \ KeV[/tex]

The energy of the recoiling electron is the change in the energy of the photons.

E = 400 KeV - 362 KeV

E = 38 KeV

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The angle between the vector and the positive x axis is approximately 60°.

The vector points into the third quadrant, which means its x and y components are both negative. Let's assume that the magnitude of the vector is represented by |A| and the magnitude of the x component is represented by |Ax|. The given condition states that |A| = 2 * |Ax|. Using this information, we can find the angle between the vector and the positive x axis.

First, we need to find the values of |A| and |Ax|. Since |A| = 2 * |Ax|, we can substitute this into the Pythagorean theorem:

|A|² = |Ax|² + |Ay|²

Now let's substitute the given information into the equation:

(2 * |Ax|)² = |Ax|² + (|Ay|)²

Expanding and simplifying the equation:

4 * |Ax|² = |Ax|² + (|Ay|)²

3 * |Ax|² = (|Ay|)²

Now, we can take the square root of both sides:

√(3 * |Ax|²) = √((|Ay|)²)

√3 * |Ax| = |Ay|

Since both |Ax| and |Ay| are negative, we can ignore the negative sign. Therefore, |Ax| = -1 and |Ay| = -√3.

Now, we have all the information we need to find the angle between the vector and the positive x axis. We can use the formula: tan(A) = |Ay|/|Ax| = (-√3)/(-1) = √3

The angle A is the inverse tangent of √3: A = atan(√3) ≈ 60°

So, the angle between the vector and the positive x axis is approximately 60°.

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

[tex]E=\frac{\sigma}{k \epsilon_0}[/tex] (1)

where

[tex]\sigma[/tex] is the surface charge density

k is the dielectric constant

[tex]\epsilon_0[/tex] is the vacuum permittivity

The area of the plates in this capacitor is

[tex]A=100 cm^2 = 100\cdot 10^{-4} m^2[/tex]

while the charge is

[tex]Q=8.9\cdot 10^{-7}C[/tex]

So the surface charge density is

[tex]\sigma = \frac{Q}{A}=\frac{8.9\cdot 10^{-7} C}{100\cdot 10^{-4} m^2}=8.9\cdot 10^{-5} C/m^2[/tex]

The electric field is

[tex]E=1.4\cdot 10^6 V/m[/tex]

So we can re-arrange eq.(1) to find k:

[tex]k=\frac{\sigma}{E \epsilon_0}=\frac{8.9\cdot 10^{-5} C/m^2}{(1.4\cdot 10^6 V/m)(8.85\cdot 10^{-12} F/m)}=7.18[/tex]

(b) [tex]7.66\cdot 10^{-7}C[/tex]

The surface charge density induced on each dielectric surface is given by

[tex]\sigma' = \sigma (1-\frac{1}{k})[/tex]

where

[tex]\sigma=8.9\cdot 10^{-5} C/m^2[/tex] is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

[tex]\sigma' = (8.9\cdot 10^{-5} C/m^2) (1-\frac{1}{7.18})=7.66\cdot 10^{5} C/m^2[/tex]

And by multiplying by the area, we find the charge induced on each surface:

[tex]Q' = \sigma' A = (7.66\cdot 10^{-5} C/m^2)(100 \cdot 10^{-4}m^2)=7.66\cdot 10^{-7}C[/tex]

Final answer:

The dielectric constant cannot be determined without the electric field value without the dielectric (E0). The induced charge on each dielectric surface can be calculated once the electric field (E) is given, using the permittivity of free space and the area of the plates.

Explanation:

To solve the student's question regarding the dielectric constant and the induced charge on a dielectric material, we need to apply concepts from electrodynamics and the properties of capacitors.

Dielectric Constant (k)

The dielectric constant (k) is given by the ratio of the electric field without the dielectric (E0) to the electric field with the dielectric (E). Here, you provided us with the electric field within the dielectric material (E) which is 1.4 × 106 V/m, but have not given us the electric field without the dielectric (E0). To find k, we would normally use the following equation:

k = E0 / E

Without knowing E0, we cannot calculate the dielectric constant directly.

Induced Charge (Qi)

We can calculate the induced charge on each surface of the dielectric (Qi) using the relation between electric field (E), surface charge density (σ), and the permittivity of free space (ε0).

E = σ / ε0

From that, the induced surface charge density is:

σi = E × ε0

Where the permittivity of free space (ε0) is approximately 8.85 × 10-12 F/m. After calculating σi, we can find Qi by multiplying σi with the plate area (A), which you've given as 100 cm2 or 0.01 m2.

