find the wavelengths of a photon and an electron that have the same energy of 20.0 evev . (the energy of the electron is its kinetic energy.)

Answers

Answer 1

The wavelength of a photon and an electron that have the same energy of 20.0 eV are as follows: Photon is 9.32 x 10^-8 m and  Electron is 2.27 x 10^-12 m.

The wavelength (λ) of a photon is given by the formula:λ = hc / E

Where h is Planck's constant (6.626 x 10^-34 J.s), c is the speed of light (2.998 x 10^8 m/s), and E is the energy of the photon.

Therefore, the wavelength of a photon that has an energy of 20.0 eV is:λ = hc / E = (6.626 x 10^-34 J.s)(2.998 x 10^8 m/s) / (20.0 eV)(1.602 x 10^-19 J/eV) = 9.32 x 10^-8 m

The energy (E) of an electron is given by the formula:

E = 1/2mv²

where m is the mass of the electron (9.11 x 10^-31 kg) and v is the velocity of the electron.

The de Broglie wavelength (λ) of an electron is given by the formula:λ = h / mv where h is Planck's constant (6.626 x 10^-34 J.s) and m is the mass of the electron.

Therefore, the velocity (v) of the electron is given by:

20.0 eV = 1/2mv^2v = sqrt(2E / m) = sqrt(2(20.0 eV)(1.602 x 10^-19 J/eV) / 9.11 x 10^-31 kg) = 3.06 x 10^6 m/s

Therefore, the de Broglie wavelength (λ) of the electron is:

λ = h / mv = (6.626 x 10^-34 J.s) / (9.11 x 10^-31 kg)(3.06 x 10^6 m/s) = 2.27 x 10^-12 m

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Related Questions

farmer has 7000 meters of fencing, and wants to enclose a rectangular plot that borders on a river. if farmer does not fence the side along the river, what is the largest area that can be enclosed? The largest area that can be enclosed is SQUARE METER

Answers

The largest area that can be enclosed is 1,221,250 square meters.

Let's assume that the length and width of the rectangular plot are 'L' and 'W', respectively. There are two widths and two lengths with fencing. We know that one of the lengths will be equal to the length of the other side along the river.

Therefore, we can have the following equation:

2L + W = 7000 - L, which can be simplified as:

3L + W = 7000 (Equation 1)

Also, the area of the rectangular plot can be expressed as:

L x W (Equation 2)

Now, we need to maximize the area of the plot by substituting Equation 1 into Equation 2.

L x W = L x (7000 - 3L)

Simplifying the above equation, we get:

L² - 3500L + area = 0 (Equation 3)

area = L x (7000 - 3L)

As we want to maximize the area, we need to find the maximum value of Equation 3. By solving this equation using the quadratic formula, we get:

L = 1750 meters area = 1,221,250 square meters

Therefore, the largest area that can be enclosed is 1,221,250 square meters.

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Find the rest energy, in terajoules, of a 15.3 g piece of chocolate. 1 TJ is equal to 102 J. rest energy TJ

Answers

The rest energy of a 15.3 g piece of chocolate is approximately 1.377 terajoules.

The rest energy of an object can be calculated using Einstein's mass-energy equivalence principle, which states that the rest energy (E) of an object is equal to its mass (m) multiplied by the speed of light squared (c^2).

The speed of light (c) is approximately 3.0 × 10^8 meters per second.

Given that the mass of the chocolate is 15.3 g, we need to convert it to kilograms before we can calculate the rest energy.

1 g = 0.001 kg

Therefore, the mass of the chocolate is 15.3 g × 0.001 kg/g = 0.0153 kg.

Now we can calculate the rest energy:

E = m * c^2

E = 0.0153 kg * (3.0 × 10^8 m/s)^2

E = 0.0153 kg * (9.0 × 10^16 m^2/s^2)

E = 1.377 × 10^15 J

To convert the rest energy to terajoules, we divide by the conversion factor:

1 TJ = 10^12 J

Rest energy (in TJ) = 1.377 × 10^15 J / (10^12 J/TJ)

Rest energy (in TJ) ≈ 1.377 TJ

Therefore, the rest energy of a 15.3 g piece of chocolate is approximately 1.377 terajoules.

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In which one of the following cases is the displacement of the object directly proportional to the elapsed time? a ball at rest is given a constant acceleration O a rocket fired from the earth's surface experiences an increasing acceleration a ball rolls with constant velocity a ball rolling with velocity vo is given a constant acceleration a bead falling through oil experiences a decreasing acceleration

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The case in which the displacement of the object is directly proportional to the elapsed time is when a ball rolls with constant velocity.

When the displacement of an object is directly proportional to the elapsed time, it means that the object is moving with a constant velocity. In this scenario, the object covers equal displacements in equal intervals of time.

1. A ball at rest is given a constant acceleration:

In this case, the ball starts from rest and experiences a constant acceleration. As a result, the velocity of the ball increases with time, and the displacement is not directly proportional to the elapsed time. The object is accelerating.

2. A rocket fired from the Earth's surface experiences an increasing acceleration:

Similar to the first case, the rocket is experiencing an increasing acceleration, which means its velocity is increasing over time. The displacement is not directly proportional to the elapsed time. The object is accelerating.

3. A ball rolls with constant velocity:

In this case, the ball is moving with a constant velocity. Since the velocity is constant, the displacement of the ball will be directly proportional to the elapsed time. The object is moving with constant velocity.

4. A ball rolling with velocity v₀ is given a constant acceleration:

When the ball is given a constant acceleration, its velocity will change over time. The displacement will not be directly proportional to the elapsed time. The object is accelerating.

5. A bead falling through oil experiences a decreasing acceleration:

In this case, the bead is experiencing a decreasing acceleration, which means its velocity is decreasing over time. The displacement is not directly proportional to the elapsed time. The object is decelerating.

Therefore, the case where the displacement of the object is directly proportional to the elapsed time is when a ball rolls with constant velocity.

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if you have 1000 lambda baryons traveling at beta =0.6 with
proper lifetime ct=8cm, whats the average distance before they
decay?

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The average distance traveled by 1000 lambda baryons before they decay is approximately 48.0 meters.

The proper lifetime of a particle, denoted as ct, is the time it takes for the particle to decay when at rest in its own frame of reference. The quantity β represents the velocity of the particles relative to the speed of light, where β = v/c.

To calculate the average distance traveled before decay, we can use the formula:

Average distance = βct

Given that β = 0.6 and ct = 8 cm, we need to convert ct to meters for consistency. 1 cm is equal to 0.01 meters.

Substituting the values into the formula:

Average distance = 0.6 * 8 cm = 0.6 * 8 * 0.01 m = 0.048 m

Since we have 1000 lambda baryons, we multiply the average distance by 1000 to account for all the particles:

Average distance for 1000 lambda baryons = 0.048 m * 1000 = 48 m

Therefore, the average distance traveled by 1000 lambda baryons before they decay is approximately 48 meters.

