Answer:
True
Explanation:
The more inertia that an object has, the more mass that it has.
find the frequency of a tuning fork that takes 2.5 x 10^-3 s to complete one oscillation.
The frequency of the tuning fork can be calculated using the formula f=1/T, where f is the frequency and T is the period of oscillation. In this case, the period is given as 2.5 x 10^-3 s. So, the frequency can be calculated as f=1/2.5 x 10^-3 = 400 Hz.
Therefore, the frequency of the tuning fork is 400 Hz, as it takes 2.5 x 10^-3 s to complete one oscillation.
To find the frequency of a tuning fork that takes 2.5 x 10^-3 s to complete one oscillation, you need to determine the number of oscillations per second.
The formula to find frequency (f) is f = 1/T, where T is the time period of one oscillation. In this case, T = 2.5 x 10^-3 s. By substituting the value of T in the formula, we get f = 1 / (2.5 x 10^-3). After calculating, we find that the frequency of the tuning fork is approximately 400 Hz.
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gravitational lensing of distant, faint irregular galaxies may be the key to
Gravitational lensing of distant, faint irregular galaxies may be the key to unlocking valuable insights and understanding various astrophysical phenomena. Gravitational lensing occurs when the gravitational field of a massive object, such as a galaxy or cluster of galaxies, bends and distorts the path of light from a more distant object behind it.
By studying the gravitational lensing effects on faint irregular galaxies, astronomers can gain information about the distribution of dark matter, the nature of dark energy, and the properties of the lensing objects themselves. It provides a unique opportunity to probe the mass distribution and dynamics of massive structures in the universe.
Furthermore, gravitational lensing can enhance the apparent brightness and resolution of distant, faint irregular galaxies, enabling us to observe and study them in more detail than would otherwise be possible. This can shed light on the formation and evolution of galaxies, the interplay between dark matter and ordinary matter, and the cosmic web of large-scale structures.
In summary, gravitational lensing of distant, faint irregular galaxies holds the potential to unravel mysteries about the nature of dark matter, dark energy, galaxy formation, and the structure of the universe itself.
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what is the ph of a solution prepared by dissolving 0.15 gram of solid cao in enough water to make 2.00 l of aqueous ca(oh)2
The pH of the solution prepared by dissolving 0.15 gram of solid CaO in enough water to make 2.00 L of aqueous Ca(OH)2 is approximately 12.5.
When CaO is dissolved in water, it reacts with water to form Ca(OH)2. This reaction is highly exothermic and releases a significant amount of heat. The resulting Ca(OH)2 is a strong base with a high pH. In this case, the Ca(OH)2 is dissolved in a large volume of water, which further dilutes the solution and lowers the pH slightly. However, since Ca(OH)2 is a strong base, the pH of the resulting solution is still quite high, around 12.5.
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What is the current in the 4 ohm resistor in the circuit in the illustration?
The current in the 4 ohm resistor is 2 amperes.
To solve this problem, we need to use Ohm's Law, which states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.
this circuit, the total resistance is 8 ohms, so the current is 4 volts / 8 ohms = 0.5 amperes. Since the current flowing through the 8 ohm resistor and the 4 ohm resistor is the same, the voltage drop across the 4 ohm resistor is 2 volts (0.5 amperes x 4 ohms), which gives us a current of 2 amperes (2 volts / 1 ohm).
Therefore, the current in the 4 ohm resistor is 2 amperes.
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A closed, rigid, 0.50 m3 tank is filled with 120 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the pressure is p2 = 4 bar. Determine the initial quality, x1, and the heat transfer, in kJ.
Assuming that the process is reversible and adiabatic, we can use the following thermodynamic relations to solve the problem. Therefore, the initial quality of water is x₁= 0.197, heat transfer is 52311 KJ
p₁[tex]V1^{k}[/tex]= p2[tex]v2^k[/tex] (where k = cp/cv is the specific heat ratio of water)
v₁/v₂ = x₁/x₂
The specific heat ratio of water can be taken as k = 1.4. The specific volume of water at state 1 can be calculated from the equation of state of water:
p₁= rho₁R T₁
where R = 0.461 kJ/kg-K is the specific gas constant of water vapor, and T1 is the initial temperature. Solving for rho1, we get:
rho₁= p₁/(R T₁) = 120000/(0.461 T₁)
The specific volume of water vapor at state 2 can be calculated from the equation of state of an ideal gas:
p₂ v₂ = m R T₂
where m is the mass of water in the tank, and T2 is the final temperature. Solving for v2, we get:
v2 = m R T₂/p₂
The quality of water at state 2 can be calculated from the definition of quality:
x₂ = (v₂- vf)/(vg - vf)
where vf and vg are the specific volumes of saturated liquid and saturated vapor at pressure p₂, respectively. These can be obtained from steam tables. For p₂ = 4 bar, we find:
vf = 0.001017 m³/kg
vg = 1.6949 m³/kg
The heat transfer during the process can be calculated from the first law of thermodynamics:
Q = m (h₂ - h₁)
where h₁ and h₂ are the specific enthalpies of water at states 1 and 2, respectively. These can also be obtained from steam tables.
