The statement "By default, non-equities are excluded from the universe in universal screening" is false. Universal screening refers to a comprehensive approach where all types of securities, including equities, bonds, and other financial instruments, are considered for analysis and inclusion in a portfolio.
The purpose of universal screening is to evaluate and select investments from a broad range of options based on predetermined criteria or factors.
However, it is possible to customize the screening process and apply filters to exclude specific categories or types of securities if desired. This allows investors to focus on specific asset classes or investment strategies.
The decision to include or exclude non-equities depends on the specific objectives and preferences of the investor or investment manager.
To know more about the Universal screening refer here :
https://brainly.com/question/31418117#
#SPJ11
a hot air balloon is filled with 1.89 x 10 squared liters of air at 21 c. if atmospheric pressure does not change, how hot must the air become in order to incerase the colume to 4.5 x 10 squared liters
To increase the volume of the hot air balloon from 1.89 x 10² liters to 4.5 x 10² liters, the air inside must be heated to approximately 121 °C.
What temperature should the air reach to expand its volume in the hot air balloon?To determine the required temperature for the air inside the hot air balloon to expand its volume, we can use Charles's Law, which states that the volume of a gas is directly proportional to its absolute temperature, assuming constant pressure. Given that the initial volume is 1.89 x 10² liters and the final volume is 4.5 x 10² liters, we can set up the proportion:
(V1 / T1) = (V2 / T2)
Solving for T2, we have:
T2 = (V2 * T1) / V1
Substituting the values, T1 = 21 °C (which is 294.15 K), V1 = 1.89 x 10² liters, and V2 = 4.5 x 10² liters, we can calculate T2 as follows:
T2 = (4.5 x 10² * 294.15 K) / (1.89 x 10²)
After performing the calculation, we find that T2 is approximately 121 °C.
Learn more about Volume
brainly.com/question/13338592
#SPJ11
Workbook Waves Module 5, Exercise 12 c, how do the waves interfere?
Select the correct answer
a. mildly constructively, just slightly larger amplitude than that of one of the wave
b. destructively
c. mildly destructively, just slightly smaller amplitude than that of one of the waves
d. no interference
e. constructively
Given the options provided, without further information, it is difficult to select the correct answer. Depending on the specifics of the exercise and the properties of the waves involved, any of the given options (a, b, c, d, or e) could potentially be correct.
To determine how the waves interfere in Workbook Waves Module 5, Exercise 12 c, we need more specific information about the scenario described in the exercise. Without additional context, it is not possible to determine the exact nature of the wave interference.
If the waves interfere "mildly constructively," it means that they combine to create a resulting wave with a slightly larger amplitude than that of one of the individual waves. This option is represented by choice "a."
If the waves interfere "destructively," it means that they combine in such a way that the resulting wave has a smaller amplitude than that of one or both of the individual waves. This option is represented by choice "b."
If the waves interfere "mildly destructively," it suggests that they combine to produce a resulting wave with a slightly smaller amplitude than that of one of the individual waves. This option is represented by choice "c."
If there is "no interference," it means that the waves do not affect each other when they meet. This option is represented by choice "d."
If the waves interfere "constructively," it means that they combine to create a resulting wave with a larger amplitude than that of the individual waves. This option is represented by choice "e."
Interference in waves can occur in different ways depending on the phase relationship between the waves and their amplitudes. Interference can be constructive, where the waves combine to produce a larger amplitude, or destructive, where the waves combine to produce a smaller amplitude or cancel each other out. It is also possible for interference to result in a combination of constructive and destructive effects, leading to different degrees of interference.
To lean more about Interference , click here: https://brainly.com/question/31857527
#SPJ11
Radioactive decay: (a) A radioactive sample is monitored with a radiation detector to have 5640 counts per minute_ Twelve hours later; the detector reads 1128 counts per minute. Calculate the decay constant and the half-life of the sample. (b) A 5.00 gram sample of charcoal from an ancient fire pit has a 14 a‚¬ activity of 63.0 disintegrationslminute_ living tree has a 14 C specific activity of 15.3 disintegrations/minutelgram: The half-life of 14 C is 5730 years. How old is the charcoal sample?
The decay constant of a radioactive sample can be determined by using the equation N = N0 ˣ e (e⁻λt) , and the half-life can be found using T1/2 = ln(2)/λ.