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A surfer floating beyond the breakers notes 17 waves per minute passing her position. If the wavelength of these waves is 38 , what is their speed? A surfer floating beyond the breakers notes 17 waves per minute passing her position. If the wavelength of these waves is 38 , what is their speed?

Answers

To calculate the speed of the waves, we can use the formula: Speed = Frequency × Wavelength

Given that the surfer notes 17 waves per minute and the wavelength is 38 units, we can substitute these values into the formula: Speed = 17 waves/minute × 38 units/wave. To determine the unit of speed, we need to know the unit of the wavelength. Let's assume the wavelength is given in meters. In that case, the unit of speed will be meters per minute. Calculating the speed: Speed = 17 waves/minute × 38 units/wave = 646 units/minute. Therefore, the speed of the waves is 646 meters per minute.

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make sure to add units if needed please :)
What is the value of each of the angles of a triangle whose sides are 151, 190, and 89 cm in length? (Hint: Consider using the law of cosines given in Appendix E.) Units The angle opposite the side of

Answers

To find the value of each angle of a triangle, we can use the law of cosines.

According to the law of cosines, for a triangle with sides of lengths a, b, and c, and the angle opposite side c denoted as C, the following equation holds:

c^2 = a^2 + b^2 - 2ab cos(C)

In this case, the sides of the triangle are given as 151 cm, 190 cm, and 89 cm. Let's denote the angles opposite these sides as A, B, and C, respectively.

Applying the law of cosines to each angle, we have:

(89 cm)^2 = (151 cm)^2 + (190 cm)^2 - 2(151 cm)(190 cm) cos(A)

(151 cm)^2 = (89 cm)^2 + (190 cm)^2 - 2(89 cm)(190 cm) cos(B)

(190 cm)^2 = (89 cm)^2 + (151 cm)^2 - 2(89 cm)(151 cm) cos(C)

Solving these equations will give us the values of angles A, B, and C in radians or degrees, depending on the unit of measurement used.

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The electric field strength 5.0 cm from a very long charged wire is 3700 n/c. What is the electric field strength 10.0 cm from the wire?

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The electric field strength 10.0 cm from the wire is 1032.25 N/C. It is given that the electric field strength at a distance of 5.0 cm from a long charged wire is 3700 N/C.

Since the charged wire is very long, its electric field is radial, and the magnitude of the electric field varies with distance r from the wire according to the equation:

E = λ/(2πεor), where λ is the linear charge density (charge per unit length), εo is the permittivity of free space (8.85 × 10−12 C2/Nm2), and 2πr is the circumference of a circle of radius r centered on the wire.

To find the electric field strength at a distance of 10.0 cm, substitute r = 10.0 cm = 0.1 m into the formula and solve for E:

E = λ/(2πεor)

E = (3700 N/C)(2π)(8.85 × 10−12 C2/Nm2)/(2 × 0.1 m)

E = 1032.25 N/C

Therefore, the electric field strength 10.0 cm from the wire is 1032.25 N/C.

The electric field strength 5.0 cm from a very long charged wire is 3700 N/C.

The electric field strength varies with distance r from the wire according to the equation: E = λ/(2πεor).

To find the electric field strength at a distance of 10.0 cm, substitute r = 10.0 cm = 0.1 m into the formula and solve for E:

E = λ/(2πεor)

E = (3700 N/C)(2π)(8.85 × 10−12 C2/Nm2)/(2 × 0.1 m)

E = 1032.25 N/C

Therefore, the electric field strength 10.0 cm from the wire is 1032.25 N/C.

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what is the wavelength from a radio station having frequency 107.7 mhz?

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The wavelength of the radio wave from the radio station with a frequency of 107.7 MHz is approximately 2.78 meters.

To calculate the wavelength of a radio wave, we can use the formula:

wavelength (λ) = speed of light (c) / frequency (f)

Where:

c is the speed of light (approximately 3.00 × 10⁸ meters per second)

f is the frequency of the radio wave

Given that the frequency of the radio station is 107.7 MHz, we need to convert it to hertz (Hz) by multiplying it by 10⁶:

f = 107.7 MHz × 10⁶ Hz/MHz = 107.7 × 10⁶ Hz

Now we can calculate the wavelength:

λ = (3.00 × 10⁸ m/s) / (107.7 × 10⁶ Hz)

λ = 2.78 meters

Therefore, the wavelength of the radio wave = 2.78 meters.

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The wavelength from a radio station having frequency 107.7 MHz can be found using the formula:

Wavelength = Speed of Light / Frequency

Using the formula Wavelength = Speed of Light / Frequency, the wavelength can be found by substituting the given values.

Speed of light = 3 × 108 m/s

Frequency = 107.7 × 106 Hz (since 1 MHz = 106 Hz)Therefore, the wavelength = (3 × 108 m/s) / (107.7 × 106 Hz)= 2.7816 m

Radio waves have different wavelengths which ranges from about 1 millimeter to 100 kilometers and frequencies ranging from about 300 GHz to 3 kHz respectively. Radio waves with higher frequencies have shorter wavelengths, and radio waves with lower frequencies have longer wavelengths.

The formula to calculate the wavelength of a radio wave is given by the equation; Wavelength = Speed of Light / Frequency.

The speed of light in a vacuum is always constant and has a value of 3 × 108 m/s. The frequency is given as 107.7 MHz. We first convert it to Hz as follows: 1 MHz = 106 Hz

Therefore, 107.7 MHz = 107.7 × 106 Hz

Now we can substitute the values in the formula:

Wavelength = Speed of Light / Frequency= 3 × 108 m/s / 107.7 × 106 Hz= 2.7816 m

Therefore, the wavelength of the radio wave from the station is 2.7816 m.

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A chinook salmon, which is a type of fish, has a maximum underwater speed of 3.58 m , but it can jump out of water with a speed of 6.26 m . To move upstream past a waterfall the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than 3.58 m ; it can then swim up the fall for the remaining distance. Because the salon must make forward progress in the water, let’s assume it can swim to the top if the water speed is 3.00 m or less. Assume the water has a horizontal speed of 1.50 m as it passes over the top ledge of the waterfall.

D) If the salmon is forced to jump so its body is at an angle that matches the velocity of the water that it enters (as shown in the video), what is the maximum height of the waterfall?

Answers

A Chinook salmon, which is a type of fish, has a maximum underwater speed of 3.58 m/s, but it can jump out of water with a speed of 6.26 m/s.

To move upstream past a waterfall the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than 3.58 m/s; it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let’s assume it can swim to the top if the water speed is 3.00 m/s or less. Assume the water has a horizontal speed of 1.50 m/s as it passes over the top ledge of the waterfall. The maximum height of the waterfall when the salmon is forced to jump so its body is at an angle that matches the velocity of the water that it enters can be determined.