Putting all these equations together and solving for x₁ and Q, we get:
v1₁= 0.50/120 = 0.00417 m³/kg
T₁= p₁/(rho1 R) = 281.7 K
v2 = m R T₂/p₂ = (120/18.015) 0.461 T₂/4 = 0.01907 T₂
x₂ = (v₂- vf)/(vg - vf) = (0.01907 T2 - 0.001017)/(1.6949 - 0.001017)
p₁ v₁^k = p₂v₂^k
[tex]p1 v1^(1.4) = p2 v2^(1.4)\\20 (0.00417)^(1.4) = 4 (0.01907 T2)^(1.4)\\ T2 = 242.9 K[/tex]
x₁ = v₁/v₂ x₂= (0.00417/0.01907) x2
= 0.242 x2 = 0.242 (0.8154) = 0.197
h₁ = 693.41 kJ/kg (from steam tables)
h₂ = h₁+ x₁(hfg) = 693.41 + 0.197 (2257.0 - 693.41)
= 1385.9 kJ/kg
Q = m (h₂ - h₁)
= 120 (1385.9 - 693.41) = 52311 kJ
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If you increase your distance by a factor of 3 from a sound source that is radiating equally in all directions. What happens to the intensity of the sound? It reduces to a. 1/3 its original value. b. 1/27 its original value. c. 1/9 its original value. d. none of the above
If you increase the distance by a factor of 3, the intensity decreases by a factor of 9. This means that the correct answer is option b.
When you increase the distance by a factor of 3 from a sound source that is radiating equally in all directions, the intensity of the sound decreases. The relationship between distance and intensity is inverse square, meaning that if you double the distance from the source, the intensity decreases by a factor of 4. Therefore, if you increase the distance by a factor of 3, the intensity decreases by a factor of 9.
This means that the correct answer is option b, where the intensity reduces to 1/27th of its original value. It is important to note that this relationship assumes that there are no obstructions or other factors that could affect the sound wave as it travels through the air.
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to establish that the universe was very, very large, edwin hubble used ________.
To establish that the universe was very, very large, Edwin Hubble used observations of galaxies and their redshifts.
Edwin Hubble, an American astronomer, made groundbreaking discoveries that revolutionized our understanding of the universe. One of his key contributions was the realization that galaxies exist beyond the boundaries of our Milky Way and that the universe is much larger than previously believed. Hubble observed numerous galaxies and noticed that their light exhibited a redshift. This redshift phenomenon occurs when light waves from distant objects stretch as the universe expands, causing the wavelengths to appear longer, shifting towards the red end of the spectrum. Hubble's observations showed a correlation between the distance of galaxies and the magnitude of their redshifts.
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an atc radar facility issues the following advisory to a pilot flying on a heading of 360°: 'traffic 10 o'clock, 2 miles, southbound...' where should the pilot look for this traffic?
The pilot should look to the left side of the aircraft, roughly at the 10 o'clock position, in order to locate the traffic being reported by the ATC radar facility.
The advisory states that the traffic is at "10 o'clock." In aviation, the direction of the clock face is used to describe the relative position of other aircraft. When the pilot is facing forward, 12 o'clock refers to straight ahead, 3 o'clock is to the right, 9 o'clock is to the left, and 6 o'clock is directly behind.
The advisory also states that the traffic is "2 miles" away. This indicates the distance between the pilot's aircraft and the traffic being reported.
Based on the ATC advisory, the pilot should look to the left side of the aircraft, approximately at the 10 o'clock position, and scan for any traffic that is southbound. It is important for the pilot to maintain situational awareness and be vigilant in order to avoid any potential conflicts with the reported traffic.
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because of the massive weight, the distance needed to stop an average train travelling at 55mph is
The distance needed to stop an average train traveling at 55 mph (miles per hour) depends on several factors, including the train's mass, speed, braking system, and track conditions. The braking capabilities of the train play a crucial role in determining the stopping distance.
On average, it is estimated that a typical freight train traveling at 55 mph may require a distance of approximately one mile (1.6 kilometers) to come to a complete stop. This estimate takes into account the train's weight, momentum, and the time required for the braking system to bring the train to a halt.
