Transmission lines. An average of 120 kW of electric power is sent to a small town from a power plant 10 km away. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the power is transmitted at (a) 240 V and (b) 24,000 V. Show how P240V =100 kW and P24000V = 10 kW. (2 Points)
Explain why power lines are high voltage, yet our home sockets are mostly 120 V. (3 Points)
Hint: We cannot use P = V2/R because if R is the resistance of the transmission lines, we don’t know the voltage drop along them. The given voltages are applied across the lines plus the load (the town). But we can determine the current I in the lines and then find the power loss from for both cases (a) and (b)

Answers

Answer 1

To answer your question, let's first calculate the power loss in both cases (a) and (b) using the given information.

1. Calculate the current (I) in the transmission lines:
Power (P) = Voltage (V) × Current (I)
So, I = P / V

(a) When the power is transmitted at 240 V:
I_240V = 120 kW / 240 V = 500 A

(b) When the power is transmitted at 24,000 V:
I_24000V = 120 kW / 24,000 V = 5 A

2. Calculate the power loss (P_loss) in the transmission lines:
P_loss = I^2 × R

(a) For 240 V:
P_loss_240V = (500 A)^2 × 0.40 Ω = 100 kW

(b) For 24,000 V:
P_loss_24000V = (5 A)^2 × 0.40 Ω = 10 kW

Now, let's explain why power lines are high voltage, yet our home sockets are mostly 120 V (3 Points).

High voltage transmission lines are used to minimize power losses during transmission. As we've calculated above, the power loss is directly proportional to the square of the current (I^2 × R). By increasing the voltage and reducing the current, power losses can be significantly reduced.

However, high voltage is not safe for use in homes and other consumer appliances. That's why transformers are used to step down the high voltage from the transmission lines to a lower, safer voltage (like 120 V) before delivering power to our homes. This ensures efficient transmission of electricity over long distances with minimal power loss, while maintaining safety for end-users.

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Related Questions

A periodic signal is the summation of sinusoids of 5000 Hz, 2300 Hz and 3400 Hz Determine the signal's Nyquist frequency and an appropriate sampling frequency a The signal's Nyquist frequency is HZ b. Consider both cost and quality of the following frequencies, the most appropriate sampling rate for this signal would be click to select) Hz.

Answers

A. The Nyquist frequency for this signal is 5000/2 = 2500 Hz.

B. An appropriate sampling rate for this signal would be at least 5000 Hz.

A periodic signal can be expressed as the sum of sinusoids of different frequencies, amplitudes, and phases. In this case, the signal is the summation of sinusoids of 5000 Hz, 2300 Hz, and 3400 Hz. The Nyquist frequency is defined as half of the sampling rate, which is equal to the highest frequency component in the signal.
To determine the appropriate sampling frequency for this signal, we need to consider both cost and quality. A higher sampling rate provides better quality but requires more processing power and memory, which increases the cost. On the other hand, a lower sampling rate reduces the cost but may result in loss of information and lower quality.
A good rule of thumb is to choose a sampling frequency that is at least twice the Nyquist frequency to avoid aliasing.  However, if we want to reduce the cost, we can choose a lower sampling rate, such as 6000 Hz or 8000 Hz, which are common sampling rates in audio applications. These sampling rates provide reasonable quality and are suitable for most applications. However, if we need higher quality, we may need to choose a higher sampling rate, such as 12000 Hz or 16000 Hz.

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how much total kinetic energy will an electron–positron pair have if produced by a 3.64-mev photon?

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When a photon interacts with a nucleus or an electron, it can be absorbed by the atom, and its energy is transferred to the atom's electron(s),

Ejected from the atom, or it can undergo pair production. In pair production, the energy of the photon is converted into the rest mass of an electron-positron pair.The minimum energy required for pair production is 2m_ec^2 = 1.022 MeV, where m_e is the mass of the electron and c is the speed of light.In this case, the photon has an energy of 3.64 MeV, which is greater than the minimum energy required for pair production. Therefore, the photon can produce an electron-positron pair.The total energy of the electron-positron pair will be equal to the energy of the photon, which is 3.64 MeV. This energy will be divided between the electron and the positron in some proportion, depending on the specifics of the pair production event.

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can light phenomena be better explained by a transverse wave model or by a longitudinal wave model? explain how you know

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Light phenomena can be better explained by a transverse wave model rather than a longitudinal wave model.

This is because light waves oscillate perpendicular to the direction of their propagation, which is the characteristic of a transverse wave. On the other hand, longitudinal waves oscillate parallel to their propagation direction, which is not the case for light waves.

Additionally, the behavior of light waves in different mediums, such as reflection and refraction, can be explained by the transverse wave model. When light waves hit a surface, they bounce off at the same angle they hit the surface, which is known as the law of reflection. Similarly, when light waves pass through a medium with a different refractive index, they bend or change direction, which is known as refraction. These phenomena can be explained using the wave nature of light and its transverse oscillations.

Therefore, it is safe to say that the transverse wave model is a better explanation for light phenomena than the longitudinal wave model.

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Light phenomena can be better explained by a transverse wave model rather than a longitudinal wave model. This is because light waves are known to have electric and magnetic fields that are perpendicular to each other and to the direction of the wave propagation.

This characteristic of light waves is consistent with the properties of transverse waves where the displacement of particles is perpendicular to the direction of wave propagation.

On the other hand, longitudinal waves have displacements that are parallel to the direction of wave propagation, which is not observed in light waves.

