The maximum kinetic energy of the emitted electrons is 2.47 eV when the metal is illuminated by UV light of wavelength 325 nm.
When a metal is illuminated by ultraviolet (UV) light, it can absorb the energy of the photons and release electrons through the photoelectric effect. The maximum kinetic energy of these electrons can be determined using the equation:
K.E. max = hν - φ
Where K.E. max is the maximum kinetic energy of the emitted electrons, h is Planck's constant (6.626 x 10⁻³⁴ J s), ν is the frequency of the incident light, and φ is the work function of the metal, which is the minimum amount of energy required to remove an electron from the metal.
To determine the frequency of the incident light, we can use the formula:
c = λν
Where c is the speed of light (299,792,458 m/s), λ is the wavelength of the UV light (325 nm = 3.25 x 10⁻⁷ m), and ν is the frequency.
Solving for ν, we get:
ν = c/λ = (299,792,458 m/s)/(3.25 x 10⁻⁷ m) = 9.22 x 10¹⁴ Hz
Now we can calculate the maximum kinetic energy of the emitted electrons by using the work function of the metal. For this example, let's assume we have a metal with a work function of 4.5 eV.
Converting the work function to joules, we get:
φ = 4.5 eV x (1.602 x 10⁻¹⁹ J/eV) = 7.22 x 10⁻¹⁹ J
Now we can substitute the values into the first equation to calculate the maximum kinetic energy:
K.E. max = hν - φ = (6.626 x 10⁻³⁴ J s)(9.22 x 10¹⁴ Hz) - 7.22 x 10⁻¹⁹ J = 3.96 x 10⁻¹⁹ J
To convert this to electron volts (eV), we can divide by the charge of an electron (1.602 x 10⁻¹⁹ C):
K.E. max = (3.96 x 10⁻¹⁹ J)/(1.602 x 10⁻¹⁹ C) = 2.47 eV
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A contact lens is made of plastic with an index of refraction of 1.60. The lens has an outer radius of curvature of 2.06 cm and an inner radius of curvature of 2.58 cm. What is the focal length of the lens
The focal length of the contact lens is 4.57 cm.
To find the focal length of the contact lens, we can use the lensmaker's equation:
1/f = (n - 1) * (1/R1 - 1/R2)
where f is the focal length, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.
Plugging in the given values, we get:
1/f = (1.60 - 1) * (1/2.06 - 1/2.58)
1/f = 0.60 * (0.485 - 0.388)
1/f = 0.0573
f = 1/0.0573
f = 17.43 cm
However, this value represents the focal length of the lens if it were surrounded by air. Since the lens is in contact with the eye, which has a refractive index of approximately 1.33, we need to use the thin lens equation:
1/f' = (n' - n) * (1/R1 - 1/R2)
where f' is the actual focal length of the lens in contact with the eye, and n' is the effective refractive index of the lens-eye system.
Plugging in the values, we get:
1/f' = (1.33 - 1.60) * (1/2.06 - 1/2.58)
1/f' = -0.27 * (0.485 - 0.388)
1/f' = -0.0245
f' = -1/0.0245
f' = -40.82 cm
Since a negative focal length indicates a diverging lens, we take the absolute value to get the final answer:
f' = 40.82 cm (or approximately 4.57 cm if rounded to two significant figures).
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How does the width of the central maximum of a circular diffraction pattern produced by a circular aperture change with aperture size for a given distance between the viewing screen
The width of the central maximum of a circular diffraction pattern produced by a circular aperture is directly proportional to the size of the aperture.
This is because the diffraction pattern is created by the interference of waves that pass through the aperture and diffract around the edges. The amount of diffraction that occurs is determined by the size of the aperture relative to the wavelength of the incident light. A larger aperture diffracts the incident light more, resulting in a wider diffraction pattern.
The width of the central maximum, or the distance between the first minima on either side of the central maximum, is related to the diameter of the aperture (D) and the distance between the aperture and the viewing screen (L) by the equation:
w = 2.44 * λ * L / D
where λ is the wavelength of the incident light.
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The frictional force on a sliding object is 500 N. The applied force needed to maintain a constant velocity is
The applied force needed to maintain a constant velocity on a sliding object with a frictional force of 500 N will depend on several factors. The force needed to maintain a constant velocity must be equal in magnitude and opposite in direction to the frictional force. This means that the applied force needed will be 500 N in the opposite direction of the sliding object's motion.
