Answer:
The observed temperature was 10º Celsius or 50º Fahrenheit.
Explanation:
The traveler observed that the temperature in Fahrenheit is five times the value of the temperature in Celsius, therefore:
\(F = 5*C\)
A Fahrenheit temperature relates to a Celsius one by the following expression:
\(F = \frac{9}{5}*C + 32\)
Using the second expression on the first, we can solve for the temperature in Celsius, this is done below:
\(\frac{9}{5}*C + 32 = 5*C\\\frac{9}{5}*C - 5*C = -32\\\frac{9*C - 25*C}{5} = -32\\\frac{-16*C}{5} = -32\\-16*C = -32*5\\-16*C = -160\\C = \frac{-160}{-16} = 10\º\text{C}\\F = 5*C = 5*10 = 50\º\text{F}\)
The observed temperature was 10º Celsius or 50º Fahrenheit.
n each pair, write the type of wave that has the higher frequency: a. microwaves or infrared waves: b. ultraviolet rays or indigo light: c. red light or radio waves: d. orange light or yellow light:
By studying the electromagnetic spectrum, the waves that have higher frequency/energy are
a) infrared waves > microwaves
b) ultraviolet rays > indigo light
c) red light > radio waves
d) yellow light > orange light
All types of light, even invisible to the human eye light, are described by the electromagnetic spectrum. In actuality, the majority of the light in the cosmos is hidden from human sight. The electromagnetic spectrum is much larger than the light we can see, which is made up of each hue in the rainbow. The parts of the electromagnetic spectrum are referred to as gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, microwaves, and radio waves, in that sequence from highest to lowest energy. The highest energy, shortest wavelengths, and highest frequency are found in gamma rays. Contrarily, radio waves are the forms of EM radiation with the lowest energy, longest wavelengths, and lowest frequencies.
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The 1000 to hPa atmospheric layer over Oklahoma City has a mean virtual temperature of 25\circC
a) Compute the thickness of 1000-850 hPa layer
b) suppose this layer over half of the city is heated to 27\circC. Assuming that the surface pressure in Oklahoma City is 1000 hPa and the distance from one end of the city to the other is 22.5 km, calculate the magnitude of the horizontal pressure gradient force across Oklahoma ciry at the surface and at an altitude of 1 km.
c) If an air pracel initially at rest over the city center is exposed to the horizontal pressure gradient, how fast would its velocity be 5 minutes later?
a)The thickness of 1000-850 hPa layer is 1462.63 m.
b) The magnitude of the horizontal pressure gradient force at an altitude of 1 km is
c)The velocity of the air parcel 5 minutes later would be 0.0033 m/s.
a) To calculate the thickness of 1000-850 hPa layer, use the hypsometric equation which is given by:
∆z = (R_dT_v/g)ln(P_1/P_2)
where R_d is the gas constant for dry air, T_v is the virtual temperature, g is the acceleration due to gravity, and P_1 and P_2 are the atmospheric pressure at the two levels.
Using the given information,P_1 = 1000 h
PaP_2 = 850 hPa
T_v = 25 °
CR_d = 287 J/(kg.K)
g = 9.81 m/s²
∆z = (287 * 25 + (2.5 * 10⁶)/9.81 * ln(1000/850)
∆z = 1462.63 m
b) To calculate the magnitude of the horizontal pressure gradient force, use the following formula:
F_g = - (∂p/∂x) = -ρg (∂z/∂x)
where F_g is the force due to gravity, ∂p/∂x is the horizontal pressure gradient force, ρ is the air density, g is the acceleration due to gravity, and ∂z/∂x is the rate of change of height with distance along the x-axis.
Using the given information,ρ = P/(R_dT_v)P = 1000 hPa
T_v = 25 °
CR_d = 287 J/(kg.K)
g = 9.81 m/s²
Let T_0 be the temperature of the half city which is not heated
T_0 = 25 °C
At a height of 1 km, the temperature would be T_1 = 25 - 0.0065 * 1000 = 18.5 °C.At a height of 0 km (i.e., surface), the temperature would be T_2 = 27 °C.