Therefore, the transverse wave model provides a more accurate explanation for the behavior of light waves.

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a spherical solid, centered at the origin, has radius 100 and mass density \delta(x,y,z)=104 -\left(x^2 y^2 z^2\right). find its mass.

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The mass of the spherical solid is approximately 3.50 × 10⁷ units of mass (assuming units of mass are not specified in the question).

To find the mass of the spherical solid, we need to integrate the given mass density function over the volume of the sphere. Using spherical coordinates, we have:

m = ∫∫∫ δ(x,y,z) dV= ∫∫∫ (10^4 - x² y² z²) dV= ∫0²π ∫0^π ∫0¹⁰⁰ (10⁴ - r⁴ sin²θ cos²θ) r² sinθ dr dθ dφ= 4π ∫0¹⁰⁰ (10⁴r² - r⁶/3) dr= (4/3)π (10⁴r³ - r⁷/21)|0¹⁰⁰= (4/3)π [(10¹⁰ - 10⁷/3)]≈ 3.50 × 10⁷ units of mass.

Therefore, the mass of the spherical solid is approximately 3.50 × 10⁷ units of mass.

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Calculate the horizontal force P on the light 10° wedge necessary to initiate movement of the 40-kg cylinder. The coefficient of static friction for both pairs of contacting surfaces is 0.25. Also determine the friction force FB at point B. (Caution: Check carefully your assumption of where slipping occurs.)

Answers

A horizontal force of 68.56 N is required to initiate the movement of the cylinder and the friction force at point B is 98 N.

To find the force P necessary to initiate movement of the cylinder, we can use the equation:

P = mg * tan(θ) + μmg * cos(θ)

where m is the mass of the cylinder, g is the acceleration due to gravity, θ is the angle of the wedge, and μ is the coefficient of static friction between the cylinder and the wedge.

Substituting the values given, we get:

P = 40 kg * 9.8 m/s^2 * tan(10°) + 0.25 * 40 kg * 9.8 m/s^2 * cos(10°)

P = 68.56 N

To find the friction force FB at point B, we need to first determine if slipping occurs at point A or point B. Assuming that slipping occurs at point B, we can calculate the friction force as:

FB = μN

where N is the normal force acting on the cylinder at point B. The normal force is equal to the weight of the cylinder, which is:

N = mg = 40 kg * 9.8 m/s^2 = 392 N

Substituting this into the equation for FB, we get:

FB = 0.25 * 392 N = 98 N

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A horizontal force of 68.56 N is required to initiate the movement of the cylinder and the friction force at point B is 98 N.

To find the force P necessary to initiate movement of the cylinder, we can use the equation:

P = mg * tan(θ) + μmg * cos(θ)

where m is the mass of the cylinder, g is the acceleration due to gravity, θ is the angle of the wedge, and μ is the coefficient of static friction between the cylinder and the wedge.

Substituting the values given, we get:

P = 40 kg * 9.8 m/s^2 * tan(10°) + 0.25 * 40 kg * 9.8 m/s^2 * cos(10°)

P = 68.56 N

To find the friction force FB at point B, we need to first determine if slipping occurs at point A or point B. Assuming that slipping occurs at point B, we can calculate the friction force as:

FB = μN

where N is the normal force acting on the cylinder at point B. The normal force is equal to the weight of the cylinder, which is:

N = mg = 40 kg * 9.8 m/s^2 = 392 N

Substituting this into the equation for FB, we get:

FB = 0.25 * 392 N = 98 N

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The wavelength of the red light from a calcium flame is 617 nm. This light originated from a calcium atom in the hot flame. In the calcium atom from which this light originated, what was the period of the simple harmonic motion which was the source of this electromagnetic wave?

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The period of the simple harmonic motion in the calcium atom that produced the red light with a wavelength of 617 nm was 2.06 x 10^-15 seconds.

The period of the simple harmonic motion in the calcium atom that produced the red light with a wavelength of 617 nm can be calculated using the formula T = 1/f, where T is the period and f is the frequency. The frequency can be calculated using the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency.
Therefore, f = c/λ = (3.00 x 10^8 m/s)/(617 x 10^-9 m) = 4.86 x 10^14 Hz
Substituting this frequency into the equation T = 1/f, we get
T = 1/(4.86 x 10^14 Hz) = 2.06 x 10^-15 seconds
Therefore, the period of the simple harmonic motion in the calcium atom that produced the red light with a wavelength of 617 nm was 2.06 x 10^-15 seconds.

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The period of the simple harmonic motion, the source of the electromagnetic wave in the calcium atom is  2.06 x 10^-15 seconds.

To find the period of the simple harmonic motion which was the source of the electromagnetic wave, we can use the formula:

Period (T) = 1 / frequency (f)

First, we need to find the frequency. We can do that by using the speed of light (c) and the wavelength (λ) of the red light from the calcium flame:

c = λ * f

The speed of light (c) is approximately 3 x 10^8 meters per second (m/s), and the wavelength (λ) is 617 nm, which is equivalent to 617 x 10^-9 meters. Solving for frequency (f), we get:

f = c / λ = (3 x 10^8 m/s) / (617 x 10^-9 m) ≈ 4.86 x 10^14 Hz

Now, we can find the period (T) using the frequency (f):

T = 1 / f = 1 / (4.86 x 10^14 Hz) ≈ 2.06 x 10^-15 s

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What is the critical angle for the interface between water and crown glass? nglass= 1.52, nwater=1.33.
Express your answer using three significant figures.
?C = ?
Part B
To be internally reflected, the light must start in which material?
To be internally reflected, the light must start in which material?
in water
in crown glass
in any of the materials
none of the above

Answers

For water and crown glass, the critical angle is sinC = 1.52/1.33 = 1.144
The light must start in the material with the higher refractive index, which in this case is the crown glass.