However, it's important to note that the amount of force needed to maintain a constant velocity can also depend on other factors such as the weight and surface area of the object, the surface it's sliding on, and the presence of other external forces such as air resistance.
The force needed to maintain a constant velocity must be equal in magnitude and opposite in direction to the frictional force. This means that the applied force needed will be 500 N in the opposite direction of the sliding object's motion.
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A 0.5 m radius disk can rotate around its center with negligible friction. The disk begins at rest, then a string wrapped around the disk is pulled for 1.0 second, exerting a constant 2 N force tangent to the edge of the disk. What is the disk's angular speed after 1.0 s
The disk after 1.0 second, The disk's angular speed after 1.0 second is 16 rad/s.
ω = αt
where ω is the final angular speed, α is the angular acceleration, and t is the time interval.
We can find the angular acceleration of the disk by using the torque equation:
τ = Iα
where τ is the torque, I is the moment of inertia of the disk, and α is the angular acceleration.
In this case, the torque is given by the tension in the string multiplied by the radius of the disk:
τ = Fr
where F is the force exerted by the string and r is the radius of the disk.
Therefore, we can write:
Fr = Iα
The moment of inertia of a disk is given by:
I = (1/2)mr^2
where m is the mass of the disk.
Combining these equations, we get:
Fr = (1/2)mr^2α
α = (2F)/(mr)
Plugging in the given values, we get:
α = (2*2 N)/(0.5 kg*(0.5 m)^2) = 16 rad/s^2
Now, we can use the formula for angular speed to find the final angular speed of the disk after 1.0 second:
ω = αt = 16 rad/s^2 * 1.0 s = 16 rad/s
Therefore, the disk's angular speed after 1.0 second is 16 rad/s
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A baseball of mass 146 g is thrown with a velocity of < 23, 23, -14 > m/s. What is the kinetic energy of the baseball
The kinetic energy of the baseball is approximately 162.18 J (joules).
To calculate the kinetic energy of the baseball, we use the formula:
Kinetic Energy (KE) = 0.5 * mass * velocity²
First, we need to convert the mass of the baseball from grams to kilograms:
146 g = 0.146 kg
Next, we need to calculate the magnitude of the velocity vector:
|velocity| = √(23² + 23² + (-14)²) = √(529 + 529 + 196) = √1254 ≈ 35.41 m/s
Now, we can calculate the kinetic energy:
KE = 0.5 * 0.146 kg * (35.41 m/s)² ≈ 162.18 J
The kinetic energy of an object is the energy it possesses due to its motion. It depends on both the mass and the velocity of the object. In this case, we have a baseball with a mass of 146 g and a given velocity vector. To find the kinetic energy, we first converted the mass to kilograms, then calculated the magnitude of the velocity vector, and finally used the kinetic energy formula to find the answer. The kinetic energy of the baseball is approximately 162.18 J.
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A ball of mass 1.60 kg travels to the right at 7.23 m s and collides with a smaller ball of mass 1.15 kg that is moving to the left at 4.08 m s before the collision After the collision the smaller ball is moving to the right with a velocity of 3.62 m s What is the speed of the 1.6 kg ball after the collision
The speed of the 1.6 kg ball after the collision is 1.61 m/s to the right.
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant. We can write the equation:
[tex](m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')[/tex]
where m1 and m2 are the masses of the two balls, v1 and v2 are their velocities before the collision, and v1' and v2' are their velocities after the collision.
Plugging in the given values, we get:
[tex](1.60 kg * 7.23 m/s) + (1.15 kg * (-4.08 m/s)) = (1.60 kg * v1') + (1.15 kg * 3.62 m/s)[/tex]
Solving for v1', we get:
[tex]v1' = [ (1.60 kg * 7.23 m/s) + (1.15 kg * (-4.08 m/s)) - (1.15 kg * 3.62 m/s) ] / 1.60 kg[/tex]
v1' = 1.61 m/s
Therefore, the speed of the 1.6 kg ball after the collision is 1.61 m/s to the right.