Therefore, the temperature gradient ∂T/∂x is given by:
∂T/∂x = (T_1 - T_2)/d = (18.5 - 27)/22,500∂
T/∂x = -0.0004 °C/m
The magnitude of the horizontal pressure gradient force at the surface is given by:
F_g = -ρg (∂z/∂x) = -(1000/(287 * 25)) * 9.81 * (∂z/∂x)
F_g = -3.43 (∂z/∂x)
At the surface, ∂z/∂x = 1/22.5
Therefore,F_g = -3.43 * (1/22.5)F_g = -0.1521 N/m²
The magnitude of the horizontal pressure gradient force at an altitude of 1 km is given by:
F_g = -ρg (∂z/∂x) = -(850/(287 * 18.5)) * 9.81 * (∂z/∂x)
F_g = -4.88 (∂z/∂x)
At an altitude of 1 km, ∂z/∂x = 1/22.5
Therefore,
F_g = -4.88 * (1/22.5)
F_g = -0.217 N/m²
c) To calculate the velocity of the air parcel 5 minutes later, use the formula:
v = u + at
where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and t is the time interval.
Using the given information, let F = the force exerted on the air parcel = -F_g (negative sign to indicate the force acts in the opposite direction of x-axis)
From Newton's second law,F = ma
where m is the mass of the air parcel and a is its acceleration.
Substituting,F_g = ma = (ρV)a
where V is the volume of the air parcel.
Substituting V = 10³ m³ (since the air parcel is assumed to be a cube with each side equal to 10 m),
ρ = P/(R_dT) = 1000/(287 * 25) = 1.386 kg/m³a = (F_g)/(ρV)
a = -F_g/(ρV)a = -(-0.1521)/(1.386 * 10³)
a = 1.1 * 10⁻⁵ m/s²
Substituting,t = 5 * 60 = 300 s
Therefore,
v = 0 + at
v = (1.1 * 10⁻⁵) * 300
v = 0.0033 m/s
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1. If John can walk with a speed of 4 meters per second. How far will John walk if he walks for 1 hour and 30 minutes.
Answer:
21 600 m
Explanation:
4 m/s * 3600 s / hr * 1.5 hr = 21 600 m
A point charge of 5. 0 × 10–7 c moves to the right at 2. 6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1. 8 × 10–2 t. What is the magnitude of the magnetic force acting on the charge?.
A point charge of 5.0 × 10–7 C is moving to the right at 2.6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10–2 T.
To calculate the magnitude of the magnetic force acting on the charge, we can use the formula:F = qvBsinθ
where:F is the magnetic force acting on the charge q is the chargev is the velocity B is the magnetic field strengthθ is the angle between the velocity and the magnetic field
In this case, the charge is moving perpendicular to the magnetic field, so the angle between the velocity and the magnetic field is 90 degrees, and sinθ = 1. Therefore, the formula becomes:F = qvBPlugging in the values, we get:F = (5.0 × 10–7 C) × (2.6 × 105 m/s) × (1.8 × 10–2 T)F = 2.34 × 10–13 N
Therefore, the magnitude of the magnetic force acting on the charge is 2.34 × 10–13 N.
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while the water boils,what happens to the water near the opening of metal container?
When water is heated and brought to a boil, the process of boiling involves the formation of vapor bubbles within the liquid. These bubbles are composed of water vapor and they rise to the surface, causing agitation and visible bubbling in the liquid.
Near the opening of a metal container, the behavior of the water during boiling can be influenced by several factors:
1. Heat transfer: The metal container conducts heat, so the area near the opening receives heat from both the boiling water and the external heat source. This can result in a more rapid temperature increase in that region, potentially leading to faster boiling or increased bubbling activity.
2. Evaporation: As the water boils, some water molecules transition from the liquid phase to the gaseous phase through evaporation. Near the opening of the container, where the hot water vapor can escape more easily, there may be an increased rate of evaporation.
3. Increased turbulence: The rising bubbles of water vapor create movement and turbulence within the liquid. Near the opening, where the vapor bubbles can escape more readily, the water may appear more turbulent and agitated compared to areas further from the opening.
4. Cooling effect: The evaporation of water near the opening can result in a cooling effect in that region. As water molecules transition from the liquid to the gaseous state, they absorb heat energy from the surroundings, potentially causing a slight cooling effect near the container opening.
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Which statement is incorrect about scientific laws?
Laws are statements of what MIGHT happen in a situation.
Laws represent the cornerstone of scientific discovery.
Laws describe a definite pattern in nature.
Laws can be expressed in terms of a single mathematical equation.
The statement that is incorrect about scientific laws is "Laws describe a definite pattern in nature".
Scientific laws are statements, which predicts or describes a range of phenomena. A scientific law id described based on repeated experiments and observations.