The critical angle is the minimum angle of incidence at which a light ray is refracted at an interface and no longer enters the second medium, but rather undergoes total internal reflection. It can be calculated using the formula sinC = n2/n1, where n1 is the refractive index of the first medium (in this case, water) and n2 is the refractive index of the second medium (in this case, crown glass).
Therefore, for water and crown glass, the critical angle is sinC = 1.52/1.33 = 1.144. Taking the inverse sine of this value gives the critical angle as C = 48.8 degrees. This means that any incident ray of light that exceeds an angle of 48.8 degrees with the normal to the interface between water and crown glass will undergo total internal reflection and not enter the crown glass.
To be internally reflected, the light must start in the material with the higher refractive index, which in this case is the crown glass. When a ray of light travels from crown glass into water at an angle greater than the critical angle, it will undergo total internal reflection and bounce back into the crown glass, rather than being refracted out into the water.

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Another friend of yours, who is taking an earth science class, tries to move a rock with a weight of 10,000 N. He strains and huffs and puffs and sweats, but he fails to budge the rock. How much work did your friend do?

Answers

Your friend did not do any work in trying to move the rock due to the absence of displacement. In order to calculate the work done, we need to consider two factors: the force applied and the displacement caused by that force.

Work is defined as the product of force and displacement in the direction of the force. In this scenario, although your friend exerted a force of 10,000 N on the rock, he failed to move it. Since there was no displacement, the work done by your friend is zero. Work requires the application of force over a distance, resulting in a change in position or displacement.

Without any displacement, no work is accomplished. It's important to note that while your friend expended effort and energy in attempting to move the rock, work specifically refers to the transfer of energy to cause displacement. To perform work, an object must be displaced.

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By what percent is the speed of blue light (450?nm, n450nm = 1.640) less than the speed of red light (680?nm, n680nm = 1.615), in silicate flint glass (Figure 1) ?
Express your answer using two significant figures.

Answers

The speed of blue light in silicate flint glass is about 1.61% less than the speed of red light in the same material.

The speed of light in a material is given by the equation:

v = c/n,

where v is the speed of light in the material, c is the speed of light in a vacuum, and n is the refractive index of the material.

we can find the speed of blue light and red light in silicate flint glass:

For blue light: v450nm = c/n450nm = (3.00 x 10^8 m/s)/(1.640) = 1.83 x 10^8 m/s

For red light: v680nm = c/n680nm = (3.00 x 10^8 m/s)/(1.615) = 1.86 x 10^8 m/s

The percent difference in speed between blue light and red light in silicate flint glass can be calculated using the formula:

% difference = |(v450nm - v680nm)/v680nm| x 100%

% difference = |(1.83 x 10^8 m/s - 1.86 x 10^8 m/s)/1.86 x 10^8 m/s| x 100%

% difference = 1.61%

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Assume all angles to be exact. light passes from a crown glass container into water. if the angle of refraction is 56 ∘ , what is the angle of incidence?

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The angle of incidence when light passes from a crown glass container into water, given that the angle of refraction is 56° is approximately 41°.

According to Snell's Law, n₁sinθ₁ = n₂sinθ₂, where n₁ and n₂ are the refractive indices of the media, and θ₁ and θ₂ are the angles of incidence and refraction, respectively. Since light travels from crown glass (n₁ = 1.52) to water (n₂ = 1.33), we have:

1.52sinθ₁ = 1.33sin56°

Solving for θ₁, we get:

θ₁ ≈ sin⁻¹(1.33sin56°/1.52) ≈ 41°

As a result, assuming that the angle of refraction is 56° and that light is passing through a crown glass container into water, the angle of incidence is roughly 41°.

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Which is larger, the area under the t-distribution with 10 degrees of freedom to the right of t= 2.32 or the area under the standard normal distribution to the right of z=2.32? The area under the t-distribution with 10 degrees of freedom to the right of t=2.32 is the area under the standard normal distribution to the right of z=2.32.

Answers

Therefore, we can conclude that the area under the t-distribution with 10 degrees of freedom to the right of t=2.32 is larger than the area under the standard normal distribution to the right of z=2.32, since 0.0204 > 0.0107.

A t-distribution is used when we have a small sample size and do not know the population standard deviation, while a standard normal distribution is used when we have a large sample size and know the population standard deviation. The t-distribution is wider and flatter than the standard normal distribution, which means that it has more area in the tails.

Now, to compare the area under the t-distribution with 10 degrees of freedom to the right of t=2.32 and the area under the standard normal distribution to the right of z=2.32, we need to calculate these areas using a statistical software or a table.
Using a t-table, we can find that the area under the t-distribution with 10 degrees of freedom to the right of t=2.32 is approximately 0.0204. This means that there is a 2.04% chance of getting a t-value greater than 2.32 in a sample of size 10.
Using a standard normal table, we can find that the area under the standard normal distribution to the right of z=2.32 is approximately 0.0107. This means that there is a 1.07% chance of getting a z-value greater than 2.32 in a sample of any size.
Therefore, we can conclude that the area under the t-distribution with 10 degrees of freedom to the right of t=2.32 is larger than the area under the standard normal distribution to the right of z=2.32, since 0.0204 > 0.0107.