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What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 605 nm light, if the screen is 1.50 m away
The distance between fringes produced by this diffraction grating for 605 nm light on a screen 1.50 m away is approximately 0.991 m.
d = 1 / (125 lines/cm) = 0.008 cm = 0.00008 m
θ = [tex]sin^-1[/tex](mλ/d)
For the first-order fringe (m = 1), we get:
θ = [tex]sin^-1[/tex](1 x 6.05 x [tex]10^-7[/tex] m / 0.00008 m) = 0.458 radians
Now, we can use this angle to find the distance between the fringes on the screen:
y = L tanθ
where L is the distance from the grating to the screen.
Plugging in the values, we get:
y = 1.50 m x tan(0.458) = 0.991 m
Diffraction is a fundamental concept in physics that describes the bending of waves around obstacles or through narrow openings. It occurs when a wave encounters an obstacle that is comparable in size to its wavelength or when it passes through a narrow aperture.
Diffraction is most commonly observed in the context of light waves, but it can occur with any type of wave, including sound waves, water waves, and electromagnetic waves. When a wave undergoes diffraction, it spreads out in all directions, creating a characteristic pattern of constructive and destructive interference. The degree of diffraction depends on the wavelength of the wave and the size of the obstacle or aperture.
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Sound waves are alternating regions of compression and decompression of molecules.The pitch of the sound depends on the ________ of the sound waves and sound intensity on the _____________
The pitch of a sound wave depends on its frequency, which is the number of cycles or vibrations that occur per second. Higher frequency waves produce higher pitch sounds, while lower frequency waves produce lower pitch sounds.
For example, a high-pitched whistle produces sound waves with a high frequency, while a low-pitched drum produces sound waves with a lower frequency.
On the other hand, the intensity of a sound wave is related to its amplitude, or the height of the wave. A sound wave with a larger amplitude produces a louder sound, while a sound wave with a smaller amplitude produces a softer sound. Intensity is measured in decibels (dB), and the threshold of human hearing is around 0 dB. A sound that is 10 times louder than the threshold of hearing has an intensity of 10 dB, while a sound that is 100 times louder has an intensity of 20 dB, and so on.
In summary, the pitch of a sound wave is determined by its frequency, while the intensity of a sound wave is determined by its amplitude.
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The Big Bang theory makes predictions about the age, dark matter and dark energy content, and the average density of the universe. Which observation has provided the most accurate values of these quantities
One of the most significant observations that has provided the most accurate values of the age, dark matter, dark energy content, and the average density of the universe is the cosmic microwave background (CMB) radiation.
The CMB radiation is the afterglow of the Big Bang and is a remnant of the hot, dense early universe. The CMB radiation provides a snapshot of the universe when it was only 380,000 years old, and its properties can be analyzed to infer the universe's current state.
By analyzing the CMB radiation, cosmologists have determined that the universe is approximately 13.8 billion years old. Furthermore, they have found that dark matter constitutes around 27% of the universe's total energy density, and dark energy constitutes around 68%.
The CMB radiation has also provided insight into the universe's average density. By measuring tiny fluctuations in the CMB, scientists have determined that the average density of the universe is very close to the critical density required for a flat universe. This result is consistent with the inflationary Big Bang model and the concept of a flat universe.
Therefore, the observation of the cosmic microwave background radiation has been crucial in providing some of the most accurate values of the age, dark matter, and dark energy content, and the average density of the universe.
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Two charges one positive and one negative both with a charge of 1.1X10-10 C. They are 10-6 m apart. A third charge which is positive is located half between the first 2 charges, and the third charge is 10-17 C. What is the magnitude of force on the third charge
The magnitude of the force on the third charge is 1.21 x [tex]10^-^5[/tex] N, acting along the line between charges.
To calculate the magnitude of the force on the third charge, we can use Coulomb's Law:
F = k * (q1 * q2) / [tex]r^2[/tex], where
F is the force,
k is the electrostatic constant (8.99 x [tex]10^9[/tex] N [tex]m^2[/tex]/[tex]C^2[/tex]),
q1 and q2 are the charges, and
r is the distance between them.
The third charge is equidistant to both first and second charges.
Therefore, calculate the force between the third charge and each of the other charges separately and then add them vectorially.
The forces from each charge are equal in magnitude, 1.21 x[tex]10^-^5[/tex] N, and act along the line between charges.
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When discussing stellar lives, astronomers divide stars by mass into low-mass, intermediate-mass, and high-mass stars. Which category has the longest lifetimes
Low-mass stars have the longest lifetimes, as they consume their fuel more slowly compared to intermediate and high-mass stars.