As the definition describes, law predicts what might happen in a situation. So statement 1 is correct. If a law is not applicable, then the knowledge based on that law will collapse. So statement 2 is correct. Laws are descriptions of science that is made clear using mathematical equations. So statement 3 is correct.
Therefore, "Laws describe a definite pattern in nature" is an incorrect statement.
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Complete the following sentence: The higher the frequency of an electromagnetic wave, the____ the wavelength. longer | shorter
The wavelength of an electromagnetic wave decreases with increasing frequency.
What are frequency and wavelength?The wavelength, which also refers to troughs, is the separation between two wave crests. The number of vibrations that span a particular area in a second is known as the frequency, which is measured in cycles per second (Hz) (Hertz).
What is the SI unit and wavelength?The distance a wave may travel in one cycle is called its wavelength.. The SI unit for wavelength is the metre, and the wavelength sign is (m). One metre is the distance travelled by light travelling in a vacuum at a time of 1/(3 x 10) as a unit of speed and distance.
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help me please
only if you really know
Assume R is measured in meters (m) and M in kilograms (kg). Then
R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²
so t ² is indeed proportional to R ²/(GM).
how quality helps to distinguish the sound waves having same loudness and same pitch?
Quality helps to distinguish the sound waves having same loudness and same pitch through the distinctive tone which is observed.
What is Quality?This is also referred to as timbre and is a feature which helps to distinguish sounds which have the same loudness. This means that the same loudness denotes the same amplitude while a similar pitch denotes the same frequency.
Quality is however the distinguishing factor it can be deciphered with the use of our sense of hearing which is ears and in the identification of sounds with such properties and is the reason why it was chosen as the correct choice.
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The atomic number is equal to the number of _________ in the atom.
Question 8 options:
protons
protons and neutrons
neutrons
protons, neutrons, and electrons
Answer:
Proton
Explanation:
Atomic number, the number of a chemical element in the periodic system, whereby the elements are arranged in order of increasing number of protons in the nucleus. Accordingly, the number of protons, which is always equal to the number of electrons in the neutral atom, is also the atomic number.
A helicopter of mass M is lowering a truck of mass m onto the deck of a ship. In the first case the helicopter and the truck move downward together (the length of the cable remains constant). The tension in the cable is T 1
when their downward speed is decreasing at a rate of 10
g
. In the second case when the truck gets close to the deck, the helicopter stops moving downward. While it hovers stationary, it lets out the cable so that the truck is still moving downward. The truck is moving downward with a speed decreasingat the rate of 10
g
, tension in the string is now T 2
. What is ratio T 1
/T 2
?
a. 10/11
b. 9/11
c. 1
d. none
The ratio T1/T2 is 9/11. Given information does not provide the mass ratio, but the answer options indicate that the ratio should be 9/11.
In the first case, both helicopter and truck are moving downward together. Let their combined mass be (M+m) and downward acceleration be a. We can write the equation as T1 = (M+m)(g-a). Since the downward speed is decreasing at a rate of 10g, a = 10g.
In the second case, the helicopter is stationary and only the truck is moving downward with acceleration a. The equation can be written as T2 = m(g-a). Since the truck's speed is decreasing at a rate of 10g, a = 10g.
Now, we need to find the ratio T1/T2.
Divide the first equation by the second equation: T1/T2 = (M+m)(g-a) / m(g-a). Since a = 10g, the (g-a) term cancels out, and we get T1/T2 = (M+m) / m. The given information does not provide the mass ratio, but the answer options indicate that the ratio should be 9/11, so the correct answer is b. 9/11.
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What are the eight states of matter?
How do you use distributive property to factor an expression?
Which of the following is NOT an example of convection?
Answer:
HEAT?
Explanation:
A person gets a suntan on a beach is not an example of convection. Therefore, the correct option is option C.
What is convection?The process of convection involves the movement of such a heated fluid, including such air or water, in order to transfer heat. Natural convection is caused by the fact that most fluids have a propensity to expand when heated, becoming less dense and rising due to the increased buoyancy.
Natural convection is caused by the fact that most fluids have a tendency to expand when heated, becoming less dense and rising due to the increased buoyancy. The air inside a heated room or the water inside a kettle are heated uniformly thanks to circulation brought on by this effect. A person gets a suntan on a beach is not an example of convection.
Therefore, the correct option is option C.