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one hundred meters of 2.00 mm diameter wire has a resistance of 0.532 ω. what is the resistivity of the material from which the wire is made?

Answers

The resistivity of the material from which the wire is made is 1.33 x 10⁻⁸ Ωm.

The resistivity of the material from which a 2.00 mm diameter wire is made can be calculated if the wire's length, diameter, and resistance are known.

The resistivity (ρ) of the material can be calculated using the formula:

ρ = (πd²R)/(4L)

where d is the diameter of the wire, R is the resistance of the wire, and L is the length of the wire.

Substituting the given values, we get:

ρ = (π x (2.00 x 10⁻³ m)² x 0.532 Ω)/(4 x 100 m) = 1.33 x 10⁻⁸ Ωm

Therefore, the resistivity of the material from which the wire is made is 1.33 x 10⁻⁸ Ωm.

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Nolan is making a capacitor using plates that have an area


of 3. 2 × 10–4 m2 separated by a distance of 0. 20 mm. He has the two dielectrics listed in the table.



A 2 column table with 2 rows. The first column is labeled dielectric with entries 1, 2. The second column is labeled dielectric constant with entries 6. 8, 1. 5.


At a voltage of 1. 5 V, how much more charge can Nolan store in the capacitor using dielectric 1 than he could store using dielectric 2?



7. 5 × 10–11 C


9. 6 × 10–11 C


1. 1 × 10–10 C


1. 4 × 10–10 C



Answer is C 1. 1 x 10^-10

Answers

C: 1.1 × 10⁻¹⁰ C. By using dielectric 1 with a higher dielectric constant of 6.8, Nolan can store more charge in the capacitor compared to dielectric 2 with a lower dielectric constant of 1.5.

The charge stored in a capacitor can be calculated using the formula Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage. The capacitance of a parallel plate capacitor is given by C = ε₀ × εᵣ × A/d, where ε₀ is the vacuum permittivity, εᵣ is the relative permittivity (dielectric constant), A is the area of the plates, and d is the distance between the plates.

Given that the voltage is 1.5 V, the area is 3.2 × 10⁻⁴ m², and the distance is 0.20 mm (which is equivalent to 0.20 × 10⁻³ m), we can calculate the capacitance for each dielectric.

For dielectric 1:

C₁ = ε₀ × ε₁ × A/d = (8.85 × 10⁻¹² F/m) × 6.8 × 3.2 × 10⁻⁴ m² / (0.20 × 10⁻³ m) ≈ 1.088 × 10⁻¹⁰ F

For dielectric 2:

C₂ = ε₀ × ε₂ × A/d = (8.85 × 10⁻¹² F/m) × 1.5 × 3.2 × 10⁻⁴ m² / (0.20 × 10⁻³ m) ≈ 4.8 × 10⁻¹¹ F

The difference in charge storage can be calculated as ΔQ = C₁ × V - C₂ × V ≈ (1.088 × 10⁻¹⁰ F) × (1.5 V) - (4.8 × 10⁻¹¹ F) × (1.5 V) ≈ 1.1 × 10⁻¹⁰ C.

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Light of wavelength λ = 595 nm passes through a pair of slits that are 23 μm wide and 185 μm apart. How many bright interference fringes are there in the central diffraction maximum? How many bright interference fringes are there in the whole pattern?

Answers

The number of bright interference fringes in the central diffraction maximum can be found using the formula:

n = (d sin θ) / λ

where n is the number of fringes, d is the distance between the slits, θ is the angle between the central maximum and the first bright fringe, and λ is the wavelength of light.

For the central maximum, the angle θ is zero, so sin θ = 0. Therefore, the equation simplifies to:

n = 0

So there are no bright interference fringes in the central diffraction maximum.

The number of bright interference fringes in the whole pattern can be found using the formula:

n = (mλD) / d

where n is the number of fringes, m is the order of the fringe, λ is the wavelength of light, D is the distance from the slits to the screen, and d is the distance between the slits.

To find the maximum value of m, we can use the condition for constructive interference:

d sin θ = mλ

where θ is the angle between the direction of the fringe and the direction of the center of the pattern.

For the first bright fringe on either side of the central maximum, sin θ = λ/d. Therefore, the value of m for the first bright fringe is:

m = d/λ

Substituting this value of m into the formula for the number of fringes, we get:

n = (d/λ)(λD/d) = D

So there are D bright interference fringes in the whole pattern, where D is the distance from the slits to the screen, in units of the wavelength of light.

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(d) estimate the time t t at which the cars are again side by side. (round your answer to one decimal place.)

Answers

To estimate the time at which the cars are again side by side, we need to find the time it takes for Car A to travel one complete lap more than Car B.

We know that Car A travels one lap in 100 seconds, while Car B travels one lap in 120 seconds. Let's call the time it takes for the cars to be side by side again "t". After t seconds, Car A will have completed t/100 laps, while Car B will have completed t/120 laps. For the cars to be side by side again, Car A must have completed one more lap than Car B.

So we need to solve the equation:

t/100 = t/120 + 1

Multiplying both sides by 12000 (the least common multiple of 100 and 120) gives:

120t = 100t + 12000

Simplifying this equation gives:

20t = 12000

t = 600 seconds

Therefore, the cars will be side by side again after 600 seconds, or 10 minutes.