However, it is important to note that this is a rough estimate and can vary based on various factors such as the train's specific configuration, the condition of the track, the effectiveness of the braking system, and the operator's skill. Additionally, trains may be equipped with additional systems such as emergency brakes to assist in stopping in shorter distances when necessary.
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a 25 mm sphere is maintained at 64 c in 20 c quiescent water find the power required to maintain the sphere
The power required to maintain the 25 mm sphere at 64°C in 20°C quiescent water is approximately 8.6 W.
To calculate the power required to maintain the sphere, we can use the following formula:
[tex]Power = (heat transfer) *(coefficient ) *(surface area) * (temperature difference)[/tex]
First, we need to find the temperature difference between the sphere and the water. The sphere is maintained at 64°C and the water is at 20°C, so the temperature difference is:
ΔT = 64°C - 20°C = 44°C
Next, we need to find the surface area of the sphere. The formula for the surface area of a sphere is:
[tex]Surface area = 4\pi r^{2}[/tex]
where r is the radius of the sphere. Since the sphere has a diameter of 25 mm, its radius is 12.5 mm (or 0.0125 m). Therefore, the surface area of the sphere is:
Surface area = 4π(0.0125 m)² = 0.0019635 m²
Now we need to find the heat transfer coefficient. This is a measure of how easily heat can be transferred between the sphere and the water. It depends on factors such as the properties of the materials and the flow rate of the water. For simplicity, let's assume a heat transfer coefficient of 100 W/(m²°C).
Finally, we can plug these values into the formula for power:
Power = 100 W/(m²°C) x 0.0019635 m² x 44°C = 8.6 W
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which one of the following statements about quasars is true? astronomers discover quasars by identifying pairs of enormous radio-emitting lobes of gas. quasars appear bright because their compact mass gravitationally lenses background light, making them appear brighter than they actually are. all quasars are active galactic nuclei, but not all active galactic nuclei are quasars. quasars generate their tremendous luminosities via nuclear fusion inside their event horizons, which is then radiated outward into space. quasars sometimes rapidly fluctuate in brightness but are nonetheless comparable in size to their host galaxies.
The statement that is true about quasars is: "All quasars are active galactic nuclei, but not all active galactic nuclei are quasars."
A quasar is a type of active galactic nucleus (AGN) that is highly luminous and characterized by a compact region at the center of a massive galaxy that emits enormous amounts of energy. Quasars are not discovered by identifying pairs of enormous radio-emitting lobes of gas, which is a characteristic of radio galaxies.
Quasars do not appear bright because they gravitationally lens background light; instead, their high luminosity is due to the accretion of matter onto a supermassive black hole at the center of the galaxy. Quasars do not generate their tremendous luminosities via nuclear fusion inside their event horizons, but rather from the frictional heating and gravitational energy released by the infalling matter. Finally, quasars can fluctuate in brightness but are not comparable in size to their host galaxies, as they are typically much smaller.
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if the hanging mass is m = 100 g, and its linear acceleration is a = 2.5 m/s2 , what is the moment of inertia i of the rotating platform?
We can use the formula for the rotational motion:
τ = Iα
The moment of inertia of the rotating platform is 0.32 kg m^2.
We can use the formula for the rotational motion:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
In this problem, we know the linear acceleration of the hanging mass, and we want to find the moment of inertia I of the rotating platform. We can relate the linear acceleration to the angular acceleration using the radius r of the platform:
a = αr
We also know the mass m and the force F acting on the hanging mass:
F = ma
The force F produces a torque τ on the platform, given by:
τ = Fr
Substituting the expressions for α and τ into the equation for rotational motion, we have:
Fr = I(a/r)
Simplifying and solving for I, we get:
I = (F/α)r^2 = (mg/α)r^2
Substituting the given values, we have:
I = (0.1 kg)(9.81 m/s^2)/(2.5 m/s^2)(0.2 m)^2
I = 0.32 kg m^2
Therefore, the moment of inertia of the rotating platform is 0.32 kg m^2.
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A light bulb is attached to a battery and lights up. Which If the following about the electric field inside the bulb filament? ats up. Which if the following statements is true a. The field must be zero because the filament is made of metal b. The field must be zero because a current is flowing c. The field must be zero because any excess charge must be zero because any excess charges are on the surface of the filament d. The field must be non-zero because the town must be non-Zero because the flowing Current produces an electric field e. The field the held must be non-zero because event will show without an applied to f. The field must be zero for reasons not given above g. The field must be non-zero for reasons not given above
The field must be non-zero because the flowing current produces an electric field.
When a current flows through a conductor, it produces a magnetic field around it. This magnetic field, in turn, produces an electric field that drives the flow of charge through the conductor. Thus, there is an electric field inside the bulb filament when a current is flowing through it. The magnitude and direction of this field depend on the amount and direction of the current, as well as the properties of the filament material. The other options are incorrect because they do not account for the effect of the current on the electric field inside the filament.