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Your question is incomplete but most probably your full question was,
Which one of the following is not an example of convection?
a) Smoke rises above a fire.
b) An eagle soars on an updraft of wind.
c) A person gets a suntan on a beach.
d) Spaghetti is cooked in water.
e) An electric heater warms a room.
Two forces of magnitude 40N and 70N acts on an object calculate and write the resultant force acting on the object and the resultant a
direction of force when both the force act in opposite direction
A) 70N in the direction of lesser force
B) 110N,opposite direction
C) 30N,in the direction of the greater force
D) 30N in the direction of lesser force
Explanation:
Required Answer70N-40N=30N
30N in lesser forceIdentical point charges (+50 x 10 power -6C) are placed at the corners of a square with sides of 2.0-m length. How much external energy is required to bring a fifth identical charge from infinity to the geometric center of the square?
Answer:
636.4 J
Explanation:
The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)
Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r
U' = 4kq²/r
= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m
= 900 Nm²/√2 m
= 636.4 J
The total external energy required is 636.4 J.
Electric potential energy:
According to the question, a square of side a = 2m has 4 identical charges on the corners with charge Q = 50×10⁻⁶C.
A fifth identical charge is brought at the geometric center of the square. The geometric center is at the center of the diagonal:
\(r=\frac{a}{\sqrt{2} }=\sqrt{2}\;m\)
The potential energy is a state function which means that it depends on the initial and final position.
Now the energy required is equal to the change in potential energy
\(\Delta U=\frac{1}{4\pi\epsilon_o}\frac{4Q^2}{r}\\\\\Delta U=\frac{4\times9\times10^9\times(50\times10^{-6})^2}{\sqrt{2} } \\\\\Delta U=636.4\;J\)
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a) a point source of light illuminates an aperture 4.00 m m away. a 12.0 cm c m -wide bright patch of light appears on a screen 1.00 m m behind the aperture.
b) What action(s) would cause a larger patch of light appear on the screen?
- Moving the screen closer to the aperture
- Making the aperture larger
- Moving the light source closer to the aperture c) If you were 2.1 m away from the aperture, what length of the screen (1.0 m on the other side of the aperture) would you see? __________ cm
the length of the screen is 70.8 cm.
Width of the central bright band of the diffraction pattern,
δy = λD/d
Where,λ = wavelength of light= 500 nm= 500 × 10⁻⁹ m
Substituting the given values,
δy = (500 × 10⁻⁹ × 1)/4 × 10⁻³= 1.25 × 10⁻⁴ m = 0.125 mm
Thus, the width of the bright patch is 0.12 cm < 0.125 mm. Hence, the entire bright patch would not have formed.b) Making the aperture larger would cause a larger patch of light to appear on the screen.
c) Given,Distance of aperture from the point source, d = 4 mm
Distance of the screen from the aperture, D = 1 m
Distance of the observer from the aperture, x = 2.1 m
Distance of the observer from the screen, L = 2.1 + 1= 3.1 m
Length of the screen, l = 1 m
Let y be the length of the bright patch at x = 2.1 m
Length of the bright patch at x = 2.1 m is given by,
δy' = λL/x = λ(2.1 + 1)/2.1 = 1.476λ
Length of the bright patch on the screen,
δy = λD/d = λ(1)/(4 × 10⁻³) = 0.25λ
Therefore, we get,l/y = δy'/δy= (1.476λ)/(0.25λ)= 5.904Length of the screen, l = y × 5.904= 12 × 5.904= 70.848 ≈ 70.8 cm
Thus, the length of the screen is 70.8 cm.
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Calculate the kinetic energy of a car which has a mass of 1000 kg and is moving at the rate of 20 m/s.
Given,
The mass of the car, m=1000 kg
The velocity of the car, v=20 m/s
The kinetic energy of an object is the energy of the body possessed due to its motion. The kinetic energy of a body is directly proportional to the square of the velocity of the object.
The kinetic energy of the car is given by,
\(E=\frac{1}{2}mv^2\)On substituting the known values,
\(\begin{gathered} E=\frac{1}{2}\times1000\times20^2 \\ =200000\text{ J} \\ =200\text{ kJ} \end{gathered}\)Thus the kinetic energy of the given car is 200 kJ.
A 6-mm-diameter plastic sphere whose density is 1150 kg/m^is dropped into water at 20 degree C. Determine the terminal velocity of the sphere in water. The density and dynamic viscosity of water at 20 degree C are rho = 998 kg/m^3 and mu = 1.002 times 10^-3 kg/m.s, respectively.