As the salmon swims up the waterfall, it will experience a force of the stream's speed. The salmon is only able to move upstream if it swims faster than the stream. Assume the water has a horizontal speed of 1.50 m/s as it passes over the top ledge of the waterfall. Because the salmon must make forward progress in the water, let’s assume it can swim to the top if the water speed is 3.00 m/s or less.Thus, the speed of the salmon relative to the waterfall when it leaps from the water is:[tex]v = 6.26 m/s - 1.50 m/s = 4.76 m/s[/tex]According to the problem statement, the salmon can swim upstream if the water speed is 3.00 m/s or less. Therefore, if the salmon is moving at 4.76 m/s relative to the water, the salmon can swim upstream if the water speed is[tex]v_w = 4.76 m/s - 3.00 m/s = 1.76 m/s[/tex]To determine the maximum height of the waterfall, we must determine the height above which the water has a speed of 1.76 m/s. Therefore, the maximum height of the waterfall when the salmon is forced to jump so its body is at an angle that matches the velocity of the water that it enters is approximately 0.16 meters.

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how to calculate the distance between a sensor and an electric harge

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In order to calculate the distance between a sensor and an electric charge, you need to know the electric field strength produced by the charge and the sensitivity of the sensor to that field strength. The calculation involves using Coulomb's Law to find the electric field strength and then using the inverse square law to determine the distance.

Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:F = k * (q1 * q2) / d^2where F is the force between the charges, k is Coulomb's constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges, and d is the distance between the charges.The electric field strength produced by the charge is given by:E = F / q2where E is the electric field strength and q2 is the test charge (the charge on the sensor).To calculate the distance between the sensor and the charge, you can use the inverse square law, which states that the intensity of a field (in this case, the electric field) is inversely proportional to the square of the distance from the source. The formula for the inverse square law is:I = I0 * (d0 / d)^2where I is the intensity of the field at distance d, I0 is the intensity of the field at distance d0, and d0 is a reference distance (usually chosen to be 1 meter). Rearranging this equation, we get:d = sqrt(I0 / I) * d0So to calculate the distance between the sensor and the charge, you need to first find the electric field strength at the sensor and the electric field strength at a reference distance (e.g. 1 meter). Then you can use the inverse square law to calculate the distance between the sensor and the charge.

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Which of the following pairs represents a medium frequency band and its common use?
Select one:
a. 3-30 MHz, CB and shortwave radio
b. 300 KHz-3MHz, AM radio
c. 144-174 MHz, TV channels
d. 30-300 KHz, cordless phones

Answers

The frequency range of 144-174 MHz represents a medium frequency band commonly used for TV channels.

a. 3-30 MHz, CB and shortwave radio: This frequency range is considered high frequency (HF) band and is commonly used for Citizens Band (CB) radio communication and shortwave radio broadcasting.

b. 300 KHz-3 MHz, AM radio: This frequency range is known as the medium frequency (MF) band and is used for AM (Amplitude Modulation) radio broadcasting.

c. 144-174 MHz, TV channels: This frequency range falls under the very high frequency (VHF) band and is commonly used for television (TV) broadcasting.

d. 30-300 KHz, cordless phones: This frequency range is part of the low frequency (LF) band and is not typically used for cordless phones. Cordless phones commonly operate in higher frequency ranges, such as the 900 MHz or 2.4 GHz bands.

Therefore, the correct pair representing a medium frequency band and its common use is option c: 144-174 MHz, TV channels.

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.A flywheel with a radius of 0.300m starts from rest and accelerates with a constant angular acceleration of 0.400rad/s2 .
A) Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start. (Answers are 0.21,0,0.21 m/s^2)
B) Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 60.0?
C) Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 120?.

Answers

The magnitudes of tangential acceleration, radial acceleration, and resultant acceleration can be computed for different angular positions of a point on the rim of the flywheel.

How can the magnitudes of tangential, radial, and resultant acceleration be calculated for different angular positions of a point on the flywheel's rim?

A) At the start (0°), the magnitude of the tangential acceleration is 0.21 m/s², the radial acceleration is 0 m/s², and the resultant acceleration is 0.21 m/s².

B) After turning through 60°, the magnitude of the tangential acceleration is 0.21 m/s², the radial acceleration is 0.12 m/s², and the resultant acceleration is 0.24 m/s².

C) After turning through 120°, the magnitude of the tangential acceleration is 0.21 m/s², the radial acceleration is -0.21 m/s², and the resultant acceleration is 0 m/s².

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Given that E = 15ax - 8az V/m at a point on the surface of a conductor, determines the surface charge density at that point. Assume that ε = £0. a. 1.50x10-10 b. 2.21x10-10 c. 1.91x10-10 d. 2.12x10-10

Answers

The surface charge density at that point with Electric field, E=15ax-8az V/m with permittivity in free space is ε=ε₀ is, σ=1.5×10⁻¹⁰ c/m². Hence, option A is correct.

The Gauss law is defined as the electric flux of the closed surface is equal to the charge enclosed by the given area. Electric flux is defined as the number of field lines crossing through a given area.

From the given area,
E = 15ax-8az V/m

ε=ε₀ (ε₀ is the permittivity in free space)=8.854×10⁻¹².

surface charge density, (σ) =?

E = σ/ε₀

σ = E×ε₀

  = (15ax-8az)×8.854×10⁻¹².

  = √(15)²+(8)²×8.854×10⁻¹².

  = 17×8.854×10⁻¹².

  = 1.50×10⁻¹⁰C/m².

Thus, the surface charge densities, σ = 1.50×10⁻¹⁰ C/m².

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PART H,I and J only pls
PART H,I and J only pls
A cylinder of volume 0.320 m³ contains 12.0 mol of neon gas at 22.8°C. Assume neon behaves as an ideal gas. (a) What is the pressure of the gas? 9.22e4 Pa (b) Find the internal energy of the gas. 4.

Answers

A cylinder of volume 0.320 m³ contains 12.0 mol of neon gas at 22.8°C. Assume neon behaves as an ideal gas. Therefore,

(a) The pressure is 9.22e4 Pa.

(b)  Internal energy is 4.42e4 J.

(c) Work done is -6.27e4 J.

(d) Temperature is 924 K

(e)  Internal energy when volume is 1 is 1.41e5 J.

Here is the explanation :

(a) The pressure of the gas can be calculated using the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given:

Volume (V) = 0.320 m³

Number of moles (n) = 12.0 mol

Temperature (T) = 22.8°C = 22.8 + 273.15 = 296.95 K

Plugging in the values:

P * 0.320 = 12.0 * R * 296.95

Simplifying and solving for P:

[tex]\[P \approx \frac{12.0 \times R \times 296.95}{0.320}\][/tex]

Using the value of the ideal gas constant, R = 8.314 J/(mol·K), we can calculate the pressure P:

[tex]\[P \approx \frac{12.0 \times 8.314 \times 296.95}{0.320} \approx 9.22 \times 10^{4} \text{ Pa}\][/tex]

Therefore, the pressure of the gas is approximately 9.22 × 10^4 Pa.