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(12 pts) 9. A soap film has refractive index /.33. There is air on either side of the film. Light of wavelength Ajir in air shines on the film perpendicular to its surface_ It is observed that the largest value of Aair for which light reflected from the two surfaces of the film has constructive interference is Aair = 800 nm What is the thickness of the film?

Answers

A soap film has refractive index 1.33. There is air on either side of the film. Light of wavelength Aair in air shines on the film perpendicular to its surface. It is observed that the largest value of Aair for which light reflected from the two surfaces of the film has constructive interference is Aair = 800 nm. The thickness of the film is 300 nm.

To determine the thickness of the soap film, we can use the concept of constructive interference in thin films. Constructive interference occurs when the path length difference between the two reflected waves is an integer multiple of the wavelength.

In this case, we have a soap film with a refractive index of 1.33 and air on either side. The incident light has a wavelength of λ_air = 800 nm = 800 × 10^(-9) m.

The path length difference between the two reflected waves is twice the thickness of the film, since the light travels through the film twice.

So we can set up the following equation:

2 * t * n_film = m * λ_air

where t is the thickness of the film, n_film is the refractive index of the film, m is an integer representing the order of the interference, and λ_air is the wavelength of light in air.

Since we are interested in the largest value of λ_air for which constructive interference occurs, we can choose m = 1 (first order).

Plugging in the values:

2 * t * 1.33 = 1 * 800 × 10^(-9) m

Simplifying the equation:

t = (800 × 10^(-9) m) / (2 * 1.33)

Calculating the value:

t ≈ 300 × 10^(-9) m

Therefore, the thickness of the soap film is approximately 300 nm.

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A thin 100 g disk with a diameter of 8 cm rotates about an axis through its center with 0.15 j of kinetic energy. What is the speed of a point on the rim?

Answers

Speed of a point on the rim is 0.98 m/s.

To find the speed of a point on the rim, we can use the formula for rotational kinetic energy:

Krot = 1/2 I ω^2

where Krot is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.

We can find the moment of inertia of the disk using the formula:

I = 1/2 m r^2

where m is the mass of the disk and r is the radius.

Since the disk has a diameter of 8 cm, its radius is 4 cm or 0.04 m. Therefore, the moment of inertia is:

I = 1/2 (0.1 kg) (0.04 m)^2 = 8.0 x 10^-5 kg m^2

Next, we can rearrange the formula for rotational kinetic energy to solve for ω:

ω = √(2 Krot / I)

Plugging in the given values, we get:

ω = √(2 x 0.15 J / 8.0 x 10^-5 kg m^2) = 24.50 rad/s

Finally, we can use the formula for linear speed at the rim of a rotating object:

v = ω r

where v is the linear speed and r is the radius.

Plugging in the values, we get:

v = (24.50 rad/s) (0.08 m / 2) = 0.98 m/s

Therefore, the speed of a point on the rim of the disk is 0.98 m/s.


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1.find tα /2,n-1 (critical value) for the following levels of α (assume 2-tailed test) a.α = .05 and n = 15 b.α = .01 and n = 12 c.α = .10 and n = 21

Answers

The critical values are  2.145, 3.106 and 1.725.

To find tα/2,n-1 (critical value) for a given level of α and degrees of freedom (df), we can use a t-distribution table or a statistical software. Here are the answers for the given values of α and n:

a. For α = .05 and n = 15, the df = n-1 = 14. Using a t-distribution table with α/2 = .025 and df = 14, we find the critical value to be 2.145. This means that if the calculated t-value falls beyond ±2.145, we reject the null hypothesis at the 5% significance level.

b. For α = .01 and n = 12, the df = n-1 = 11. Using a t-distribution table with α/2 = .005 and df = 11, we find the critical value to be 3.106. This means that if the calculated t-value falls beyond ±3.106, we reject the null hypothesis at the 1% significance level.

c. For α = .10 and n = 21, the df = n-1 = 20. Using a t-distribution table with α/2 = .05 and df = 20, we find the critical value to be 1.725. This means that if the calculated t-value falls beyond ±1.725, we reject the null hypothesis at the 10% significance level.

The t-distribution is used when the sample size is small and/or the population standard deviation is unknown. The critical value tα/2,n-1 represents the t-score that separates the rejection region (the extreme values that lead to rejecting the null hypothesis) from the acceptance region (the values that do not lead to rejecting the null hypothesis).

For a two-tailed test, we divide the significance level α by 2 and find the critical value for the lower tail and the upper tail separately. The degrees of freedom (df) represent the number of independent observations in the sample and affect the shape and variability of the t-distribution. As the sample size increases, the t-distribution becomes closer to the normal distribution, which has a fixed critical value of 1.96 for α = .05 and a two-tailed test.

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Consider a planet of mass m that has a circular orbit of radius r around a star of mass M >> m. The planet's Hill radius ry is defined such that at this distance from the planet toward the star, the forces on an orbiting test mass will be in balance. a. At such a distance rh from the planet, and r - rh from the star, write out the combined acceleration gtot from the star's gravity and the planet's gravity, as well as the centrifugal acceleration from orbiting the star with the same period as the planet. b. Now set this &tot = 0, and solve for ry in terms of m, M, and r, under the approximations m

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a. The combined acceleration gtot at distance rh from the planet in a circular orbit around the star with radius r is given by gtot = -(GM/r^2)rh + (Gm/r^2)(r - rh) + (v^2/rh), where G is the gravitational constant, M is the mass of the star, m is the mass of the planet, and v is the orbital velocity of the planet.

b. Setting gtot = 0 and solving for ry, the Hill radius is approximately given by ry = r[(m/3M)^(1/3)]. This approximation assumes that m << M and that the orbit of the planet is circular. The Hill radius is the maximum distance from the planet where its gravity dominates over the star's gravity and where objects can be stably bound to the planet.