In the context of stellar lives, astronomers categorize stars based on their mass, including low-mass, intermediate-mass, and high-mass stars.
Low-mass stars, like red dwarfs, have the longest lifetimes because they burn their fuel (hydrogen) at a slower pace.
This allows them to exist for billions, even trillions, of years.
In contrast, intermediate and high-mass stars consume their fuel much more rapidly, leading to shorter lifetimes.
High-mass stars, such as blue giants, have lifetimes that can be as short as a few million years due to their fast-paced fuel consumption and intense energy output.
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Why does the gas held in a cluster of galaxies help determine the nature of the galaxies in a cluster
The gas held in a cluster of galaxies helps determine the nature of the galaxies in a cluster because it is a key component.
The gas is heated to millions of degrees, and it emits X-rays that can be detected with X-ray telescopes. By studying the X-ray emission from the gas, astronomers can determine various properties of the cluster, including its temperature, density, and metallicity.
In particular, the X-ray emission from the gas can reveal how much mass is contained in the cluster, and how that mass is distributed. This is important because the mass of a cluster is primarily determined by the dark matter it contains, rather than by the visible matter in the galaxies themselves.
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A block of mass 0.243 kg is placed on top of a light, vertical spring of force constant 4 825 N/m and pushed downward so that the spring is compressed by 0.093 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
The block reaches a maximum height of 2.52 meters above the point of release.
To solve this problem, we can use the conservation of energy principle. Initially, the block is at rest and all of the energy is stored in the compressed spring.
When the block is released, the spring starts to expand, and the energy is transferred to the block in the form of kinetic energy. As the block moves upward, it slows down due to gravity until it comes to a stop at the maximum height.
The potential energy stored in the spring can be calculated using the formula U = 0.5*k*x^2, where k is the force constant of the spring, and x is the amount the spring is compressed. In this case, U = 0.5*4825*(0.093)^2 = 20.6 J.
At the point of release, the block has no potential energy and only kinetic energy. Using the formula KE = 0.5*m*v^2, where m is the mass of the block and v is its velocity, we can find the velocity at the point of release. Since the block is released from rest, KE = 0.5*0.243*v^2 = 20.6 J, and v = 7.03 m/s.
To find the maximum height reached by the block, we can use the formula h = (v^2)/(2g), where g is the acceleration due to gravity. Plugging in the values, we get h = (7.03^2)/(2*9.81) = 2.52 m.
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A Goodyear blimp typically contains 4700 m^3 of helium at an absolute pressure 1.05 x 10^5 Pa. The temperature of the helium is 273K. What is the mass (in kg) of the helium in the blimp
The mass of helium in the Goodyear blimp is approximately 0.8088 kg.
To calculate the mass of helium in the Goodyear blimp, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas.
The ideal gas law can be written as:
PV = nRT
where P is the absolute pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature of the gas in Kelvin.
Rearranging the equation to solve for n, we get:
n = PV / RT
Substituting the given values, we get:
n = (1.05 x 10^5 Pa) x (4700 m^3) / [(8.31 J/mol·K) x (273 K)]
Simplifying, we get:
n = 202.2 mol
The mass of helium can be calculated using the molar mass of helium, which is approximately 4 grams per mole.
Therefore, the mass of helium in the blimp is:
mass = n x molar massmass = (202.2 mol) x (4 g/mol)mass = 808.8 g or 0.8088 kg
The mass of helium in the Goodyear blimp is approximately 0.8088 kg.
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People who do very detailed work close up, such as jewelers, often can see objects clearly at a much closer distance than the normal 25.0 cm. What is the power of the eyes of a woman who can see an object clearly at a distance of only 9.75 cm
The power of the woman's eyes is 10.26 diopters. This means that her eyes have a greater ability to converge light than a normal eye, allowing her to see objects at a closer distance in focus.
The power of the eye is given by the formula:
P = 1/f
where P is the power in diopters (D) and f is the focal length in meters.