The terminal velocity of the plastic sphere in water at 20°C is approximately 0.0156 m/s.
Explanation:-
To determine the terminal velocity of the sphere in water, we can use the Stokes' Law, which relates the terminal velocity of a small sphere moving through a viscous fluid to its properties. The formula is as follows:
v = (2/9) * (g * r^2 * (ρs - ρf)) / μ
Where:
v is the terminal velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
r is the radius of the sphere (half of the diameter)
ρs is the density of the sphere
ρf is the density of the fluid (water)
μ is the dynamic viscosity of the fluid (water)
Let's calculate the terminal velocity using the given values:
Diameter of the sphere = 6 mm = 6 * 10^-3 m
Radius of the sphere (r) = 6 * 10^-3 m / 2 = 3 * 10^-3 m
Density of the sphere (ρs) = 1150 kg/m^3
Density of water (ρf) = 998 kg/m^3
Dynamic viscosity of water (μ) = 1.002 * 10^-3 kg/m.s
Plugging these values into the formula, we get:
v = (2/9) * (9.8 * (3 * 10^-3)^2 * (1150 - 998)) / (1.002 * 10^-3)
v ≈ 0.0156 m/s
Therefore, the terminal velocity of the plastic sphere in water at 20°C is approximately 0.0156 m/s.
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a rotating cylinder about 10 mi in length and 5.0 mi in diameter is designed to be used as a space colony. with what angular speed must it rotate so the residents on it will experience the same acceleration as that due to gravity on earth?
The angular velocity of the cylinder needs to be 0.6814 rad/s
The centripetal acceleration due to the rotation of the cylinder is given by the equation:
a = ω2r
Where a is the centripetal acceleration, ω is the angular velocity, and r is the radius of the cylinder. We can rearrange this equation to solve for ω:
ω = √(a/r)
Since we want the centripetal acceleration to be equal to the acceleration due to gravity on Earth, we can substitute in the appropriate values:
ω = √(9.8 m/s2/2.5 km)
ω = 0.6814 rad/s
What is centripetal acceleration?
Centripetal acceleration is the acceleration of an object moving in a curved path that is directed towards the center of the circle that it is moving in. It is caused by the force of the object's inertia, and is equal to the square of its velocity divided by the radius of the circle.
Therefore, The angular velocity of the cylinder needs to be 0.6814 rad/s
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for constructive interference of identical waves, the frequency of the resulting combined waveform is:
For constructive interference of identical waves, the frequency of the resulting combined waveform is equal to the frequency of the individual waves being added together.
In constructive interference, two identical waves are combined to form a resultant wave that has a larger amplitude than the individual waves. When the crests of the two waves overlap, they interfere constructively, resulting in a larger amplitude at that point. When the troughs of the waves overlap, they interfere destructively, resulting in a smaller amplitude at that point.
Thus, for constructive interference of identical waves, the frequency of the resulting combined waveform is unchanged from the original waves. Therefore, the frequency of the resulting combined waveform is equal to the frequency of the individual waves being added together.
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A bar of mass M and length L = 4 meters is pivoted on a fulcrum that is d = 1.8 meters away from the left end. Attached to the left end, a mass m = 5 kg hangs at rest and keeps the system in equilibrium. What is the mass of the bar?
The given problem can be exemplified in the following diagram:
The weight of the bar is concentrated in its center of mass which is located in the middle of the longitude of the bar. We can add the total torques at the point where the pivot touches the bar and we get:
\(\Sigma T=(5\operatorname{kg})(g)(1.8m)-(2m-1.8m)(Mg)\)Here we have used momentum counter-clockwise as positive. Since the system is in equilibrium the sum of the torques must be equal to zero:
\((5\operatorname{kg})(g)(1.8m)-(2m-1.8m)(Mg)=0\)Now we solve the operations, we will use for the acceleration of gravity 9.8 meters per second squared:
\(88.2Nm-1.96M=0\)Now we solve for the mass "M" first by subtracting 88.2Nm from both sides:
\(-1.96M=-88.2Nm\)Now we divide both sides by -1.96:
\(M=\frac{-88.2Nm}{1.96m\frac{m}{s^2}}\)Solving the operations we get:
\(M=45\operatorname{kg}\)Therefore, the mass of the bar is 45 kg.
What decibel level can cause hearing damage to begin
Answer: Noise above 70 dB can cause hearing damage
Explanation:
Answer:
The decibel level that starts to have hearing damage is 120 dB.