(b) The internal energy of an ideal gas can be given by the equation:

[tex]\begin{equation}U = \frac{3}{2}nRT[/tex]

where U is the internal energy, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given the same values as before, we can substitute them into the equation:

[tex]\[U = \frac{3}{2} \times 12.0 \times 8.314 \times 296.95 \approx 4.42 \times 10^{4} \text{ J}\][/tex]

Therefore, the internal energy of the gas is approximately 4.42 × 10^4 J.

(c) The work done on the gas during an expansion at constant pressure can be calculated using the equation:

W = P * ΔV

where W is the work done, P is the pressure, and ΔV is the change in volume.

Given:

Initial volume (V₁) = 0.320 m³

Final volume (V₂) = 1.000 m³

Pressure (P) = 9.22 × 10⁴ Pa

ΔV = V₂ - V₁ = 1.000 m³ - 0.320 m³ = 0.680 m³

Plugging in the values:

[tex]\[W = (9.22 \times 10^{4} \text{ Pa}) \times (0.680 \text{ m}^3) \approx -6.27 \times 10^{4} \text{ J}\][/tex]

The negative sign indicates work done on the gas.

Therefore, the work done on the gas during the expansion is approximately -6.27 × 10⁴ J.

(d) To find the temperature of the gas at the new volume, we can rearrange the ideal gas law equation:

PV = nRT

Solving for T:

[tex]\[T = \frac{PV}{nR}\][/tex]

Given:

Pressure (P) = 9.22 × 10⁴ Pa

Volume (V) = 1.000 m³

Number of moles (n) = 12.0 mol

Ideal gas constant (R) = 8.314 J/(mol·K)

Plugging in the values:

[tex][T = \frac{9.22 \times 10^{4} \text{ Pa} \times 1.000 \text{ m}^3}{12.0 \text{ mol} \times 8.314 \frac{\text{J}}{\text{mol K}}}][/tex]

T ≈ 924 K

Therefore, the temperature of the gas at the new volume is approximately 924 K.

(e) The internal energy of the gas when its volume is 1

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Complete question :

A cylinder of volume 0.320 m³ contains 12.0 mol of neon gas at 22.8°C. Assume neon behaves as an ideal gas. (a) What is the pressure of the gas? 9.22e4 Pa (b) Find the internal energy of the gas. 4.42.e4 J (c) Suppose the gas expands at constant pressure to a volume of 1.000 m³. How much work is done on the gas? -6.27e4 J (d) What is the temperature of the gas at the new volume? 9.24e2 K (e) Find the internal energy of the gas when its volume is 1.000 m³. 1.38e5 J (f) Compute the change in the internal energy during the expansion. 9.40e4 (g) Compute AU - W. 15.6e4 J (h) Must thermal energy be transferred to the gas during the constant pressure expansion or be taken away? This answer has not been graded yet. (1) Compute Q, the thermal energy transfer. J (j) What symbolic relationship between Q, AU, and W is suggested by the values obtained?

find the frequency in terahertz of visible light with a wavelength of 621 nm in vacuum.

Answers

The frequency of visible light with a wavelength of 621 nm in vacuum is approximately 483.3 THz.

Visible light is electromagnetic radiation, which means it has both electric and magnetic components and moves at the speed of light. It has a wavelength between 400 to 700 nanometers (nm) and a frequency range between 405 THz to 790 THz.

The formula to find the frequency of electromagnetic waves is:

[tex]f = c / λ[/tex]

Where, f is the frequency of the wave,c is the speed of light in vacuum, andλ is the wavelength of the wave.

In the given question, the wavelength of visible light is 621 nm. Therefore, the frequency of visible light with a wavelength of 621 nm in vacuum can be calculated as:

f = c / λ

= (3 x 10^8 m/s) / (621 x 10^-9 m)

= 4.833 x 10^14 Hz

= 483.3 THz

Thus, the frequency of visible light with a wavelength of 621 nm in vacuum is approximately 483.3 THz.

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when its 80 kwkw engine is generating full power, a small single-engine airplane with mass 750 kgkg gains altitude at a rate of 2.5 m/sm/s.

Answers

When the 80 kW engine of the small single-engine airplane is generating full power, the airplane with a mass of 750 kg gains altitude at a rate of 2.5 m/s.

The power generated by the engine is equal to the rate of work done, which is given by the equation Power = Force × Velocity.
In the case of the airplane gaining altitude, the force is equal to the weight of the airplane, which is given by Weight = mass × gravitational acceleration.
Assuming the gravitational acceleration is approximately 9.8 m/s², we can calculate the weight of the airplane. Then, by rearranging the power equation, we can solve for the velocity.
By substituting the known values of power (80 kW), weight (mass × gravitational acceleration), and the given altitude rate (2.5 m/s) into the equations, we can determine the velocity at which the airplane is climbing.

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Following is the complete answer: When its 80 kW engine is generating full power, a small single-engine airplane with mass 750 kg gains altitude at a rate of 2.5 m/s. Part A What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.) Express your answer as a percentage to two significant figures.

Part A An astronaut on another planet drops a 1.kg rock from rest. The astronaut notices that the rock falls 2 meters straight down in one second. On this planet, how much does the rock weigh? 1 N 5N 10 N 4N

Answers

On earth, the value of acceleration due to gravity is approximately 9.81 m/s^2.  Therefore, if the mass of an object is 1 kg, its weight will be 9.81 N. nearest value is 10N therefore option c

Part A:

Given that an astronaut on another planet drops a 1 kg rock from rest. The astronaut notices that the rock falls 2 meters straight down in one second. We are required to find how much does the rock weigh on this planet.

As per the given information,

Acceleration due to gravity (g) = 2 m/s^2Mass (m) = 1 kg

The formula for weight (W) of an object is given as:

W = m × g

Substituting the given values in the above equation, we get

W = 1 kg × 2 m/s^2W = 2 N

Therefore, the rock weighs 2 N on this planet.

Part B:

The weight of an object is the force acting on it due to gravity. It is a vector quantity, which means it has both magnitude and direction. The weight of an object is directly proportional to its mass and the acceleration due to gravity. It can be measured in Newtons (N).

On earth, the value of acceleration due to gravity is approximately 9.81 m/s^2.

However, the value of acceleration due to gravity is different on different planets, which means the weight of an object will also differ on different planets. For example, on the moon, the value of acceleration due to gravity is approximately 1.62 m/s^2.

Therefore, if the mass of an object is 1 kg, its weight will be 1.62 N on the moon.