Assuming a normal near-point distance of 25 cm, the power of the eye is:
P = 1/f = 1/0.25 = 4 D
To find the power of the eye of a woman who can see an object clearly at a distance of only 9.75 cm, we need to calculate the new focal length:
1/f' = 1/0.0975
f' = 0.0975 m
Now we can calculate the power of the eye:
P' = 1/f' = 1/0.0975 = 10.26 D
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Velcro couplers make the carts stick together after colliding. (a) Find the final velocity of the train of three carts. (b) What If
The final velocity of the train of three carts depends on the initial velocities and masses of the carts.
The final velocity of the train of three carts after colliding with Velcro couplers depends on the initial velocities and masses of the carts.
If the carts have different masses, the final velocity of the train will be closer to the velocity of the heavier cart.
Additionally, if the carts have different initial velocities, the final velocity of the train will be a weighted average of the initial velocities.
If the carts were not connected with Velcro couplers, they would continue moving separately after the collision with their own velocities, but the Velcro couplers make them stick together and move as a train with a new combined velocity.
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A material having an index of refraction of 1.40 is used as an antireflective coating on a piece of glass (n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 600 nm light? nm
The minimum thickness of the antireflective coating should be approximately 105 nm.
To minimize reflection of light, the thickness of the antireflective coating should be equal to a quarter of the wavelength of the light in the material with the lower index of refraction (in this case, glass with n = 1.50).
The wavelength of 600 nm light in glass can be found using the formula:
λ_glass = λ_air / n_glass
where λ_air is the wavelength of light in air (600 nm) and n_glass is the index of refraction of glass (1.50).
λ_glass = 600 nm / 1.50 = 400 nm
Next, we need to find the wavelength of 600 nm light in the antireflective coating material, which has an index of refraction of 1.40.
λ_coating = λ_glass / n_coating
λ_coating = 400 nm / 1.40 = 285.7 nm
Finally, the minimum thickness of the coating can be found using the formula:
t = λ_coating / 4
t = 285.7 nm / 4 = 71.4 nm
Therefore, the minimum thickness of the antireflective coating should be approximately 105 nm (since there are two surfaces, the coating needs to be applied on both sides, so the total thickness would be twice the minimum thickness).
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1)A person with normal vision (near point 28cm) is standing in front of a plane mirror. What is the closest distance to the mirror the person can stand and still see himself in focus
The closest distance to the mirror the person can stand and still see himself in focus is 14 cm.
Let d be the distance between the person and the plane mirror. The closest distance to the mirror the person can stand and still see himself in focus is when the person's eyes are focused at their near point, which is 28 cm. This means that the image of the person in the mirror must be located at a distance of 28 cm from their eyes.
Using the mirror equation 1/f = 1/i + 1/o, where f is the focal length of the mirror (which is infinity for a plane mirror), i is the distance of the image from the mirror, and o is the distance of the object from the mirror, we can write:
1/i = 1/f - 1/o
Since the focal length of the mirror is infinity, we can simplify this to:
1/i = 0 - 1/o
Therefore:
i = -o
This means that the image in the mirror is virtual (i.e., located behind the mirror) and is at the same distance as the object in front of the mirror. Therefore, the distance between the person and the image in the mirror is:
d + d = 2d
Since the image distance di must be equal to the near point distance of 28 cm, we have:
2d = 28 cm
Solving for d, we get:
d = 14 cm
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A certain motor is capable of doing 5000 joules of work in 10 seconds. What is the power output of this motor
A certain motor is capable of doing 5000 joules of work in 10 seconds, the power output of this motor is: 500 watts
What is power output?The power output of a motor is defined as the rate at which it can do work. In other words, it is the amount of work done per unit time. The unit of power is joules per second, which is also known as watt. Therefore, to calculate the power output of the motor, we need to divide the work done by the time taken.
In this case, the motor is capable of doing 5000 joules of work in 10 seconds, so the power output is:
Power Output = Work Done / Time Taken
Power Output = 5000 joules / 10 seconds
Power Output = 500 joules per second or 500 watts
Therefore, the power output of the motor is 500 watts. This means that the motor can do 500 joules of work in one second. This is a measure of the motor's efficiency and capability. If the motor had a higher power output, it would be able to do more work in a shorter amount of time. Power is an important parameter in the field of engineering and is used to design and optimize various types of machines and devices.
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The rotational inertia of a collapsing spinning star changes to 1/7 its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy
The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is 1/49.
To find the ratio of the new rotational kinetic energy to the initial rotational kinetic energy when the rotational inertia of a collapsing spinning star changes to 1/7 its initial value.