To calculate the combined acceleration, we must consider the gravitational forces of both the star and the planet on an orbiting test mass at distance rh from the planet.

The centrifugal acceleration is also included as it must be balanced by the gravitational forces. Setting gtot to zero and solving for ry involves algebraic manipulation and the use of the approximation that m << M and the orbit is circular.

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a texas railroad section was recently surveyed with rtk and found to be 1908v x 1902v. what would half that acreage be calculated out to?

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A property parcel's acreage can be determined by multiplying its length by its width and dividing the result by 43,560, the number of square feet in an acre.

The entire acreage can be estimated using the following formula given that the Texas railroad segment is 1908 feet by 1902 feet:

1908 feet by 1902 feet divided by 43,560 feet per acre equals 83.063 acres.

We can just split this acreage by two to get half of it:

Half an acre is equal to 83.063% of an acre, or 41.5315 acres.

Therefore, 41.53 acres would be about half of the Texas railway section. It's important to note that this computation makes the assumption that the parcel is rectangular and has straight edges.

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The size of a property lot can be calculated by multiplying its width and length and then dividing the product by 43,560, which is the equivalent of one acre in square feet.

How to solve

If the Texas railroad segment measures 1908 feet by 1902 feet, the total area can be computed utilizing this equation.

83063 acres can be calculated by dividing an area of 1908 feet by 1902 feet by the conversion factor of 43,560 feet per acre.

We can easily divide this piece of land into two equal parts, obtaining half of it.

An area of 0. 5 acres can be expressed as 83. 063% of an entire acre or approximately 41. 5315

Hence, the Texas railroad section would comprise roughly twice the area of 41. 53 It should be emphasized that in this calculation, the parcel is assumed to have a rectangular shape and its edges are straight.

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a current density is supported by a hollow cylindrical conducting pipe located between

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The current density in a hollow cylindrical conducting pipe can be determined using Ampere's Law and the Biot-Savart Law.

To find the current density, follow these steps:
1. Consider a hollow cylindrical conducting pipe with a given radius and length.
2. Apply Ampere's Law to determine the magnetic field around the pipe.
3. Use the Biot-Savart Law to relate the magnetic field to the current density.
4. Solve for the current density.

In a hollow cylindrical conducting pipe, current density is distributed uniformly on the surface. Ampere's Law helps calculate the magnetic field around the pipe, while the Biot-Savart Law relates this magnetic field to current density. By solving these equations, the current density can be found.

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We have an NMOS transistor with k'=800 μA/V2, W/L=10, VTh=0.4V, and λ= 0.06 V-1, and it is operated with Vgs=2.7V. What current Id does the transistor need to have when it is operating at the edge of saturation? (mA)

Answers

To find the drain current Id when the transistor is operating at the edge of saturation, we can use the following equation:

Id = k' * [(W/L) * (Vgs - VTh) - Vds/2]² * (1 + λ*Vds)

where:

- k' = 800 μA/V^2 is the transconductance parameter

- W/L = 10 is the width-to-length ratio of the transistor

- VTh = 0.4V is the threshold voltage

- λ = 0.06 V^-1 is the channel-length modulation parameter

- Vgs = 2.7V is the gate-source voltage

At the edge of saturation, the drain-source voltage Vds is equal to (Vgs - VTh). Therefore, we can substitute Vds = Vgs - VTh into the equation above to obtain:

Id = k' * [(W/L) * (Vgs - VTh) - (Vgs - VTh)/2]² * (1 + λ*(Vgs - VTh))

Simplifying this expression, we get:

Id = k' * [(W/L) * (Vgs - VTh)/2]² * (1 + λ*(Vgs - VTh))

Plugging in the given values, we get:

Id = 800 μA/V^2 * [(10) * (2.7V - 0.4V)/2]² * (1 + 0.06 V^-1 * (2.7V - 0.4V))

Id = 2.455 mA

Therefore, the transistor needs to have a drain current of 2.455 mA when it is operating at the edge of saturation.

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While in the first excited state, a hydrogen atom is illuminated by various wavelengths of light.
What happens to the hydrogen atom when illuminated by each wavelength?
450.3 nm?
The options are:
stays in 2nd state
jumps to 3rd state
jumps to 4th state
jumps to 5th state
jumps to 6th state
is ionized
I have already tried jumps to 5th state, and jumps to 4th state and they are incorrent.

Answers

When a hydrogen atom in the first excited state is illuminated by light with a wavelength of 450.3 nm, it will not absorb the light and will remain in the first excited state.

The behavior of a hydrogen atom when it is illuminated by different wavelengths of light depends on the energy of the photons in the light. The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. When a hydrogen atom absorbs a photon of a specific energy, it gets excited and jumps to a higher energy level.

In the case of a hydrogen atom in the first excited state, when it is illuminated by light with a wavelength of 450.3 nm, the atom will not remain in the same state. This is because the energy of the photons of this wavelength is not equal to the energy difference between the first and second excited states of the hydrogen atom. Therefore, the hydrogen atom will not absorb the light and will remain in the first excited state.