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The materials that made up the solar nebula can be categorized into these four general types. Rank these materials from left to right based on the temperature at which each would condense into a solid, from highest to lowest. Note: For a substance that does not condense at all, rank it as very low temperature.
From highest to lowest condensation temperature, the four general types of materials that made up the solar nebula are refractory metals and silicates, volatile metals and ices, water, and hydrogen and helium.
Refractory metals and silicates, such as iron, nickel, and silicon, have the highest condensation temperature and would solidify first in the cooling solar nebula. Volatile metals and ices, like zinc and carbon dioxide, have a lower condensation temperature and would condense next. Water has a lower condensation temperature and would come after volatile metals and ices, while hydrogen and helium have the lowest condensation temperature and would not condense until very low temperatures were reached. These materials formed the building blocks for the planets in the solar system.
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8. the ballast in some of the older florescent luminaires was said to have a low power factor. what device was added to the circuit to improve the power factor?
To improve the power factor, a device called a "power factor correction capacitor" can be added to the circuit.
In older fluorescent luminaires with magnetic ballasts, the ballast could have a low power factor, which means that the input power to the ballast may not be fully utilized by the lamp.
The power factor correction capacitor works by introducing capacitive reactance into the circuit, which cancels out the inductive reactance of the ballast. This reduces the overall inductance of the circuit and increases the power factor, resulting in a more efficient use of electrical power.
The power factor correction capacitor is typically installed in parallel with the ballast in the fluorescent luminaire circuit. It can be connected directly to the ballast or to the incoming power supply, depending on the specific design of the luminaire. By improving the power factor, the power factor correction capacitor can help reduce energy consumption and improve the overall efficiency of the lighting system.
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The fastest pitch ever thrown was 105 mph, or 47 m/s. Wow! The moon's gravity is 16. 7% that of earth. If this pitcher threw the ball straight up on earth, it would reach a height of 112. 59 meters. How high would it go on the moon? This is a conversation of energy program. Kinetic energy = Potential energy. Truncate
The ball would reach a height of approximately 68.88 meters on the moon.
How to find height?To find the height the ball would reach on the moon, equate the kinetic energy of the ball on Earth to the potential energy it reaches at its maximum height.
On Earth:
Kinetic energy (KE) = Potential energy (PE)
The kinetic energy of the ball can be calculated using the formula:
KE = (1/2) × mv²
Let's assume the mass of the ball is m, and the velocity is v.
On Earth, the kinetic energy of the ball is:
KE_earth = (1/2) × m × (47 m/s)²
The potential energy of the ball at its maximum height on Earth is given by:
PE_earth = m × g × h_earth
where g = acceleration due to gravity on Earth (9.8 m/s²) and h_earth = height reached on Earth (112.59 meters).
Therefore, the equation:
KE_earth = PE_earth
(1/2) × m × (47 m/s)² = m × 9.8 m/s² × 112.59 meters
Now, calculate the height the ball would reach on the moon using the given information that the moon's gravity is 16.7% that of Earth's.
On the Moon:
The acceleration due to gravity on the Moon (g_moon) is 16.7% of Earth's gravity:
g_moon = 0.167 × 9.8 m/s²
The potential energy of the ball at its maximum height on the Moon is given by:
PE_moon = m × g_moon × h_moon
where h_moon = height we need to calculate.
Since the kinetic energy of the ball is conserved (KE_earth = KE_moon), we can set up the equation:
(1/2) × m × (47 m/s)² = m × g_moon × h_moon
Now solve for h_moon:
(1/2) × (47 m/s)² = g_moon × h_moon
h_moon = [(1/2) × (47 m/s)²] / g_moon
Substituting the values:
h_moon = [(1/2) × (47 m/s)²] / (0.167 × 9.8 m/s²)
h_moon = 68.88 meters
Therefore, the ball would reach a height of approximately 68.88 meters on the moon.
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The minimum size of a meteoroid that is capable of surviving its passage through Earth's atmosphere and hitting the ground is about as big as
a. a car.
b. a house.
c. a basketball.
d. a grain of sand.
e. your fist.
The minimum size of a meteoroid that is capable of surviving its passage through Earth's atmosphere and hitting the ground is about as big as, option d) a grain of sand.
When meteoroids enter Earth's atmosphere, they experience high temperatures and pressures due to friction with the air. Smaller meteoroids tend to burn up completely before reaching the ground, while larger ones may break up or explode in the atmosphere.