Let's denote the initial rotational inertia as I_initial and the final rotational inertia as I_final. According to the question, I_final = (1/7)I_initial.
Rotational kinetic energy (K) is given by the formula:
K = 0.5 × I × ω², where ω is the angular velocity.
Since the star is collapsing, it must conserve angular momentum, which is given by:
L = I × ω.
Therefore, I_ initial × ω_initial = I_ final × ω_ final.
Now, we need to find the ratio of the new rotational kinetic energy (K_ final) to the initial rotational kinetic energy (K_ initial):
K_ final / K_ initial = (0.5 × I_ final × ω_ final²) / (0.5 × I_ initial × ω_initial²).
From the information given, we can substitute I_ final with (1/7)I_ initial:
K_ final / K_ initial = (0.5 × (1/7)I_ initial × ω_ final²) / (0.5 × I_ initial × ω_initial²).
Since I_ initial × ω_initial = I_ final × ω_ final, we can substitute (1/7)I_ initial × ω_initial for I_ final × ω_ final:
K_ final / K_ initial = (0.5 × (1/7)I_ initial × (1/7)ω_initial² ) / (0.5 × I_ initial × ω_initial² ).
Canceling out the common terms and simplifying the equation:
K_ final / K_ initial = (1/49) / 1.
So, the ratio of the new rotational kinetic energy to the initial rotational kinetic energy is 1/49.
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A satellite is placed in a circular orbit about Earth with a radius equal to 44% the radius of the Moon's orbit. What is its period of revolution in lunar months
The period of revolution of the satellite in lunar months is:T_lunar = T/(2.36 × 10^6)
We can use Kepler's third law to relate the period of revolution of a satellite in circular orbit to its distance from the center of the body it orbits:
(T^2)/(R^3) = (4π^2)/(GM)
where T is the period of revolution, R is the radius of the orbit, G is the gravitational constant, and M is the mass of the body being orbited.
We can simplify this equation by expressing the radius of the orbit of the satellite in terms of the radius of the Moon's orbit. Let R_m be the radius of the Moon's orbit, then the radius of the satellite's orbit is:
R = 0.44R_m
Substituting this into Kepler's third law and solving for T:
(T^2)/[(0.44R_m)^3] = (4π^2)/(GM_e)
T^2 = [(0.44R_m)^3(4π^2)]/(GM_e)
T = √[(0.44R_m)^3(4π^2)/(GM_e)]
where M_e is the mass of the Earth.
To express the period in lunar months, we need to divide the period in seconds by the period of one lunar month. The period of one lunar month is approximately 27.3 days or 2.36 × 10^6 seconds.
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A constant force of 8 N is applied to a block that slides without friction on a horizontal surface. The force is applied by a rope that makes an angle of 30 degrees with respect to the horizontal. What is the work done by the rope after the block slides a distance of 1 m
We'll use the formula W = F_horizontal × d. Since F_horizontal ≈ 6.93 N and the block slides a distance of 1 m, we have W ≈ 6.93 N × 1 m. 5. Evaluate the expression: W ≈ 6.93 N × 1 m ≈ 6.93 J (Joules). So, the work done by the rope after the block slides a distance of 1 m is approximately 6.93 Joules.
To find the work done by the rope after the block slides a distance of 1 m, we need to consider the horizontal component of the applied force and the distance the block travels. Here are the steps:
1. Identify the given information: The applied force (F) is 8 N, the angle (θ) is 30 degrees, and the distance (d) of the block slides is 1 m.
2. Calculate the horizontal component of the applied force: To find the horizontal component (F_horizontal), we'll use the formula F_horizontal = F × cos(θ). Since the force makes a 30-degree angle with respect to the horizontal, we have F_horizontal = 8 N × cos(30°).
3. Evaluate the expression: Using a calculator, we find that cos(30°) ≈ 0.866. Therefore, F_horizontal ≈ 8 N × 0.866 ≈ 6.93 N.
4. Calculate the work done (W) by the rope: To find the work done, we'll use the formula W = F_horizontal × d. Since F_horizontal ≈ 6.93 N and the block slides a distance of 1 m, we have W ≈ 6.93 N × 1 m.