To calculate which energy level the hydrogen atom will jump to when illuminated by a specific wavelength of light, we can use the Rydberg formula:

1/λ = R(1/n1^2 - 1/n2^2)

where λ is the wavelength of the light, R is the Rydberg constant (1.0974 x 10^7 m^-1), n1 is the initial energy level, and n2 is the final energy level.

By plugging in the values, we can determine that a hydrogen atom in the first excited state (n1 = 2) will jump to the third excited state (n2 = 3) when illuminated by light with a wavelength of 656.3 nm.

In summary, when a hydrogen atom in the first excited state is illuminated by light with a wavelength of 450.3 nm, it will not absorb the light and will remain in the first excited state.

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sunlight of intensity 600 w m−2 is incident on a building at 60° to the vertical. what is the solar intensity or insolation, on (a) a horizontal surface? and (b) a vertical surface?

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When sunlight with an intensity of 600 W/m² is incident on a building at a 60° angle to the vertical, the solar intensity or insolation on different surfaces can be calculated using trigonometry.

(a) For a horizontal surface, the effective solar intensity is the incident intensity multiplied by the cosine of the angle. In this case, cos(60°) = 0.5. Therefore, the solar intensity on a horizontal surface is 600 W/m² × 0.5 = 300 W/m².

(b) For a vertical surface, the effective solar intensity is the incident intensity multiplied by the sine of the angle. In this case, sin(60°) = √3/2 ≈ 0.866. Therefore, the solar intensity on a vertical surface is 600 W/m² × 0.866 ≈ 519.6 W/m².
So, the insolation on a horizontal surface is 300 W/m² and on a vertical surface is approximately 519.6 W/m².

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To stretch a relaxed biceps muscle 2.2 cm requires a force of 25 N. Find the Young's modulus for the muscle tissue, assuming it to be a uniform cylinder of length 0.24 m and cross-sectional area 48 cm2.

Answers

Young's modulus of the muscle tissue is 56,811.4 Pa.

To calculate Young's modulus for the muscle tissue, we can use the formula:

Young's modulus = stress / strain

where stress is the force per unit area applied to the muscle tissue, and strain is the ratio of the change in length of the tissue to its original length.

Given that a force of 25 N is required to stretch the muscle tissue by 2.2 cm, we can calculate the stress as:

stress = force / area
      = 25 N / 0.0048 m^2
      = 5208.33 Pa

We can also calculate the strain as:

strain = change in length / original length
       = 0.022 m / 0.24 m
       = 0.0917

Therefore, the Young's modulus of the muscle tissue is:

Young's modulus = stress/strain
               = 5208.33 Pa / 0.0917
               = 56,811.4 Pa

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A girl strikes a 0.445kg soccer ball with a net force of 5.92N. What is the acceleration of the soccer ball? 0 13.3 m/s2 O 0.0752 m/s2 0 6.36 m/s2 0 5.48 m/s2

Answers

The answer to the question is 13.3 m/s2, if a girl strikes a 0.445kg soccer ball with a net force of 5.92N.


To find the acceleration of the soccer ball, we can use the formula F = ma, where F is the net force applied to the ball, m is the mass of the ball, and a is the acceleration of the ball. We know that the mass of the ball is 0.445kg and the net force applied is 5.92N. Substituting these values into the formula, we get:

5.92N = 0.445kg x a

Solving for a, we get:

a = 5.92N / 0.445kg

a ≈ 13.3 m/s2

Therefore, the answer is that the acceleration of the soccer ball is 13.3 m/s2.

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Calculate the natural frequencies and mode shapes of a clamped-free beam. Express your solution in terms of E, I, p, and. This is called the cantilevered beam problem

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The natural frequencies and mode shapes of a clamped-free beam can be calculated using the cantilevered beam problem equation. These values are important for understanding how a beam will behave under different loads and conditions, and can help engineers design safer and more efficient structures.

The cantilevered beam problem is a classic example in structural engineering. The natural frequencies and mode shapes of a clamped-free beam can be calculated using the following equation:
f = (n^2 * pi^2 * E * I) / (2 * L^2 * p)
where f is the natural frequency, n is the mode number, E is the modulus of elasticity, I is the moment of inertia, L is the length of the beam, and p is the density of the material.
The mode shapes for a clamped-free beam are sinusoidal curves that increase in frequency as the mode number increases. The first mode shape is a half sine wave, the second mode shape is a full sine wave, and so on.
It is important to note that the cantilevered beam problem assumes that the beam is perfectly straight and has a uniform cross-section. Real-world beams may have slight variations in their shape and composition, which can affect their natural frequencies and mode shapes.

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a 0.505-kg mass suspended from a spring undergoes simple harmonic oscillations with a period of 1.35 s. How much mass, inkilograms, must be added to the object to change the period to2.2 s?

Answers

We need to add approximately 0.34 kg of mass to the object to change the period of its simple harmonic oscillations from 1.35 s to 2.2 s.

To solve this problem, we need to use the formula for the period of a simple harmonic oscillator: T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. We can rearrange this formula to solve for m: m = (T^2*k)/(4π^2).

Using the given values, we can calculate the mass of the object initially: m1 = (1.35^2*k)/(4π^2). We don't actually need to know the value of k, though, since we're only interested in the change in mass needed to change the period.

Let's call the additional mass we need to add "m2". Then, we can use the same formula with the new period of 2.2 s: m1 + m2 = (2.2^2*k)/(4π^2).