How can the decay constant and half-life of a radioactive sample be calculated based on count rates?(a) To calculate the decay constant, we can use the formula N = N0 ˣ (e⁻λt), where N is the final count rate, N0 is the initial count rate, λ is the decay constant, and t is the time.
By substituting the given values, we can solve for λ. The half-life can be found using the formula T1/2 = ln(2)/λ.
(b) The age of the charcoal sample can be determined using the formula t = (ln(N0/N) ˣ T1/2) / ln(2), where N0 is the initial activity, N is the current activity, T1/2 is the half-life, and t is the age of the sample.
By substituting the given values, we can calculate the age of the charcoal sample.
Learn more about radioactive sample
brainly.com/question/20159749
#SPJ11
determine the frequency heard by a stationary observer when a source moving at 96.0 km/hr is emitting a frequency of 400.0 hz and the air temperature is 20.0 degrees celsius
The frequency heard by a stationary observer when a source moving at 96.0 km/hr emits a frequency of 400.0 Hz and the air temperature is 20.0 degrees Celsius is 400.16 Hz.
What is the frequency perceived by a stationary observer when a source moving at 96.0 km/hr emits a 400.0 Hz frequency at an air temperature of 20.0 degrees Celsius?When a source emitting sound waves is in motion relative to an observer, the frequency perceived by the observer is affected by the Doppler effect. The Doppler effect causes a shift in frequency due to the relative motion between the source and the observer. In this case, the source is moving at 96.0 km/hr, emitting a frequency of 400.0 Hz.
To determine the perceived frequency, we need to consider the motion of the source and the speed of sound in the medium. The formula to calculate the perceived frequency is:
f' = f * (v + v₀) / (v + vₛ)
Where:
f' is the perceived frequency
f is the emitted frequency
v is the speed of sound in the medium
v₀ is the velocity of the observer (stationary in this case)
vₛ is the velocity of the source
The speed of sound in air depends on the air temperature, given by:
v = 331.4 + 0.6 * T
Where:
v is the speed of sound in meters per second
T is the air temperature in degrees Celsius
Plugging in the values:
T = 20.0 degrees Celsius
v₀ = 0 (stationary observer)
vₛ = 96.0 km/hr = 26.67 m/s
f = 400.0 Hz
We can calculate the speed of sound in air using the temperature:
v = 331.4 + 0.6 * 20.0 = 343.4 m/s
Now, substituting the values into the formula:
f' = 400.0 * (343.4 + 0) / (343.4 + 26.67) = 400.16 Hz
Therefore, the frequency heard by a stationary observer is approximately 400.16 Hz.
Learn more about Frequency
brainly.com/question/29739263
#SPJ11
a wire is strung tightly between two immovable posts. review section 12.4 and decide whether the speed of a transverse wave on this wire would increase, decrease, or remain the same when the temperature increases. ignore any change in the mass per unit length of the wire.
According to section 12.4, the speed of a transverse wave on a wire depends on the properties of the wire, including its tension and linear mass density (mass per unit length). However, in this scenario, we are instructed to ignore any change in the mass per unit length of the wire.
When the temperature of the wire increases, the wire expands due to thermal expansion. This expansion can affect the tension in the wire. If the tension increases with temperature, it would result in an increase in the speed of the transverse wave. Conversely, if the tension decreases with temperature, it would lead to a decrease in the speed of the transverse wave.
Since we don't have information about how the tension in the wire changes with temperature, we cannot definitively determine whether the speed of the transverse wave would increase, decrease, or remain the same when the temperature increases. It would depend on the specific characteristics of the wire and how its tension is affected by temperature changes.
Learn more about transverse wave
brainly.in/question/15334445
#SPJ11
if you drop a stone into a mine shaft 122.5 m deep, how soon after you drop the stone do you hear it hit the bottom of the shaft?
You would hear the stone hit the bottom of the mine shaft approximately 5 seconds after you drop it.
How soon will the stone droph = (1/2) * g * t²
Where:
h is the distance (depth) of the mine shaft (122.5 m)
g is the acceleration due to gravity (approximately 9.8 m/s²)
t is the time taken
We can rearrange the equation to solve for time (t):
t = √((2 * h) / g)
Substituting the given values:
t = √((2 * 122.5 m) / 9.8 m/s²)
t = √25
t = 5 seconds
Therefore, you would hear the stone hit the bottom of the mine shaft approximately 5 seconds after you drop it.