However, some small meteoroids, typically less than a millimeter in size, can make it to the ground. These meteoroids are often called micrometeorites and are collected from places such as rooftops and polar ice caps. Despite their small size, they can provide valuable information about the composition of the early solar system. Larger meteoroids, such as those that produce meteorites, are typically at least the size of a basketball or larger.
Therefore,the minimum size of a meteoroid that is capable of surviving its passage through Earth's atmosphere and hitting the ground is about as big as, option d) a grain of sand.
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if the maximum intensity is 3.90 w/m2 at the antenna, what is the intensity at ±5.50∘ from the center of the central maximum at the distant antenna? express your answer in watts per square meter.
To determine the intensity at ±5.50∘ from the center of the central maximum at the distant antenna, we need to consider the concept of diffraction patterns produced by antennas or apertures.
In a diffraction pattern, the intensity decreases as we move away from the central maximum. The intensity can be described by the equation:
Intensity = (Max Intensity) * (sin(θ)/θ)^2
where θ is the angle from the center of the central maximum.
Given that the maximum intensity is 3.90 W/m², we can calculate the intensity at ±5.50∘ using the above formula:
Intensity = (3.90 W/m²) * (sin(5.50∘)/5.50∘)^2
Calculating this expression gives us the intensity at ±5.50∘ from the center of the central maximum at the distant antenna. Please note that the specific value of the intensity depends on the diffraction characteristics of the antenna or aperture, which may vary in different scenarios.
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10 J of heat are removed from a gas sample while it is being compressed by a piston that does 20 J of work. What is the change in the thermal energy of the gas?
The change in the thermal energy of the gas is -30 J.
According to the first law of thermodynamics, the change in thermal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). In this case, the heat removed from the gas sample is 10 J and the work done on the gas by the piston is 20 J, so we can write:
ΔU = Q - W
ΔU = -10 J - 20 J
ΔU = -30 J
The negative sign indicates that the thermal energy of the gas has decreased, which makes sense since heat was removed from the system. Therefore, the change in the thermal energy of the gas is -30 J. It's important to note that this calculation assumes that there are no other forms of energy transfer (such as through radiation) and that the gas behaves ideally.
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you are in the middle of a large field. you walk in a straight line for 100 m, then turn left and walk 100 m more in a straight line before stopping. when you stop, you are 100 m from your starting point. by how many degress did you tur
The correct answer is E) This is impossible. You cannot walk 200m and be only 100m away from where you started. To determine the number of degrees you turned, we can use basic geometry.
You started in the middle of a field and walked 100m in a straight line. This forms a right triangle, where one side is the distance you walked (100m) and the other side is the distance from the starting point to the final position (100m).
Since the triangle is a right triangle, we can use the Pythagorean theorem to find the length of the hypotenuse, which represents the distance you are from the starting point.
Using the Pythagorean theorem:
[tex]c^2 = a^2 + b^2[/tex]
[tex]c^2 = 100^2 + 100^2[/tex]
[tex]c^2 = 20000[/tex]
c = √20000
c ≈ 141.42 m
From this, we can see that the distance from the starting point is approximately 141.42m, not 100m. Therefore, the given situation is impossible.
The correct answer is E) This is impossible. You cannot walk 200m and be only 100m away from where you started.
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Full Question ;
You are in the middle of a large field. You walk in a straight line for 100m, then turn left and walk 100m more in start line before stopping. When you stop, you are 100m from the starting point.By how many degrees did you turn?
A) 90
B) 120
C) 30
D) 180
E) This is impossible. You cannot walk 200m and be the only 100m away from where you started.
the resistance on a delta wired three-phase 240 vac 15 kw electric duct heater. if one element is open, what would the measurements be
If one element in a delta-wired three-phase 240 VAC 15 kW electric duct heater is open, the resistance between any two phases would increase by a factor of three.
In a delta-wired three-phase system, each element is connected between two phases, forming a closed loop. If one element is open, the circuit is broken, and the resistance between the two phases connected to the open element increases by a factor of three. This is because, in a delta-wired system, the total resistance is the sum of the individual resistances of each element, and with one element open, the total resistance increases significantly.
In terms of measurements, we could use a multimeter to measure the resistance between any two phases before and after the element is opened. The resistance between the two phases connected to the open element would increase by a factor of three, while the resistance between the other two phases would remain the same. It is important to note that with one element open, the total power output of the heater would be reduced, as the open element would not be contributing to the heating.
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Complete Question:
the resistance on a delta wired three-phase 240 vac 15 kw electric duct heater. if one element is open, what would the measurements be?
A car is travelling along a country road that resembles a roller coaster truck. If car
travels with uniform speed, the force exerted by road on the car is maximum at:
a) A b) B c) C d) equal at all position A, B & C
A car is travelling along a country road that resembles a roller coaster truck. If car travels with uniform speed, the force exerted by road on the car is maximum at A.