The weight of an object can be determined using a spring balance or a weighing scale. The spring balance works on the principle of Hooke's law, which states that the force applied on a spring is directly proportional to its extension. When an object is suspended from a spring balance, the spring extends due to the force of gravity acting on the object. The weight of the object can be calculated by measuring the extension of the spring. The weighing scale works on the principle of measuring the force applied on a rigid body due to the weight of an object. When an object is placed on a weighing scale, its weight exerts a force on the rigid body, which is then measured by the scale.

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A 24.0 kg child plays on a swing having support ropes that are 1.80 m long. A friend pulls her back until the ropes are 45.0 degree from the vertical and releases her from rest. What is the potential energy for the child as she is released, compared with the potential energy at the bottom of the swing? How fast will she be moving at the bottom of the swing? How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

Answers

The potential energy for the child as she is released is 82.1 J, she will be moving at a speed of 4.01 m/s at the bottom of the swing, and the work done by the tension in the ropes as the child swings from the initial position to the bottom is 193 J.

A 24.0 kg child is playing on a swing having support ropes that are 1.80 m long. A friend pulls her back until the ropes are 45.0 degree from the vertical and releases her from rest. The potential energy for the child as she is released, compared with the potential energy at the bottom of the swing is given by;`U = mgh``U = 24.0 kg × 9.81 m/s^2 × (1.8 m - 1.8m cos 45°)`On solving this equation, we get `U = 82.1 J`

The potential energy at the bottom of the swing is equal to kinetic energy at the top of the swing since there is no external work done on the system. Therefore, the kinetic energy of the child when she is at the bottom of the swing is equal to the potential energy of the child when she is released.

Kinetic energy at the bottom of the swing is given by;`K = (1/2)mv^2``82.1 J = (1/2) × 24.0 kg × v^2``v = 4.01 m/s`The work done by the tension in the ropes as the child swings from the initial position to the bottom is given by;`W = ∆K = Kf - Ki``W = (1/2)mvf^2 - (1/2)mvi^2``W = (1/2) × 24.0 kg × (4.01 m/s)^2 - (1/2) × 24.0 kg × 0 m/s``W = 193 J`

Therefore, the potential energy for the child as she is released is 82.1 J, she will be moving at a speed of 4.01 m/s at the bottom of the swing, and the work done by the tension in the ropes as the child swings from the initial position to the bottom is 193 J.

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A car of mass 1000kg is travelling down a steep hill. The brakes fail and the driver uses a horizontal sand-filled safety road to stop the car.
The car enters the sand at a speed of 10 m / s and experiences a constant stopping force of
2500N.
How far does the car travel in the sand before coming to rest?
A 2.0 m
B 4.0 m
C 20 m
D 40 m

Answers

The car travels a distance of C. 20 meters in the sand before coming to rest.

To determine how far the car travels in the sand before coming to rest, we can use the principle of work-energy.

The work-energy principle states that the work done on an object is equal to its change in kinetic energy. In this case, the work done by the stopping force on the car will be equal to the initial kinetic energy of the car.

The work done is given by the equation:

Work = Force × Distance

Since the force is constant at 2500 N, and the work done is equal to the initial kinetic energy, we have:

2500 N × Distance = (1/2) × mass × velocity²

Substituting the given values:

2500 N × Distance = (1/2) × 1000 kg × (10 m/s)²

2500 N × Distance = 50000 J

Distance = 50000 J / 2500 N

Distance = 20 m

Therefore, the car travels a distance of 20 meters in the sand before coming to rest. By equating the work done by the stopping force to the initial kinetic energy of the car, we found that the car travels a distance of 20 meters in the sand before coming to rest. Therefore, Option C is correct.

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In the greenhouse effect, far infrared radiation is earth's surface and absorbed and reemitted by gases in the atmosphere. These gases have _____ in concentration over the past century.
a. Increased
b. Decreased
c. Remained constant
d. Varied unpredictably

Answers

In the greenhouse effect, far infrared radiation is earth's surface and absorbed and reemitted by gases in the atmosphere. The issue of global warming has received a lot of attention and has prompted a lot of research to better understand its impacts and how we can mitigate it. Therefore, the correct answer is (a) increased.

The greenhouse effect is defined as a phenomenon in which the atmosphere of the earth traps the sun's warmth on the surface of the planet. It is known that far-infrared radiation is emitted by the Earth's surface and absorbed and re-emitted by gases present in the atmosphere. These gases include water vapor, carbon dioxide, and methane, among others.

The concentration of these gases in the atmosphere has increased over the past century.  a. Increased The amount of carbon dioxide in the atmosphere, for example, has increased by over 30% in the last 100 years.

This rise in greenhouse gases has contributed to global warming, as the Earth's temperature rises in response to the additional trapped heat. As a result, the issue of global warming has received a lot of attention and has prompted a lot of research to better understand its impacts and how we can mitigate it. Therefore, the correct answer is (a) increased.

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Constants The magnetic field in a EM wave traveling in free space has an rms strength of 24.5 nt. Part A How long does it take to deliver 355 J of energy to 1.05 cm? of a wall that it hits perpendicularly? Express your answer using three significant figures. IV ALQ R O O ?

Answers

The time taken by the wave to deliver energy to a wall of 1.05 cm² area is 0.753 μs. The RMS strength of the magnetic field in an EM wave traveling in free space is 24.5 nt and energy delivered to 1.05 cm2 of a wall is 355J.

Solution: Part A The energy delivered by an electromagnetic wave per second per unit area is given as

Poynting vector= [tex][E × B]/μ0[/tex]

Here,E is the electric field strength, B is the magnetic field strength and

μ0 is the permeability of free space. If the energy delivered to the area, dA, is dE, in time dt, then from the above equation

Poynting vector=[tex]dE/dt × dA[/tex]

On integration, the total energy delivered by the wave over time t is given as[tex]E= 1/μ0 × ∫p dt[/tex]

Since the Poynting vector, [tex]|P|= EB/μ0[/tex] and the strength of the magnetic field in the EM wave is given as B = Brms

Hence the Poynting vector is given as

[tex]|P|= ErmsBrms/μ0[/tex]

= Erms²/377 watts/m²

The energy delivered to an area, dA, in time dt is given by

[tex]dE= P dt × dA[/tex]

= Erms²/377 × dt × dA

The energy delivered to an area A in time dt is given by

dE = Erms²/377 × A × dt

The total energy delivered to an area, A, in time t is given by

E = ∫dE = Erms²/377 × A × ∫dtE

= Erms²/377 × A × t

Thus, the time duration of an EM wave to deliver energy, E, to an area, A, is given by

t = 377 E / (Erms)² × A

Here,E = 355J

A = 1.05 cm²

= 1.05 × 10⁻⁴ m²Brms

= 24.5 nT

= 24.5 × 10⁻⁹ Tt

= 377 × 355 / (24.5 × 10⁻⁹)² × 1.05 × 10⁻⁴

= 7.53 × 10⁻⁷ seconds

= 0.753 μs (rounded to three significant figures)

Answer: The time taken by the wave to deliver energy to a wall of 1.05 cm² area is 0.753 μs.