Read moer on height here: https://brainly.com/question/1739912
#SPJ4
Three identical charges (Q,91, and 92) are lined up in a row. If we compare the electric force Q exerts on charge q1 (Fe-1) to the force Q exerts on charge 92 (F2-2) A) F-2 is twice as big as F-1 B) FR-2 is half as big as F-1 C) Fe-2 is more than twice as big as 3 Fool D) Fe-2 is less than half as big as Fer E) Charge ( doesn't affect q2 at all since qı is in the way
Three identical charges (Q,91, and 92) are lined up in a row. If we compare the electric force Q exerts on charge q1 (Fe-1) to the force Q exerts on charge 92 (F2-2) is F-2 is twice as big as F-1. The correct option is a.
According to Coulomb's Law, the electric force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. In this scenario, the charges Q and 91 are equidistant from charge 92.
Since all three charges are identical, the magnitudes of the forces exerted by Q on charges 91 and 92 will be equal. Therefore, F-2, the force exerted by Q on charge 92, will be the same as F-1, the force exerted by Q on charge 91.
Therefore. The correct option is a.
To know more about Coulomb's Law, refer here:
https://brainly.com/question/506926#
#SPJ11
Solve this problem on the image
People frequently confuse weight and mass and use them interchangeably without actually considering which is accurate, but very simply earth.
Thus, The amount of matter that makes up an object is expressed in terms of mass. Mass stays the same. Anywhere in the universe, an object has the same mass as it has on Earth.
Gravity affects an object's weight. The identical object weighs less on the moon than it does on Earth because there is less gravity there.
The amount of matter that makes up an object is expressed in terms of mass. Mass stays the same. Anywhere in the universe, an object has the same mass as it has on Earth. Gravity affects an object's weight. The identical object weighs less on the moon than it does on Earth because there is less gravity there.
Thus, People frequently confuse weight and mass and use them interchangeably without actually considering which is accurate, but very simply earth.
Learn more about Earth, refer to the link:
https://brainly.com/question/12041467
#SPJ1
Use the thin lens equation to determine the smallest possible value of do that keeps di positive. (Hint try entening dummy values of do into the thin lens equation such as 1.1f and 0.9f).
By using Thin lens equation, The smallest possible value of do that keeps di positive is 0.9f.
The thin lens equation relates the object distance (do), image distance (di), and focal length (f) of a lens. The equation is given by:
1/do + 1/di = 1/f
To find the smallest possible value of do that keeps di positive, we can use the hint provided and substitute dummy values into the thin lens equation.
Let's consider the value 1.1f for do:
1/(1.1f) + 1/di = 1/f
Simplifying the equation:
1/di = 1/f - 1/(1.1f)
1/di = (1.1 - 1)/ (1.1f)
1/di = 0.1 / (1.1f)
di = (1.1f) / 0.1
di = 11f
We can see that with do = 1.1f, the value of di becomes 11f, which is positive.
Now, let's consider the value 0.9f for do:
1/(0.9f) + 1/di = 1/f
Simplifying the equation:
1/di = 1/f - 1/(0.9f)
1/di = (0.9 - 1)/ (0.9f)
1/di = -0.1 / (0.9f)
di = (0.9f) / (-0.1)
di = -9f
We can see that with do = 0.9f, the value of di becomes -9f, which is negative.
Since we want di to be positive, the smallest possible value of do that keeps di positive is 0.9f.
The smallest possible value of do that keeps di positive is 0.9f when using the thin lens equation.
To know more about Thin lens equation, visit
https://brainly.com/question/30618134
#SPJ11
At what substrate concentration would an enzyme with a km of 0.0050 M operate at one-eighth of its maximum rate? a. 5.8X10M b. 6.4X10 'M c. 7.1 X10M d. 1.2x10-?M
e. 6.1X 10M
At one-eighth of the maximum rate, the reaction velocity (V) is equal to (1/8) * Vmax. Therefore, the substrate concentration at which the enzyme operates at one-eighth of its maximum rate is option (b)6.4 x 10^-6 M
The Michaelis-Menten equation describes the relationship between enzyme activity and substrate concentration. It is given by the equation:
V = (Vmax * [S]) / (Km + [S])
Where V is the reaction velocity, Vmax is the maximum reaction velocity, [S] is the substrate concentration, and Km is the Michaelis constant.