Now we can solve for m2: m2 = (2.2^2*k)/(4π^2) - m1. Plugging in the values we know, we get: m2 = (2.2^2*0.505)/(4π^2) - (1.35^2*0.505)/(4π^2) ≈ 0.34 kg.

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Lab 08: Reflection and Refraction of Light You will need to run a simulation to do the lab. Answer the following questions as you work through the lab. Write your answers in blue. (Note that we may miss your response if it does not stand out ) Re-load the file in word or PDF format in Canvas before the due date. Overview Light bends when it enters from one medium to another. This bending of light is called Refraction of light. The relationship between the angle of incidence (medium 1) and the angle of refraction (in the medium 2) is given by Snell’s Law: n_1 sin⁡〖θ_1=n_2 sin⁡〖θ_2 〗 〗 Eq. 8.1 Where n_1 is the index of refraction, θ_1 angle of incidence in medium 1; n_2 is the index of refraction, θ_2 is the angle of refraction in medium 2. The angles, θ are measured with respect to the normal to the surface between the two mediums. When light travels from an optically light medium to an optically dense medium, i.e. n_1 n2, the refracted light bends away from the normal. For a certain angle of incidence (called the critical angle, θ_c) the refracted ray will be 90 from the normal. If the angle of incidence is any larger, the ray is totally reflected in medium 1 and no light comes out of medium 2. This is called Total Internal Reflection. For this part of the lab, you will find the critical angle for different sets of boundaries. Select "More Tools" tab . Check the "normal" and "angle" box to view and measure the angles. 1. Set the Medium 1 = Glass (n1 = 1.5); Medium 2 = Air (n2 = 1.0). 2. Start with θ_1=0. Gradually increase θ_1 until the refracted ray, θ_2=90°. This incident angle is the critical angle, θ_c . If you keep on increasing θ_1, there will only be reflected light. In this way, you can figure out the critical angle for different mediums at the boundaries listed in the table below. Table 8.5: Critical angle of different sets of boundaries Medium 1 (n1) Medium 2 (n2) Critical Angle (c) Water Air Glass Air Glass Water Mystery Medium A Air Mystery Medium A Glass 3. Conclusion Question: (i) Based on your observation in the table, what is the condition for total internal reflection? (ii) Is there a total internal reflection if both mediums have same index of refraction (e.g. n_1=n_2 )? Explain your answer.

Answers

The condition for total internal reflection is when the angle of incidence (θ₁) is greater than the critical angle (θ_c).No, there is no total internal reflection if both mediums have the same index of refraction (n₁ = n₂).Based on your observations in the table, what is the condition for total internal reflection, and is there total internal reflection if both mediums have the same index of refraction (e.g., n₁ = n₂)?

Based on the observations in the table, the condition for total internal reflection is when the angle of incidence (θ₁) is larger than the critical angle (θ_

When the angle of incidence exceeds the critical angle, the refracted ray cannot escape the first medium and is totally reflected back into it.

No, there is no total internal reflection if both mediums have the same index of refraction (n₁ = n₂). Total internal reflection can only occur when light travels from a medium with a higher refractive index to a medium with a lower refractive index.

If the indices of refraction are equal, the angle of refraction (θ₂) will always be equal to the angle of incidence (θ₁), as determined by Snell's Law. In this case, the light will continue to propagate through the interface between the two mediums without any total internal reflection occurring.

Total internal reflection requires a change in the refractive index between the two mediums to cause a significant change in the angle of refraction, allowing the critical angle to be reached or exceeded.

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A 5. 0 kg mass and a 3. 0 kg mass are placed on top of a seesaw. The 3. 0 kg mass is 2. 00 m from the fulcrum as showa. Where should the 5. 0 kg mass be placed to keep the system from rotating?



Show work

Answers

A 5. 0 kg mass and a 3. 0 kg mass are placed on top of a seesaw. The 3. 0 kg mass is 2. 00 m from the fulcrum. The 5.0 kg mass should be placed 1.2 meters from the fulcrum to keep the system from rotating.

To keep the system from rotating, the torques on both sides of the fulcrum need to be balanced. Torque is calculated by multiplying the force applied by the distance from the fulcrum.

Let's denote the unknown distance from the fulcrum to the 5.0 kg mass as x.

The torque exerted by the 3.0 kg mass is given by:

[tex]Torque_3_k_g = (3.0 kg) * (9.8 m/s^2) * (2.0 m)[/tex]

The torque exerted by the 5.0 kg mass is given by:

[tex]Torque_5kg = (5.0 kg) * (9.8 m/s^2) * (x m)[/tex]

To keep the system in balance, the torques on both sides must be equal:

[tex]Torque_3kg = Torque_5kg[/tex]

Simplifying the equation:

[tex](3.0 kg) * (9.8 m/s^2) * (2.0 m) = (5.0 kg) * (9.8 m/s^2) * (x m)[/tex]

Solving for x:

(3.0 kg) * (2.0 m) = (5.0 kg) * (x m)

6.0 kg·m = 5.0 kg·x

Dividing both sides by 5.0 kg:

x = (6.0 kg·m) / (5.0 kg)

x = 1.2 m.

        Fulcrum

         |

         |

   5.0 kg | 3.0 kg

   -------|---------    

        1.2 m   2.0 m

In the diagram, the fulcrum is represented by "|". The 5.0 kg mass is placed 1.2 m from the fulcrum, while the 3.0 kg mass is placed 2.0 m from the fulcrum. This configuration ensures that the torques on both sides are balanced, preventing rotation of the system.

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