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

v²=490m²/s²

v=22.14m/s which can be approximated to 22m/s

a voltage in the secondary coil

Answerr

Explanation:

b

Explanation:

Let's begin by explaining that the Work [tex]W[/tex] done by a Force [tex]F[/tex] refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a distance [tex]y[/tex].  When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

[tex]W=F.y[/tex] (1)

In the case of this rectangular aquarium (see figure attached), we know its initial volume [tex]V_{i}[/tex] (which is the same volume occupied by water, because the tank is full) is:

[tex]V_{i}=4m.1m.1m=4m^{3}[/tex] (2)

In addition, we know the force needed to pump a small amount of water is:

[tex]F=m_{water}.g[/tex] (3)

On the other hand, we know the density of the water [tex]\rho_{water}[/tex] is given by the following equation:

[tex]\rho_{water}=\frac{m_{water}}{V}[/tex] (4)

Where [tex]m_{water}[/tex] is the mass of water and [tex]V=4m.1m.dy[/tex] is the volume of a thin "sheet" of water.

Finding [tex]m_{water}[/tex]:

[tex]m_{water}=\rho_{water}.V[/tex]  (5)

Substituting (5) in (3):

[tex]F=\rho_{water}.V.g[/tex]   (6)

And substituting (6) in (1):

[tex]W=\rho_{water}.V.g.y[/tex] (7)

Now, we are asked to find the work needed to pump half of the water out of the aquarium. So, if the aquarium is [tex]1m[/tex] deep, the half is [tex]0.5m[/tex]:

[tex]W=(1000\frac{kg}{m^{3}})(4m.1m.dy)(9.8\frac{m}{s^{2}})y[/tex] (8)

[tex]W=39200\frac{kg.m^{6}}{s^{2}}ydy[/tex] (9)

Well, in order to solve this, we have to write the definite integral from y=0 mto y=0.5m:

[tex]W=39200\frac{kg.m^{6}}{s^{2}}\int\limits^{0.5}_0 {y} \, dy[/tex] (10)

Knowing [tex]\int\limits^b_a {f(y)} \, dy= F(b)-F(a)[/tex]:

[tex]W=39200\frac{kg.m^{6}}{s^{2}}(\frac{(0.5m)^{2}}{2}-\frac{(0m)^{2}}{2})[/tex] (11)

[tex]W=4900kg\frac{m^{2}}{s^{2}}[/tex] (12)

[tex]W=4900J[/tex] >>>>This is the work needed to pump half of the water out of the aquarium

Work is the product of force and displacement. The amount of work needed to pump half of the water out of the aquarium is 4900 J.

Work done can be defined as the amount of force needed to displace an object from one location to another.

[tex]W = F \cdot ds[/tex]

Given to us

Volume of the aquarium, V = 4 x 1 x s = 4m³

Height of the aquarium, s = 1 m

Acceleration due to gravity, g = 9.8 m/s²

Density of water, ρ = 1000 kg/m³

We know about work done, it is given as,

[tex]W = F \cdot ds[/tex]

We also know that force can be written as,

[tex]F = m \cdot a[/tex]

Also, the mass can be written as,

[tex]m = \rho \times V[/tex]

Therefore, work can be written as,

[tex]W = F\cdot ds\\\\W = m \cdot a \cdot ds\\\\W = (\rho \times V \times a )ds\\\\W = \int (\rho \times V \times a )ds[/tex]

As we need to pump half of the water, therefore, from below the tank to half the distance

[tex]W = \int_0^{\frac{1}{2}} (\rho \times V \times a )ds\\\\W = \int_0^{\frac{1}{2}} (\rho \times (4 \times 1 \times s) \times a )ds\\\\W = (\rho \times 4 \times a )[s^2]_0^{\frac{1}{2}\\\\[/tex]

Substitute all the values,

[tex]W = 1000\times 4 \times g \times[(\dfrac{0.5}{2}^2) -0^2]\\\\W = 1000\times 4 \times 9.8 \times 0.125\\\\W = 4900\ J[/tex]

Hence, the amount of work needed to pump half of the water out of the aquarium is 4900 J.