Option A is correct.
How do we know?We can determine the force exerted by the road on the car by taking into consideration the direction of the net force acting on the car which is the force exerted by the road on the car is due to the normal force, which is perpendicular to the road surface.
The car is at the bottom of the roller coaster-like track at position A. We notice that the normal force from the road acts in the upward direction which is in opposition to the gravitational force on the car.
The force exerted by the road on the car is maximum at position A, hence the net force on the car is directed upward.
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what would be the energy of photons having the same wavelength as these electrons? express your answer in electron volts.
To calculate the energy of photons having the same wavelength as the electrons, we'll use the following steps:
1. Convert the wavelength of electrons to wavelength of photons using de Broglie's equation:
λ = h / (m * v), where λ is the wavelength, h is Planck's constant (6.626 × 10^-34 Js), m is the electron mass (9.109 × 10^-31 kg), and v is the electron velocity.
2. Calculate the frequency of the photons using the speed of light:
ν = c / λ, where ν is the frequency, c is the speed of light (2.998 × 10^8 m/s), and λ is the wavelength obtained in step 1.
3. Calculate the energy of the photons in joules using Planck's equation:
E = h * ν, where E is the energy, h is Planck's constant, and ν is the frequency obtained in step 2.
4. Convert the energy from joules to electron volts (eV) using the conversion factor:
1 eV = 1.602 × 10^-19 J
Once you have the electron velocity (v), you can follow these steps to find the energy of the photons in electron volts.
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A square tile having 50 cm side is designed to withstand a pressure of 200n/m sq. Find out if a force of 45n can be applied on tile
A square tile that has a 50 cm side is designed to withstand a pressure of 200 N/m². We have to determine if a force of 45 N can be applied to the tile.
The formula for pressure is:
Pressure = Force/Area, where area is in square meters.
To determine the maximum force that can be applied to the tile, we need to first convert the side length of the tile from centimeters to meters.50 cm = 0.5 m
The area of the tile is:
Area = (side length)²Area = (0.5 m)²Area = 0.25 m²
Using the formula for pressure: Pressure = Force/Area200 N/m² = F/0.25 m²
Multiplying both sides by 0.25 m²:50 N = F
Therefore, the maximum force that can be applied to the tile is 50 N, which is greater than the force of 45 N.
Therefore, a force of 45 N can be applied to the tile.
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mourning doves have a small patch of iridescent feathers. the color is produced by a 330-nm-thick layer of keratin (n
Mourning doves are known for their distinctive cooing and gentle demeanor. These birds are also recognized for their unique physical characteristics, including a small patch of iridescent feathers.
The iridescent color of the feathers is produced by a thick layer of keratin that measures 330 nanometers. This layer reflects light in a way that creates a colorful, shimmering effect.
It is interesting to note that mourning doves are not the only birds that exhibit iridescence in their feathers. Many other species, including peacocks and hummingbirds, have iridescent feathers that are used to attract mates or establish dominance. In the case of mourning doves, the iridescent feathers are thought to play a role in courtship displays and other social behaviors.
Despite their small size, the iridescent feathers of mourning doves are a remarkable example of the beauty and complexity of the natural world. These delicate birds serve as a reminder of the importance of protecting and preserving our natural environment for future generations to enjoy.
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The decay chain of uranium includes radon, a noble gas. When uranium in the soil decays to radon, it may seep into houses; this can be a significant source of radiation exposure. Most of the exposure comes from the decay products of radon, but some comes from alpha decay of the radon itself. If radon in the air in your home is at the maximum permissible level, the gas in your lungs will have an activity of about 0.22 Bq. Each decay generates an alpha particle with 5.5 MeV of energy, and essentially all that energy is deposited in lung tissue. Over the course of 1 year, what will be the dose equivalent in Sv to the approximately 0.90 kg mass of your lungs?
Dose equivalent = Absorbed dose x Quality factor = 6.10 x 10^-6 J/kg x 20 = 1.22 x 10^-4 Sv
So, the dose equivalent in Sv to the approximately 0.90 kg mass of your lungs over the course of 1 year is approximately 1.22 x 10^-4 Sv.