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A tank is filled with an ideal gas at 400 degrees K and pressure of 1.00 atm.

The tank is heated until the pressure of the gas in the tank doubles. What is the temperature of the gas?

Answers

The temperature of the gas after the pressure has doubled is 800 K.

Given that a tank is filled with an ideal gas at 400 K and pressure of 1.00 atm. The tank is heated until the pressure of the gas in the tank doubles.

The ideal gas law can be used to solve the problem.

The ideal gas law states that PV=nRT, where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature of the gas. Since the gas is an ideal gas, it follows that the number of moles of the gas is constant.

We can therefore write that P1V1/T1 = P2V2/T2, where P1 = 1.00 atm, V1 is the initial volume of the gas, T1 = 400 K, P2 = 2.00 atm, and V2 is the final volume of the gas.

Rearranging the equation, we get T2 = T1P2V1/P1V2. Since V2 = V1/2, we can substitute this into the equation to obtain T2 = 400 K * 2.00 atm * V1/(1.00 atm * V1/2) = 800 K.

The temperature of the gas after the pressure has doubled is 800 K.

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explain why we do not get a lunar and solar eclipse every month.

Answers

We do not get a lunar and solar eclipse every month because of the fact that the Moon's orbital plane is not aligned with the Earth's orbit around the Sun.

In order for a lunar or solar eclipse to occur, there must be an alignment between the Earth, the Moon, and the Sun. During a lunar eclipse, the Earth passes between the Sun and the Moon, casting a shadow on the Moon. Meanwhile, during a solar eclipse, the Moon passes between the Sun and the Earth, blocking out the Sun's light. However, the Moon's orbit is tilted at an angle of about 5 degrees to the Earth's orbit around the Sun. As a result, the Moon does not always pass through the Earth's shadow during a full moon (lunar eclipse) or align perfectly with the Sun during a new moon (solar eclipse). This is why lunar and solar eclipses are relatively rare occurrences.

Every month, the Moon goes through its phases as it orbits the Earth. At the new moon, the Moon is between the Earth and the Sun, but it does not necessarily block out the Sun's light because the Moon's orbit is tilted slightly. Likewise, at the full moon, the Moon is on the opposite side of the Earth from the Sun, but it does not always pass through the Earth's shadow because of the same tilt. So, lunar and solar eclipses can only occur when the Moon is in just the right position relative to the Sun and Earth. The occurrence of a lunar or solar eclipse is also dependent on the geometry of the three bodies; they have to be in alignment. Additionally, Earth's atmosphere plays a role in the occurrence of solar and lunar eclipses. If the atmosphere is filled with smoke or dust, or if the Earth's atmosphere is very clear, this can impact the visibility of the eclipses. Ultimately, the rarity of eclipses is due to the complex interplay of many factors, including the Moon's orbit, the Earth's orbit around the Sun, and the geometry of the three bodies.

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A cylinder is inscribed in a right circular cone of height 6.5 and radius (at the base) equal to 4.5. What are the dimensions of such a cylinder which has maximum volume?
Radius ?
Height ?

Answers

The dimensions of the cylinder with maximum volume are:Radius = 4.5, Height = 6.5

the cylinder is inscribed in the cone, the height of the cylinder is equal to the height of the cone, which is 6.5.To find the radius of the cylinder, we need to consider similar triangles formed by the cone and the cylinder. The radius of the cone at the base is 4.5, and the height of the cone is 6.5. The radius of the cylinder will be a fraction of the radius of the cone: By using the similar triangles, we can set up the following equation: r / 4.5 = h / 6.5.

Simplifying the equation, we get: r = (4.5 * h) / 6.5
Since we know the height of the cylinder is equal to the height of the cone, we can substitute h = 6.5 into the equation:
r = (4.5 * 6.5) / 6.5. r = 4.5 .Therefore, the radius of the cylinder is 4.5.The height of the cylinder is the same as the height of the cone, which is 6.5. So, the dimensions of the cylinder with maximum volume are:
Radius = 4.5, Height = 6.5

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calculate the amount of work done to move 1 kg mass from the surface of the earth to a point 10⁵ km from the centre of the earth

Answers

The  quantum of work done to move a 1 kg mass from the  face of the earth to a point 10 ⁵ km from the centre of the earth is4.92 x 10 ⁸J.    

The  quantum of work done to move 1 kg mass from the  face of the earth to a point 10 ⁵ km from the centre of the earth can be calculated using the gravitational implicit energy formula.

The gravitational implicit energy is the  quantum of work done by an external force in bringing an object from  perpetuity to a point in space where it can be  told  by  graveness. When an object is moved from the  face of the earth to a point 10 ⁵ km from the centre of the earth, the gravitational implicit energy of the object increases.  

The formula for gravitational implicit energy is given by  U = - GMm/ r  where U is the gravitational implicit energy  G is the universal gravitational constant  M is the mass of the earth  m is the mass of the object  r is the distance between the object and the centre of the earth.  

We know that the mass of the object is 1 kg,  the mass of the earth is    and the distance from the centre of the earth to a point 10 ⁵ km down is           Plugging these values into the formula, we get         thus, the  quantum of work done to move a 1 kg mass from the  face of the earth to a point 10 ⁵ km from the centre of the earth is 4.92 x 10 ⁸J.

the mass of the earth is [tex]5.97 * 10^2^4 kg[/tex],

and the distance from the centre of the earth to a point 10⁵ km away is:

[tex]= 6.38 * 10^6 + 10^5 km[/tex]

[tex]= 6.48 * 10^6 km[/tex]

[tex]= 6.48 * 10^9 m[/tex].

Plugging these values into the formula, we get

[tex]U = -6.67 * 10^-^1^1 * 5.97 * 10^2^4 * 1 / 6.48 * 10^9[/tex]

   [tex]= -4.92 * 10^8 J[/tex]

Therefore, the amount of work done to move a 1 kg mass from the surface of the earth to a point 10⁵ km from the centre of the earth is 4.92 x 10⁸ J.

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If the intensity of sound is increased by a factor of 30 , by how many decibels is the sound level increased ?
a. 12 dB
b. 14.77 dB
c. 10 dB
d. 13 dB

Answers

By how many decibels is the sound level increased if the intensity of sound is increased by a factor of 30?The sound level is increased by 15 dB when the intensity of sound is increased by a factor of 30.