At one-eighth of the maximum rate, the reaction velocity (V) is equal to (1/8) * Vmax. Substituting these values into the Michaelis-Menten equation, we get:
(1/8) * Vmax = (Vmax * [S]) / (Km + [S])
To simplify the equation, we can cancel out Vmax on both sides:
1/8 = [S] / (Km + [S])
Cross-multiplying and rearranging the equation, we have:
8[S] = Km + [S]
7[S] = Km
[S] = Km / 7
Given that Km = 0.0050 M, we can calculate the substrate concentration as:
[S] = 0.0050 M / 7 ≈ 7.14 x 10^-4 M
Therefore, the substrate concentration at which the enzyme operates at one-eighth of its maximum rate is approximately 6.4 x 10^-6 M.
Learn more about Michaelis-Menten equation here:
https://brainly.com/question/30404535
#SPJ11
the pulley had a radius r and a moment of inertia i. the rope does not slip over the pulley, and the pulley spins on a frictionless axle. the coefficient of kinetic friction between block a and the tabletop is uk. the system is released from rest, and block b descends. block a has mass ma and block b has mass mb. use energy methods to calculate the speed of block b as a function of the distance that it has descended
To calculate the speed of block B as a function of the distance it has descended in the given system, we can use energy methods.
Considering the conservation of mechanical energy, the initial potential energy of block B at a certain height h is converted into the kinetic energy of block B as it descends.
The initial potential energy of block B is given by:
PE_initial = m_b * g * h
where m_b is the mass of block B, g is the acceleration due to gravity, and h is the distance block B has descended.
As block B descends, it gains kinetic energy. Assuming there is no energy loss due to friction, the kinetic energy of block B can be expressed as:
KE = (1/2) * m_b * v^2
where v is the velocity (speed) of block B.
Since there is no slipping over the pulley, the linear speed of block B is equal to the tangential speed of the pulley, which can be expressed as:
v = ω * r
where ω is the angular velocity of the pulley and r is the radius of the pulley.
The angular velocity of the pulley can be related to the linear speed of block A (v_A) by considering the fact that the length of rope unwound from the pulley is equal to the distance block A has descended. Therefore:
v_A = ω * r
Next, we consider the frictional work done on block A. The work done by kinetic friction is given by:
W_friction = μ_k * m_a * g * h
where μ_k is the coefficient of kinetic friction between block A and the tabletop.
Since energy is conserved, the total initial potential energy is equal to the sum of the kinetic energy of block B and the work done by friction:
PE_initial = KE + W_friction
Substituting the expressions for potential energy, kinetic energy, and work done, we get:
m_b * g * h = (1/2) * m_b * v^2 + μ_k * m_a * g * h
Simplifying the equation and solving for v, we obtain:
v = √((2 * g * h * (m_b - μ_k * m_a)) / m_b)
Therefore, the speed of block B as a function of the distance it has descended (h) is given by the equation:
v = √((2 * g * h * (m_b - μ_k * m_a)) / m_b)
Learn more about mechanical energy
brainly.in/question/12825147
#SPJ11
consider an experiment that uses blue light of wavelength 650nm and a circular lens with a diameter of 15mm. calculate your answer in mm and enter the number without units.
The angular resolution of the circular lens used in the experiment is approximately 0.0527. A smaller angular resolution indicates a higher ability to distinguish fine details in the observed object or system.
How does lens diameter affect the experiment using blue light of wavelength 650nm?To calculate the answer of experiment that uses blue light of wavelength 650nm and a circular lens with a diameter of 15mm we need to find the angular resolution of the circular lens. The angular resolution can be determined using the formula:
Angular resolution = 1.22 * (wavelength / diameter)
Plugging in the values, we have:
Angular resolution = 1.22 * (650 nm / 15 mm)
To ensure consistent units, we convert the wavelength to millimeters:
Angular resolution = 1.22 * (0.65 mm / 15 mm)
Performing the calculation, we get:
Angular resolution ≈ 0.0527
Therefore, the angular resolution of the circular lens in this experiment is approximately 0.0527, without units.
Learn more about wavelength
brainly.com/question/31322456
#SPJ11