To calculate the dose equivalent in Sv to the approximately 0.90 kg mass of your lungs, we need to use the following formula: Dose equivalent (Sv) = Absorbed dose (Gy) x Quality factor (Q)
First, we need to find the absorbed dose in gray (Gy), which is a unit of energy absorbed per unit of mass. We know that each decay generates an alpha particle with 5.5 MeV of energy, and essentially all that energy is deposited in lung tissue. To convert MeV to joules (J), we can use the following conversion factor:
1 MeV = 1.602 x 10^-13 J
So, the energy deposited in lung tissue by each decay is:
5.5 MeV x 1.602 x 10^-13 J/MeV = 8.81 x 10^-13 J
The activity of radon in the air in your home is at the maximum permissible level, which means that there are 0.22 disintegrations per second (Bq) in your lungs. This means that in one year (3.15 x 10^7 seconds), there will be:
0.22 Bq x 3.15 x 10^7 s = 6.93 x 10^6 disintegrations
Therefore, the absorbed dose in lung tissue over the course of 1 year is:
Absorbed dose = 8.81 x 10^-13 J/disintegration x 6.93 x 10^6 disintegrations = 6.10 x 10^-6 J/kg
Next, we need to find the quality factor (Q), which takes into account the type of radiation and its ability to cause biological damage. For alpha particles, the quality factor is 20. Therefore:
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Water flows in a smooth 0.2m diameter pipeline that is 65m long. The Reynolds number is 10 . Determine the flow rate and pressure drop. After many years of use, minerals deposited on the pipe cause the same pressure drop to produce only one-half the original flow rate. Estimate the size of the relative roughness elements for the pipe.
Flow rate: Using the Reynolds number equation for laminar flow, we can solve for the flow rate Q:
Re = (rho * v * D) / mu, where rho is the density of water, v is the velocity, D is the diameter of the pipe, and mu is the dynamic viscosity of water.
Re-Number = 10^4
Diameter = 0.2 m
Density of Water, rho = 1000 kg/m^3
Dynamic viscosity of Water, mu = 0.001 Ns/m^2
Solving for the velocity v: v = Re * mu / (rho * D) = 10^4 * 0.001 / (1000 * 0.2) = 0.05 m/s
Solving for the flow rate: Q = pi * D^2 / 4 * v = pi * (0.2)^2 / 4 * 0.05 = 0.00157 m^3/s
Pressure drop: We can use the Hagen-Poiseuille equation to calculate the pressure drop:
delta P = 32 * mu * L * Q / (pi * D^4)
delta P = 32 * 0.001 * 65 * 0.00157 / (pi * (0.2)^4) = 16.5 Pa
Relative roughness: After many years of use, the minerals deposited on the pipe cause the same pressure drop to produce only one-half the original flow rate. Let's assume that the flow rate is reduced by a factor of 1/2. We can then use the Darcy-Weisbach equation to solve for the relative roughness:
delta P / L = f * rho * v^2 / (2 * D)
f = 64 / Re (for laminar flow)
For the reduced flow rate, we have:
Q' = 0.5 * Q = 0.5 * 0.00157 = 0.000785 m^3/s
v' = Q' / (pi * D^2 / 4) = 0.000785 / (pi * (0.2)^2 / 4) = 0.198 m/s
Re' = (rho * v' * D) / mu = (1000 * 0.198 * 0.2) / 0.001 = 39,600
Using the Darcy-Weisbach equation, we can solve for the relative roughness epsilon/D:
delta P / L = f * rho * v^2 / (2 * D)
delta P' / L = f' * rho * (v')^2 / (2 * D)
Since delta P' = delta P / 2 and Q' = Q / 2, we have:
f' = f / 4 = 16 / Re = 16 / 39600 = 0.000404
Solving for epsilon/D:
delta P / L = f * rho * v^2 / (2 * D)
epsilon / D = 2 * delta P / (f' * rho * v^2) - 1
epsilon / D = 2 * 16.5 / (0.000404 * 1000 * 0.198^2) - 1
epsilon / D = 0.0057
Therefore, the size of the relative roughness elements for the pipe is approximately 0.57%.
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1. In a bike race, Lance Armstrong traveled at a speed of 25 m/s. How long did it
take him to travel 350 m?
2. What is your speed if you bike 50 meters in 15 seconds?
3. If Lance can go from 2 m/s to 30 m/s in 12 seconds, what is his acceleration?
The first two exercises are of the uniform rectilinear movement (MRU) is a type of movement that is characterized by a constant speed and a rectilinear trajectory. In other words, an object moving with MRU always moves at the same speed and in a straight line.
The formula to calculate the speed in the MRU is simple: v = d/t, where:
v is the speed
d is the distance traveled
t is the elapsed time
It is also possible to use the formula s = vt to calculate the distance traveled, where s is the distance and t is the time.
It is important to note that velocity in the MRU is a vector that has a magnitude (the velocity itself) and a direction (the direction of motion). In the MRU, the direction of movement is always constant and coincides with the direction of the trajectory.
1. In a bike race, Lance Armstrong traveled at a speed of 25 m/s. How long did it take him to travel 350 m?First we get the data:
V = Velocity = 25 m/s
T = time = ?