The relation between sound intensity (I) and sound level (L) is given by:L = 10 log (I/I0)where I0 is the threshold of hearing, which is the reference intensity level.Using this equation, we can find the increase in sound level when the intensity is increased by a factor of 30 as follows:Let L1 be the original sound level and I1 be the original intensity level. Let L2 be the new sound level and I2 be the new intensity level. Then we have:L2 = 10 log (I2/I0)L1 = 10 log (I1/I0)Since the intensity is increased by a factor of 30, we have:I2 = 30 I1Substituting this into the equation for L2, we get:L2 = 10 log (30 I1/I0)L2 = 10 (log 30 + log (I1/I0))L2 = 10 (1.477 + L1)Note that log 30 = 1.477 (approx).Therefore, the sound level is increased by 10 (1.477) = 14.77 dB when the intensity is increased by a factor of 30.

Sound level is a measure of the intensity of sound and is expressed in decibels (dB). Decibels are used because the human ear is sensitive to sound over a wide range of intensities, from the threshold of hearing to the threshold of pain. The decibel scale is logarithmic, which means that a small increase in intensity is represented by a large increase in sound level. For example, an increase in sound level from 60 dB to 70 dB represents a ten-fold increase in intensity.Therefore, the sound level will increase by a certain amount. We can use the relation between sound intensity and sound level to find out how much the sound level will increase.Let L1 be the original sound level and I1 be the original intensity level. Let L2 be the new sound level and I2 be the new intensity level. Then we have:L2 = 10 log (I2/I0)L1 = 10 log (I1/I0)Since the intensity is increased by a factor of 30, we have:I2 = 30 I1Substituting this into the equation for L2, we get:L2 = 10 log (30 I1/I0)L2 = 10 (log 30 + log (I1/I0))L2 = 10 (1.477 + L1)Note that log 30 = 1.477 (approx).Therefore, the sound level is increased by 10 (1.477) = 14.77 dB when the intensity is increased by a factor of 30.

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A hollow spherical shell with mass 2.05 kgkg rolls without slipping down a slope that makes an angle of 30.0 ∘∘ with the horizontal.
Find the minimum coefficient of friction μμmu needed to prevent the spherical shell from slipping as it rolls down the slope.

Answers

The minimum coefficient of friction needed to prevent the spherical shell from slipping as it rolls down the slope is 0.31.

Mass of hollow spherical shell, m = 2.05 kg. Angle of slope with the horizontal, θ = 30°. The forces acting on the spherical shell are: Weight, W = mg. Normal force, N = mg cosθForce parallel to the slope, f = mg sinθ. Force of friction, f'. Let R be the radius of the spherical shell. For the shell to not slip on the slope, the force of friction should be equal to the force parallel to the slope and acting on the shell.

Therefore, we have; f' = f (Minimum coefficient of friction needed)mg sinθ = f' = μNμ = (mg sinθ) / (mg cosθ)μ = tanθμ = tan30°μ = 0.31. Hence, the minimum coefficient of friction needed to prevent the spherical shell from slipping as it rolls down the slope is 0.31.

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Your velocity is given by v(t) = t^2 + 6 in m/sec, with t in seconds. Estimate the distance, s, traveled between t=0 and t=5. Use an overestimate with data every one second

Answers

The overestimated distance traveled between t=0 and t=5 is 158 meters.

To estimate the distance traveled, we can use the trapezoidal rule to approximate the area under the curve of the velocity function v(t). The trapezoidal rule divides the interval [0, 5] into subintervals with a width of 1 second and approximates each subinterval as a trapezoid. The formula for the trapezoidal rule is ∫[a,b] f(x) dx ≈ ∑[(i=1 to n)] [f(x_i-1) + f(x_i)] * Δx / 2, where Δx is the width of each subinterval.

Using this formula, we can calculate the overestimated distance traveled:

s ≈ [f(0) + 2f(1) + 2f(2) + 2f(3) + 2f(4) + f(5)] * Δt / 2

≈ [0 + 2(1^2 + 6) + 2(2^2 + 6) + 2(3^2 + 6) + 2(4^2 + 6) + (5^2 + 6)] * 1 / 2

≈ 158 meters.

This provides an overestimate of the distance traveled between t=0 and t=5.

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Once a carousel is in motion, the constant movement of the carousel horse around the
center of the circle can BEST be described as:

A) Acceleration; Change in speed
B) Velocity; Speed plus direction
C) Acceleration; Chang in velocity
D) Speed; Distance traveled over time

Answers

The constant movement of the carousel horse around the center of the circle can BEST be described as velocity; speed plus direction.

The correct answer to the given question is option B.

Velocity is the vector that describes how fast and in what direction something moves. It has a magnitude (the speed of the movement) and a direction (the direction of the movement). The motion of the carousel horse is circular and, as a result, has a constant speed (distance travelled over time) and a direction (tangential to the circumference of the circle). Therefore, it can be described as velocity.

Acceleration is a measure of how fast the velocity is changing, and in the case of the carousel, the horse is not changing direction or speed, so it is not experiencing any acceleration. Finally, speed and distance traveled over time are related but do not describe the direction of the motion.

Since the motion of the carousel horse is circular, speed and distance traveled over time alone do not provide a complete description. Thus, the best answer is velocity; speed plus direction.

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how fast must you be approaching a red traffic light ( λ = 675 nm ) for it to appear yellow ( λ = 575 nm )? express your answer in terms of the speed of light.

Answers

So, to observe the red traffic light as yellow, the observer must approach the light with a speed of 0.148 times the speed of light.

When the observer approaches the red traffic light with a speed, the light appears shifted towards the blue end of the spectrum. The apparent frequency and wavelength shift is calculated using the Doppler effect equation.

The Doppler shift is given by the relation f′= f (v+vO)/c

where, f' is the observed frequency, f is the frequency of the wave, v is the speed of the observer, v O is the speed of the source and c is the speed of the wave.

For the red traffic light,

f= c/λ = 4.44 × 10^14 Hzλ

= 675 nm

For the yellow traffic light,

f = c/λ

= 5.22 × 10^14 Hzλ

= 575 nm

As we know that the light appears yellow when the red light shifts 575 nm.

Therefore, the observer should be approaching the light with a speed given by the relation as,

∆f/f = v/c⇒ ∆λ/λ

= v/c⇒ v

= c (∆λ/λ)

= c [(λ_0 - λ)/λ_0 ]

Where,λ is the wavelength of the shifted light (λ = 575 nm),λ0 is the wavelength of the unshifted light (λ0 = 675 nm)

Therefore,

v = c [(675 - 575)/675]⇒ v

= 0.148c

So, the observer must approach the red traffic light at a speed of 0.148 times the speed of light to observe it as yellow.

An observer, when approaching a red traffic light, experiences a shift in the light's wavelength towards the blue end of the spectrum. This apparent frequency and wavelength shift is given by the Doppler effect equation.

The Doppler shift can be expressed using the relation,

f′= f (v+vO)/c

where, f' is the observed frequency, f is the frequency of the wave, v is the speed of the observer,v O is the speed of the source and c is the speed of the wave.