Learn more about Work:

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(a) [tex]3.77\cdot 10^5 MeV, 1.51\cdot 10^6 MeV, 3.39\cdot 10^3 GeV[/tex]

The energy levels of an electron in a box are given by

[tex]E_n = \frac{n^2 h^2}{8mL^2}[/tex]

where

n is the energy level

[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant

[tex]m=9.11\cdot 10^{-31}kg[/tex] is the mass of the electron

[tex]L=1.0\cdot 10^{-15} m[/tex] is the size of the box

Substituting n=1, we find the energy of the ground state:

[tex]E_1 = \frac{1^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=6.03\cdot 10^{-8}J[/tex]

Converting into MeV,

[tex]E_1 = \frac{6.03\cdot 10^{-8} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =3.77\cdot 10^5 MeV[/tex]

Substituting n=2, we find the energy of the first excited state:

[tex]E_2 = \frac{2^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=2.41\cdot 10^{-7}J[/tex]

Converting into MeV,

[tex]E_2 = \frac{2.41\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =1.51\cdot 10^6 MeV[/tex]

Substituting n=3, we find the energy of the second excited state:

[tex]E_3 = \frac{3^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=5.43\cdot 10^{-7}J[/tex]

Converting into GeV,

[tex]E_3 = \frac{5.43\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-9} GeV/eV =3.39\cdot 10^3 GeV[/tex]

(b) [tex]1.10 \cdot 10^{-18} m[/tex]

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

[tex]E=E_2 - E_1 = 2.41\cdot 10^{-7}J - 6.03\cdot 10^{-8} J=1.81\cdot 10^{-7} J[/tex]

And the energy of the electromagnetic radiation is

[tex]E=\frac{hc}{\lambda}[/tex]

where c is the speed of light; so, re-arranging the formula, we find the wavelength:

[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.81\cdot 10^{-7}J}=1.10 \cdot 10^{-18} m[/tex]

(c) [tex]6.59 \cdot 10^{-19} m[/tex]

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

[tex]E=E_3 - E_2 = 5.43\cdot 10^{-7} J - 2.41\cdot 10^{-7}J =3.02\cdot 10^{-7} J[/tex]

Using the same formula as before, we find the corresponding wavelength:

[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.02\cdot 10^{-7}J}=6.59 \cdot 10^{-19} m[/tex]

(d) [tex]4.12 \cdot 10^{-19} m[/tex]

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

[tex]E=E_3 - E_1 = 5.43\cdot 10^{-7} J - 6.03\cdot 10^{-8}J =4.83\cdot 10^{-7} J[/tex]

Using the same formula as before, we find:

[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.83\cdot 10^{-7}J}=4.12 \cdot 10^{-19} m[/tex]

Answer:56 N

Explanation:

Weight is equal to mass times the acceleration of gravity.  On earth, that's 9.8 m/s².

W = mg

W = (5.7 kg) (9.8 m/s²)

W = 55.86 N

Since the mass is precise to 2 significant figures, we should round our answer to 2 significant figures: W = 56 N.

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

This is what Einstein proposed:  

Light behaves like a stream of particles called photons with an energy  [tex]E[/tex]:

[tex]E=h.f[/tex] (1)  

So, the energy [tex]E[/tex] of the incident photon must be equal to the sum of the Work function [tex]\Phi[/tex] of the metal and the kinetic energy [tex]K[/tex] of the photoelectron:  

[tex]E=\Phi+K[/tex] (2)  

Where [tex]\Phi[/tex] is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal.  

In this case [tex]\Phi=2eV[/tex]  and [tex]K_{1}=4eV[/tex]

So, for the first light source of wavelength [tex]\lambda_{1}[/tex], and  applying equation (2) we have:

[tex]E_{1}=2eV+4eV[/tex]   (3)  

[tex]E_{1}=6eV[/tex]   (4)  

Now, substituting (1) in (4):  

[tex]h.f=6eV[/tex] (5)  

Where:  

[tex]h=4.136(10)^{-15}eV.s[/tex] is the Planck constant

[tex]f[/tex] is the frequency  

Now, the frequency has an inverse relation with the wavelength

[tex]\lambda_{1}[/tex]:  

[tex]f=\frac{c}{\lambda_{1}}[/tex] (6)  

Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum  

Substituting (6) in (5):  

[tex]\frac{hc}{\lambda_{1}}=6eV[/tex] (7)  

Then finding [tex]\lambda_{1}[/tex]:  

[tex]\lambda_{1}=\frac{hc}{6eV } [/tex] (8)  

[tex]\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}[/tex]  

We obtain the wavelength of the first light suorce [tex]\lambda_{1}[/tex]:  