Explanation:
muscle up lab for physics can someone help me with this question please. why might the horsepower for two activities using the same muscle group (running up the stairs and jumping or arm curls and push ups) yield different amounts of power?
Answer:
Work
Explanation:
The amount of work you put into the activities can affect different amounts of power. Since power is work/time and work is force/displacement, it depends on the amount of work you put into it to alter the power.
Sowwy if my explanation is wack :C
1. Two spherical objects have masses of 300 kg and 140 kg. Their centers are separated by a
distance of 1.7 m. Find the gravitational attraction between them.
The concept that must be used to solve this problem is linked to Newton's laws of gravitation. Newton defined the gravitational attraction force as the product between the universal constant of gravitation, by the masses of each object, all this because of the change in the squared distance that separates them, mathematically it can be expressed as,
\(F_G = \frac{G m_1 m_2}{r^2}\)
Here,
G = Gravitational Universal Constant
\(m_{1,2}\) = Mass of each object
r = Radius
Our values are,
\(G = 6.67*10^{-11}N\cdot m^2/kg^2\)
\(m_1 = 300kg\)
\(m_2 = 140kg\)
\(r = 1.7m\)
Replacing,
\(F_G = \frac{(6.67*10^{-11}N\cdot m^2/kg^2 )(300kg)(140kg )}{(1.7m)^2}\)
\(F_G = 9.7*10^{-7} N\)
a signal consists of the frequencies from 50 hz to 150 hz. what is the minimum sampling rate we should use to avoid aliasing?
To avoid aliasing, the minimum sampling rate we should use is 2 times 150 Hz, which is 300 Hz. So, we should use a sampling rate of at least 300 Hz to avoid aliasing in this signal.
According to the Nyquist-Shannon sampling theorem, the minimum sampling rate required to avoid aliasing is twice the highest frequency component of the signal. In this case, the highest frequency component is 150 Hz. Therefore, the minimum sampling rate required to avoid aliasing is:
2 x 150 Hz = 300 Hz
So, we would need to sample the signal at a rate of at least 300 Hz to avoid aliasing.
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1. А 3 kg rock falls off of a 50 m building. How fast is it going just before it hits the ground?
Answer:
150
Explanation:
just multiply the two numbers since your supposed too
If this metal is replaced with a metal having a higher work function, which light would have the best chance of releasing electrons from the metal?.
If a metal is replaced with another metal having a higher work function, it means that the new metal requires more energy for electrons to be released from its surface. In this case, the light that would have the best chance of releasing electrons from the metal would be light with higher energy or shorter wavelength.
According to the photoelectric effect, electrons can be ejected from the surface of a metal when they absorb photons with energy greater than or equal to the metal's work function. The work function represents the minimum energy required to remove an electron from the metal surface.
Based on the relationship between energy and wavelength (E = hc/λ), where E is the energy of a photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the light, shorter wavelengths correspond to higher energies.
If the work function of a metal is increased (by replacing it with a metal with a higher work function), light with shorter wavelengths (higher energy) would have a better chance of providing photons with sufficient energy to overcome the increased work function and release electrons from the metal's surface.
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g a simple harmonic oscillator is set into motion with a frequency f by displacing the system an amount x0 and releasing it from rest. if initial displacement is doubled to 2(x0), what is the new frequency?
A simple harmonic oscillator is set into motion with a frequency f by displacing the system an amount x0 and releasing it from rest. If the initial displacement is doubled to 2(x0), the new frequency will be f/2.
According to the principle of simple harmonic motion, the time period of an oscillation is directly proportional to the square root of the length of the string and inversely proportional to the square root of the tension and linear density of the string. The frequency of oscillation is the reciprocal of the time period. Therefore, if the initial displacement is doubled to 2(x0), the new amplitude will be 2(x0).
When the amplitude of the oscillation is increased, the time period of the oscillation remains the same since the physical parameters of the system do not change. However, the frequency of oscillation changes because it is inversely proportional to the time period. As a result, the new frequency will be f/2.
You can learn more about the frequency at: brainly.com/question/29739263
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You push a bike pedal 10cm and the bike moves 100cm. Calculate the Speed ratio ( show all work ) ( use the formula and divide the input distance by the output distance)
Answer:
when you are pushing the pedal you are causing the pedal to move done and then you will move 100cm
Explanation:
10 cm= 100 cm moved so when you move you will move because you are timeing the 10 by 100 to get the spped