The frequency and wavelength of the red and yellow traffic lights are,

f= c/λ

= 4.44 × 10^14 Hz,

λ = 675 nm and

f = c/λ

= 5.22 × 10^14 Hz,

λ = 575 nm.

Since we know that the light appears yellow when the red light shifts by 575 nm, the observer must be approaching the light with a velocity given by the following relation:

∆f/f = v/c⇒ ∆λ/λ

= v/c⇒ v

= c (∆λ/λ_0 ) where λ_0 is the wavelength of the unshifted light (λ_0 = 675 nm)

Therefore,

v = c [(675 - 575)/675]⇒ v

= 0.148c

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Which of the following statements appropriately interprets the diagram above? O a. The final allocation of grain is point F. Oh Bruno will chooco Collegati Bushels of grain 4 0 00 O a. The final allocation of grain is point F. O b. Bruno will choose an allocation that maximises the sum of his and Angela's utility. Oc. Angela is indifferent between points, A, C and E. Od. This diagram shows how institutional changes affect the allocation of resources. O e. The final allocation of grain occurs at a point on Angela's biological survival constraint. Name formula mol. Eq mw mmol amount imine intermediate 1. 00 280 mg nabh4 70 mg thf - - - 10 ml product Future value Jack and Jil are saving for a rainy day and decide to put $70 away in their local bank every year for the next 30 years. The local Up-the-Hill Bank will pay them 8% on their account a. If Jack and J put the money in the account faithfully at the end of every year, how much will they have in it at the end of 30 years?b. Unfortunately, Jack had an accident in which he sustained head injuries after only 10 years of savings. The medical bill has come to $1.000. Is there enough in the rainy day fund to cover ? Let X exp(1) and Y exp(2) beindependent random variables. Find the function of density of Z =X/Y and calculate P[X < Y ]. Score: 90.32%, 31.61 of 35 points Points: 0.37 of t Save Homework: Chapter #3 - Homework A sample of -grade classes was studied in an article One of the variables collected was the class size in terms of student-to-faculty ratio. The student-to-faculty ratios of the 84 fifth-grade classes sampled have a mean of 16 05 and a standard deviation of 1.24 Complete parts (a) through (d) bellow a. Construct the graph shown bele 1 2-3 = 13 13 *** = 18.09 x-2 = 14:37 X+2 = 19.33 3-* = 1561 * +36 = 20.57 (Type integers or decimals. Do not round) b. Apply Property 1 of the empirical rule to make pertinent statements about the observations in the sample fifth-grade classes sampled have student-to-faculty ratios between 15.61 and 18.09 Type integers or decimals De not round) Help me solve this View an example Get more help - 3 An object of height 2.3 cm is placed 27 cm in front of a diverging lens of focal length 17 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length. A 50% Part (a) Find the location of the final image, in centimeters beyond the converging lens. A 50% Part (b) What is the magnification of the final image? Include its sign to indicate its orientation with respect to the object. by 1828, andrew jackson had established the political apparatus of the democratic party. T/F Check the boxes of the points where the graph has a local minimum. Then check where it has a local maximum 0abc1dsxCheck the boxes of the points where the graph has an absolute maximumO A. aO B. bO C.cO D.dO E.e In early 2022, inflation in the U.S. was higher than inflation in Australia. (i) (5 marks) What is relative PPP? Under relative PPP alone, what should this inflation differential do to the value of the U.S. dollar (USD) in terms of Australian dollars (AUD), holding nominal interest rates constant. That is, what happens to the Australian nominal exchange rate? (ii) (5 marks) What is the Fisher effect? Under the Fisher effect alone, what should this inflation differential do to the U.S.-Australian nominal interest rate differential? (iii) (5 marks) What is the real exchange rate? If the effect you have noted in part (5)(b)(i) translates into a corresponding change in the real exchange rate, what now would be the expected effect of this inflation differential on the U.S.-Australian nominal interest rate differential? Show how you figure this out. Which of the following statements is true of self-determination theory? It stems from the incentive approaches to motivation. It suggests that different cultures have different basic psychological needs. It suggests that competence, autonomy, and relatedness are genetically determined. It was proposed by Walter Cannon and Philip Bard. under certain conditions, it's okay to drive with just your parking or fog lights on. Bramble Ltd. had the following investment portfolio at January 1, 2020: Fair Value at Dec. 31, 2019 Investment Quantity Cost per Share Earl Corp. 990 $15.00 $10.60 Josie Corp. 850 20.00 15.60 Asher Corp. 500 8.10 7.20 During 2020, the following transactions took place: 1. On March 1, Josie Corp. paid a $2 per share dividend. 2 On April 30, Bramble sold 300 shares of Asher Corp. for $10.00 per share. 3. On May 15, Bramble purchased 200 more Earl Corp. shares at $16.00 per share. 4. At December 31, 2020, the shares had the following market prices per share: Earl Corp. $17.00; Josie Corp. $19.00; and Asher Corp. $8.00. During 2021, the following transactions took place: 5. On February 1, Bramble sold the remaining Asher Corp. shares for $7.00 per share. 6. On March 1, Josie Corp. paid a $2 per share dividend. 7. On December 21, Earl Corp. declared a cash dividend of $3 per share to be paid in the next month. 8. At December 31, 2021, the shares had the following market prices per share: Earl Corp. $19.00 and Josie Corp. $21.00. Bridgeport Ltd. Statement of Financial Position 2020 LA 69 2021 Bridgeport Ltd. Statement of Comprehensive Income $X + Bridgeport Ltd. Statement of Changes in Accumulated Other Comprehensive Income 2020 2020 $X 2021 $ $X + 2021 $X Bridgeport Ltd. Statement of Comprehensive Income $X + Bridgeport Ltd. Statement of Changes in Accumulated Other Comprehensive Income 2020 2020 $X 2021 $ $X + 2021 $X Net Income The net income 69 2020 change. 69 2021 Which of the following areas is sourcing NOT attempting tocontinuously improve?a. marketingb. costc. qualityd. delivery this question is related to the intriduction to the business law.Craig works full time for BrAd Ad Solutions as a creative designer. After he hadbeen employed for 12 months, Craig asked to take four weeks paid annual leave.Brenda told Craig that it was the firms policy that all staff should only receivethree weeks paid annual leave and this was a term of the employment contractsigned by Craig when he first commenced. Craig did sign the contract on his firstday at work, but he did not read the terms carefully.(a) How many weeks annual leave should Craig receive and what is the authorityfor your answer?(b) Will the clause in the employment contract be enforceable against Craig withrespect to annual leave? Why do you think Mercedes decided to build the plant from the ground up in Alabama rather than buy an existing plant in, say, Detroit? Explain and give examples Which of the following groups is not mentioned in the case as having conflicted with Wendys over its nonparticipation in the FFP? Multiple Choice mayors the media university workers local legislators university students