[tex]\lambda_{1}=2.06(10)^{-7}m[/tex]   (9)

Now, we are told the second light source [tex]\lambda_{2}[/tex]  has the double the wavelength of the first:

[tex]\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)[/tex]   (10)

Then: [tex]\lambda_{2}=4.12(10)^{-7}m[/tex]   (11)

Knowing this value we can find [tex]E_{2}[/tex]:

[tex]E_{2}=\frac{hc}{\lambda_{2}}[/tex]   (12)

[tex]E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}[/tex]   (12)

[tex]E_{2}=3.011eV[/tex]   (13)

Knowing the value of [tex]E_{2}[/tex] and [tex]\lambda_{2}[/tex], and knowing we are working with the same work function, we can finally find the maximum kinetic energy [tex]K_{2}[/tex] for this wavelength:

[tex]E_{2}=\Phi+K_{2}[/tex] (14)  

[tex]K_{2}=E_{2}-\Phi[/tex] (15)  

[tex]K_{2}=3.011eV-2eV [/tex]  

[tex]K_{2}=1.011 eV[/tex]  This is the maximum kinetic energy for the second light source

Answer:

532 J

Explanation:

If there are no frictional forces/air resistance involved in the problem, then the mechanical energy of the system is conserved.

This means that:

[tex]E_i = E_f[/tex]

where E_i is the initial mechanical energy and E_f is the final mechanical energy. The mechanical energy is the sum of potential energy U and kinetic energy K:

[tex]U_i + K_i = U_f + K_f[/tex]

At the highest point, the speed of the swing is zero, so

[tex]K_i = 0[/tex]

while at the bottom point, the potential energy is zero (if we take the bottom point of the swing as reference level), so

[tex]U_f =0[/tex]

This means that the previous equation becomes

[tex]U_i = K_f[/tex]

and since

[tex]U_i = 532 J[/tex]

the kinetic energy at the bottom of its swing is

[tex]K_f = 532 J[/tex]

1) 13.8 V

We can use the transformer equation:

[tex]\frac{N_p}{N_s}=\frac{V_p}{V_s}[/tex]

where we have

[tex]N_p = 300[/tex] is the number of turns in the primary coil

[tex]N_s=18[/tex] is the number of turns in the secondary coil

[tex]V_p=230.0 V[/tex] is the voltage in the primary coil

[tex]V_s = ?[/tex] is the voltage in the secondary coil

Solving for Vs, we find

[tex]V_s = \frac{N_s}{N_p}V_p=\frac{18}{300}(230.0 V)=13.8 V[/tex]

2) 5.17 A

For an ideal transformer, the power in input is equal to the power in output, so we can write:

[tex]P_{in} = P_{out}\\V_p I_p = V_s I_s[/tex]

where

[tex]V_p=230.0 V[/tex] is the voltage in the primary coil

[tex]V_s = 13.8 V[/tex] is the voltage in the secondary coil

[tex]I_p=0.31 A[/tex] is the current in the primary coil

[tex]I_s = ?[/tex] is the current in the secondary coil

Solving for Is, we find

[tex]I_s = \frac{I_p V_p}{V_s}=\frac{(0.31 A)(230.0 V)}{13.8 V}=5.17 A[/tex]

10.8 seconds is the correct answ

Answer:

[tex]6.03\cdot 10^{-12} m[/tex]

Explanation:

First of all, we need to find the initial wavelength of the photon.

We know that its energy is

[tex]E=301 keV = 4.82\cdot 10^{-14}J[/tex]

So its wavelength is given by:

[tex]\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m[/tex]

The formula for the Compton scattering is:

[tex]\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)[/tex]

where

[tex]\lambda[/tex] is the original wavelength

h is the Planck constant

m is the electron mass

c is the speed of light

[tex]\theta=77.5^{\circ}[/tex] is the angle of the scattered photon

Substituting, we find

[tex]\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m[/tex]

The fundamental frequency of the auditory canal, which acts like a tube closed at one end, can be calculated using the speed of sound and the length of the canal. With a typical length of 2.26 cm and a speed of sound of 348 m/s, the fundamental frequency is approximately 3846 Hz.

The student is asking about the fundamental frequency of the auditory canal, which is a tube that resonates at a specific frequency when sound waves travel through it and strike the eardrum. To calculate this frequency, we consider that the auditory canal acts like a tube closed at one end with the closed end being the eardrum.