D = distance = 350 m
We already know that the MRU formula is V = d/t. But we must calculate the time, we clear the time, then:
t = d/v
We continue, substitute data and solve for time, then
t = d/v
t = (350 m)/(25 m/s)
t = 14 m
To travel 350 meters, it takes me 14 minutes.
2. What is your speed if you bike 50 meters in 15 seconds?To solve this, we first get the data:
V = speed = ?
D = distance = 50 m
T = time = 15 s
We know that the MRU formula is V = d/t. It is also the same formula for calculating velocity.
How should we calculate speed? It is not necessary to clear the formula, we substitute data and solve, then
V = d/t
V = (50 m)/(15 s)
V = 3.33 m/s
If I ride a bicycle 50 meters in 15 seconds, my speed is 3.33 meters per second (m/s).
As for the third exercise, it is an exercise of a rectilinear motion uniformly accelerated (MRUA), since Lance's speed increases constantly in time. A rectilinear motion uniformly accelerated (MRUA) is a type of motion in which an object moves in a straight line and its velocity changes uniformly in time due to constant acceleration. In this type of movement, the acceleration is constant, which means that the speed increases or decreases at a constant rate in each unit of time.
In the exercise presented, the change in speed of Lance is described, which increases from 2 m/s to 30 m/s in a time period of 12 seconds. Since the velocity changes uniformly in time, this motion can be determined to be an MRUA.The MRUA is a fundamental concept in physics and is used to describe many phenomena, from free-falling objects to moving vehicles. Uniformly accelerated rectilinear motion is described mathematically by a series of equations, relating position, velocity, acceleration, and time.
One of the most important equations in the MRUA is the velocity equation, which relates the final velocity (Vf), the initial velocity (Vi), the acceleration (a) and the time (t). The velocity equation can be expressed as:
Vf = Vi + a * tThis equation shows how the final velocity of an object in an MRUA depends on its initial velocity, acceleration, and elapsed time.
Another important equation in the MRUA is the position equation, which relates the final position (x), the initial position (x0), the initial velocity (Vi), the acceleration (a) and the time (t).
The position equation can be expressed as:
x = x0 + Vi × t + 1/2 * a × t²This equation shows how the position of an object in an MRUA changes as a function of time, initial velocity, acceleration, and initial position.
3. If Lance can go from 2 m/s to 30 m/s in 12 seconds, what is his acceleration?First we get the data, this is the first step to start solving:
Vf = Final speed = 12 m/s
Vo = Initial velocity = 2 m/s
t = time = 12 s
a = acceleration = ?
The velocity equation can be expressed as:
Vf = Vi + a * t.
We use this formula to clear the formula to calculate the acceleration,
a = (Vf - Vo)/tWe continue solving, now we substitute our data and solve:
a = (Vf - Vo)/t
a = (30 m/s - 2 m/s)/(12 s)
a = 2.33 m/s²
Its acceleration is 2.33 meters per second squared (m/s²).
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an airplane flies at 550 km=hr at 1500 m altitude on a standard day. the plane climbs to 15,000 m and flies at 1200 km=h. calculate the mach number of flight in both cases.
The airplane's Mach number is 0.45 when flying at 1500 m altitude and 1.12 when flying at 15,000 m altitude.
The mach number is the ratio of the speed of an object to the speed of sound in the medium through which it is moving. At sea level on a standard day, the speed of sound is approximately 1225 km/h. At an altitude of 1500 m, the speed of sound decreases slightly, but for simplicity, we will assume that it remains the same.
At 1500 m altitude, the airplane's speed is 550 km/h, which is approximately 0.45 times the speed of sound (Mach 0.45). When the plane climbs to 15,000 m altitude and travels at 1200 km/h, the speed of sound is approximately 1075 km/h. Therefore, the plane's Mach number is approximately 1.12 (1200/1075).
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In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 11,500 N to the ball for a time of 0.60 x 10–s. Part A Assuming that this force is constant, what is the magnitude J of the impulse on the ball? Enter your answer numerically in newton seconds using two significant figures.
The magnitude of the impulse on the ball can be calculated by multiplying the force applied to the ball by the time for which the force is applied. Impulse (J) = Force (F) * Time (Δt)
Given that the force applied by the batter is 11,500 N and the time of contact is 0.60 × 10^(-3) s, we can substitute these values into the equation:
J = 11,500 N * 0.60 × 10^(-3) s
J = 6.90 N⋅s
Therefore, the magnitude of the impulse on the ball is 6.90 N⋅s.
Impulse is a vector quantity with both magnitude and direction. However, since the direction is not specified in this problem, we only calculate its magnitude.
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