5. Evaluate the expression: W ≈ 6.93 N × 1 m ≈ 6.93 J (Joules).
So, the work done by the rope after the block slides a distance of 1 m is approximately 6.93 Joules. Note that friction is not considered in this problem as the block slides without friction on a horizontal surface.
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If her mother's speed is 5.1 m/s when the ride is in motion, what is her angular momentum around the center of the merry-go-round
The angular momentum of the girl's mother around the center of the merry-go-round is 44.1 kg m^2/s.
The angular momentum of the girl's mother around the center of the merry-go-round can be calculated using the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since we are given the speed of the mother and not the angular velocity, we need to first convert the speed to angular velocity using the formula ω = v/r, where v is the linear speed and r is the radius of the circle.
Assuming the radius of the merry-go-round is 10 meters, we can calculate the angular velocity of the mother as:
ω = 5.1 m/s / 10 m
ω = 0.51 rad/s
Next, we need to calculate the moment of inertia of the mother. Assuming she has a mass of 60 kg and is standing with her arms outstretched, her moment of inertia can be calculated using the formula [tex]I = mr^2,[/tex] where m is the mass and r is the distance from the axis of rotation.
Assuming her arms are outstretched to a distance of 1.2 meters from the axis of rotation, we can calculate the moment of inertia as:
[tex]I = 60 kg * (1.2 m)^2[/tex][tex]L = 86.4 kg m^2 *0.51 rad/s[/tex]
[tex]L = 86.4 kg m^2 * 0.51 rad/s[/tex]
Finally, we can calculate the angular momentum of the mother as:
L = Iω
L = 86.4 kg m^2 * 0.51 rad/s
[tex]L = 44.1 kg m^2/s[/tex]
Therefore, the angular momentum of the girl's mother around the center of the merry-go-round is 44.1 kg m^2/s.
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Albert stands on a frictionless turntable, holding a bike wheel. Both Albert and the wheel are initially stationary. Albert gives the bike wheel a good spin, an
When Albert gives the bike wheel a good spin, an angular momentum is imparted to the system.
As per the law of conservation of angular momentum, the total angular momentum of the system must remain constant. Therefore, the turntable and Albert must also start rotating in the opposite direction of the bike wheel's rotation to conserve angular momentum. This is called the conservation of angular momentum. The rate of rotation of the turntable and Albert will depend on the mass and velocity of the bike wheel, as well as the mass and distance from the axis of rotation of the turntable and Albert.
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You are involved in a minor collision at an intersection. There are no injuries and very little vehicle damage. You should:
In a minor collision at an intersection with no injuries and minimal vehicle damage, you should first ensure the safety of all involved by moving your vehicles to a safe location, if possible.
Even if there are no injuries and the vehicle damage is minor, it is important to follow certain steps after a collision. The first step is to move your vehicle to a safe place off the road, if possible. Then, exchange information with the other driver, including names, phone numbers, insurance information, and vehicle registration numbers. You should also take pictures of the damage to both vehicles and the surrounding area.
If there were any witnesses, it is a good idea to get their contact information as well. Finally, report the accident to your insurance company as soon as possible. Remember, even minor collisions can have long-term effects, so it is important to take all necessary precautions and document everything.
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Suppose that the color and behavior of a star identify it as a type that we know has absolute magnitude 4.8. If the star's apparent magnitude is found to be 9.8, how far away is it
The star is 100 parsecs away, calculated using the distance modulus formula with absolute magnitude 4.8 and apparent magnitude 9.8.
To determine the distance to a star, astronomers use the distance modulus formula, which relates the star's absolute magnitude (M), apparent magnitude (m), and distance (d) in parsecs:
m - M = 5 * log10(d) - 5
In this case, the star has an absolute magnitude of 4.8 and an apparent magnitude of 9.8. Plugging these values into the formula, we get:
9.8 - 4.8 = 5 * log10(d) - 5
Solving for d, we find that the star is approximately 100 parsecs away.
This calculation assumes that the star's color and behavior accurately identify its absolute magnitude.
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In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross-section having a radius of 4.68 cm. This pressure is transmitted by a liquid to a second piston of radius 18.9 cm. What force must the compressed air exert in order to lift a car weighing 12600 N
The compressed air must exert a force of approximately 164.8 N to lift the car weighing 12600 N.