To find the fundamental frequency (f1), we must use the formula for a tube closed at one end:
f1 = v / (4 * L), where v is the speed of sound and L is the length of the auditory canal. Given that the speed of sound (v) is 348 m/s and the typical length (L) of an adult's auditory canal is about 2.26 cm or 0.0226 m, the fundamental frequency can be calculated as follows:

f1 = 348 m/s / (4 * 0.0226 m)

After calculating this expression, we find:

f1 = 348 / (4 * 0.0226)

f1 ≈ 3846 Hz

This result indicates that the fundamental frequency is within the range that the human ear is most sensitive to, specifically in the 2000-5000 Hz range, partly explaining the ear's acute sensitivity to certain sounds.

Doctors practice antiseptic medicine to D) prevent introducing infection-causing microbes to the patient. Managing bacterial infections requires careful use of antibiotics to avoid killing beneficial bacteria and prevent antibiotic resistance.

The reason why doctors practice antiseptic medicine is to avoid introducing microbes that could infect the patient. This is an essential part of infection control in medical settings. While antibiotics are valuable in treating bacterial infections by killing the bacteria causing the illness, they also impact the good bacteria that protect the body from infection. When deciding on using antimicrobial drugs, doctors must consider the type of bacterial infection and the potential resistance of bacteria to available antibiotics.

Antibiotic resistance is a significant concern in medicine, as it can lead to infections that are harder to treat, requiring longer and more expensive hospital stays. It emerges from the bacteria's adaptation through natural selection, effectively making certain drugs less effective over time.

Answer:

Third Option

[tex]B = -0.5A[/tex]

Explanation:

If we have a vector A = ax + by we know that by definition

cA = cax + cby

Where c is a constant.

In this case we have two vectors

[tex]A = 7.6\^x -9.2\^y\\\\B = -3.8\^x + 4.6\^y[/tex]

You may notice that vector B has an opposite direction to vector A.

You may also notice that | Ax | is the double of | Bx | and | Ay | is double of |By |

That is to say

[tex]3.8 +3.8 = 7.6\\\\4.6 +4.6 = 9.2[/tex]

So the equation that relates to vectors A and B is:

[tex]B = -0.5A[/tex].

You can verify this relationship by performing the operation

[tex]B = -0.5A[/tex]

[tex]-3.8\^x + 4.6\^y = -0.5(7.6\^x -9.2\^y)\\\\\-3.8\^x + 4.6\^y = -3.8\^x + 4.6\^y[/tex]

The answer is:

The third option,

[tex]B=-0.5A[/tex]

To solve the problem, we need to remember the following vector property:

Vector multiplied by a scalar or constant number:

Multiplying a vector by a scalar number results in multiplying each of the components of the vector (x, y and z) by the scalar or constant number (including its sign).

So, we are given the following vectors:

[tex]A=(7.6,-9,2)\\B=(-3.8,4.6)[/tex]

So, writing an equation that involves both given vectors, we have:

[tex]B=-0.5A\\\\(-3.8,4.6)=-0.5((7.6,-9,2))=((-0.5*7.6),(-0.5*-9.2)\\\\(-3.8,4.6)=(-3.8,4.6)[/tex]

Hence, we have that the answer is the third option,

[tex]B=0.5A[/tex]

Have a nice day!

Answer: 0.1 m/s

Explanation:

m=1

s=10

1/10=0.1

The average velocity of the object is 0.1 m/s

From the question,

We are to determine the object's average velocity

Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t)

That is,

[tex]Average\ velocity = \frac{x_{f}-x_{0} }{t_{f}-t_{0} }[/tex]

Where [tex]x_{f}[/tex] is the final position

[tex]x_{0}[/tex] is the initial position

[tex]t_{f}[/tex] is the final time

and [tex]t_{0}[/tex] is the initial time

From the question,

[tex]x_{f} = 5 \ m[/tex]

[tex]x_{0} = 0\ m[/tex]

[tex]t_{f} = 50 \ s[/tex]

[tex]t_{0} = 0 \ s[/tex]

Putting the parameters into the formula, we get

[tex]Average\ velocity = \frac{5-0}{50-0}[/tex]

[tex]Average\ velocity = \frac{5}{50}[/tex]

[tex]Average\ velocity = 0.1 \ m/s[/tex]

Hence, the average velocity of the object is 0.1 m/s

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Answer:

higher

Explanation:

A bigger voltage will result in a faster movement of charge.