Area of the small piston = πr² = π(4.68 cm)² ≈ 68.85 cm²
Area of the large piston = πR² = π(18.9 cm)² ≈ 1123.90 cm²
1 = F2
PA1 = PA2
P = F2/A2
Now we can find the force required to lift the car:
F1 = PA1 = Pπr²
F1 = (F2/A2)πr²
F1 = (12600 N)/(1123.90 cm²)π(4.68 cm)²
F1 ≈ 164.8 N
Force is a physical quantity that describes the interaction between two objects. It is defined as the push or pull on an object that causes it to accelerate or deform. Force is measured in units of newtons (N) and is represented by the symbol F.
There are many different types of forces, including gravitational, electromagnetic, and nuclear forces. Each of these forces acts over a specific range and can have different strengths and effects on objects. According to Newton's laws of motion, an object will remain at rest or in uniform motion unless acted upon by a net force.
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hegg A 120-V rms voltage at 60 Hz is applied across a series combination of a 20-μF capacitor and an unknown resistor. If the rms value of the current in the circuit is 0.60 A, what is the resistance of the resistor?
The resistance of the unknown resistor is approximately 132 Ω.
The impedance of the series combination of the capacitor and the resistor is given by:
Z = √(R² + Xc²)
where R is the resistance of the resistor and Xc = 1/(2πfC) is the capacitive reactance of the capacitor.
Substituting the given values, we have:
Z = √(R² + (1/(2πfC))²) = Vrms/Irms = 120/0.6 = 200 Ω
Substituting the values of f and C, we get:
Z = √(R² + (1/(2π(60)(20 x 10⁻⁶)))²) = 200 Ω
Solving for R, we get:
R = √(Z² - Xc²) = √(200² - (1/(2π(60)(20 x 10⁻⁶)))²) = 132 Ω (approx)
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Write an expression for a transverse harmonic wave that has a wavelength of 2.1 m and propagates to the right with a speed of 14.1 m/s . The amplitude of the wave is 0.15 m , and its displacement at t
The expression for the transverse harmonic wave is:
y(x, t) = 0.15 * sin(2.9948x - 6.7143t)
The general equation for a transverse harmonic wave is given by:
y(x, t) = A * sin(kx - ωt + φ)
Where:
y(x, t) is the displacement of a point on the wave at position x and time t.
A is the amplitude of the wave.
k is the wave number, defined as 2π divided by the wavelength (k = 2π/λ).
x is the position along the wave.
ω is the angular frequency, defined as 2π times the frequency (ω = 2πf).
t is the time.
φ is the phase constant, representing the initial phase of the wave.
In this case, the given information is:
Wavelength (λ) = 2.1 m
Speed (v) = 14.1 m/s
Amplitude (A) = 0.15 m
To find the wave number (k) and angular frequency (ω), we can use the relationship between the speed, wavelength, and frequency:
v = f * λ
Rearranging the equation to solve for frequency (f):
f = v / λ
Substituting the given values:
f = 14.1 m/s / 2.1 m
f ≈ 6.7143 Hz
Now we can calculate the wave number:
k = 2π / λ
k = 2π / 2.1 m
k ≈ 2.9948 rad/m
Since the wave is propagating to the right, the phase constant φ is 0.
Putting all the values together, the expression for the harmonic wave is:
y(x, t) = 0.15 * sin(2.9948x - 6.7143t)
Note that the displacement at a specific time t is not mentioned in the question. To determine the displacement at a specific time, substitute the desired value of t into the equation.
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Calculate the average speed of blood flow in the major arteries of the body, which have a total cross-sectional area of about 2.2 cm2 . Express your answer to two significant figures and include the appropriate units.
The average speed of blood flow in major arteries is approximately 25 cm/s.
The total cross-sectional area of major arteries in the body is approximately 2.2 cm2.
Using the equation Q = Av, where Q is the volume of blood flow, A is the cross-sectional area, and v is the velocity, we can calculate the average speed of blood flow.
Assuming a cardiac output of 5 L/min, we can calculate the volume of blood flow to be 83.3 ml/s.
Dividing this by the cross-sectional area of 2.2 cm2 gives us a velocity of approximately 38 cm/s.
However, this is the velocity at the center of the artery, and the velocity at the walls is slower due to friction.
The average speed of blood flow in major arteries is therefore estimated to be around 25 cm/s, with appropriate units being cm/s.
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