Ultraviolet is correct answer .UV radiation is the type of electromagnetic energy that ozone would block in the stratosphere.
The ozone layer in the stratosphere blocks ultraviolet (UV) radiation, which is a form of electromagnetic radiation. This is important because UV radiation can be harmful to living organisms, including humans, causing skin damage and increasing the risk of skin cancer.
The majority of the sun's dangerous ultraviolet (UV) radiation, especially UV-B rays, are absorbed by ozone in the stratosphere, which serves as a natural filter. In addition to causing skin cancer, cataracts, and other health issues, UV radiation can be dangerous to living things. The majority of UV-B radiation falls within the 200–300 nanometer (nm) band of UV light that is absorbed by ozone. This is one of the reasons the ozone layer is crucial for life on our planet because it shields it from the damaging effects of UV radiation. Ozone in the stratosphere does not appreciably absorb other types of electromagnetic radiation, such as cosmic rays, infrared, microwaves, and visible light.
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a hollow sphere has mass , radius , and moment of inertia about an axis through its center of mass. it is initially rolling on a horizontal plane with center of mass velocity and angular velocity . for all of the questions on this page you should assume that there is no slipping.
A hollow sphere has a moment of inertia I = (2/3)MR^2, where M is the mass, and R is the radius.
In the given scenario, the hollow sphere is initially rolling without slipping on a horizontal plane, with center of mass velocity (v) and angular velocity (ω).
The relationship between linear velocity and angular velocity can be described as v = Rω, due to the no-slip condition. Since the moment of inertia of a hollow sphere is given by the formula I = (2/3)MR^2, we can use this information to further analyze the sphere's motion.
Summary: For a hollow sphere rolling without slipping on a horizontal plane, its moment of inertia is I = (2/3)MR^2, and its linear and angular velocities are related by the equation v = Rω.
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A 750-Hz plane sound wave in seawater (13 degree C) is normally incident on a stone seawall. The sound field is found to have a maximum at the wall; the SPL there is 190 dB re 1 muPa. The SPL 0.5 m from the wall is 177 dB. (a) Use the information given about the sound field to find the impedance of the wall. (b) Find the SPL of the incident wave.
(a) The impedance of the seawall is approximately 2.13 MRayl.
(b) The SPL of the incident wave is approximately 147 dB re 1 μPa.
How to measure sound pressure levels?(a) The impedance of the seawall can be found using the measured sound pressure levels at the wall and at a distance from the wall, as follows:
First, we can use the sound pressure level (SPL) at 0.5 m from the wall to calculate the sound pressure amplitude (p0) of the incident wave:
SPL0.5m = 177 dB
p0 = 20 μPa (for a reference pressure of 1 μPa)
Next, we can use the maximum SPL at the wall to find the sound pressure amplitude at the wall (p_wall):
SPLwall = 190 dB
p_wall = 200 μPa
The impedance of the seawall (Z_wall) can then be calculated using the formula:
Z_wall = p_wall / v + vρ
where v is the speed of sound in seawater at 13°C (approximately 1532 m/s), and ρ is the density of seawater at 13°C (approximately 1025 kg/m³):
Z_wall = (200 μPa) / (1532 m/s) + (1532 m/s)(1025 kg/m³) = 2.13 MRayl (megarayls)
Therefore, the impedance of the seawall is approximately 2.13 MRayl.
(b) To find the SPL of the incident wave, we can use the formula for sound intensity (I):
I = p^2 / (2ρv)
where p is the sound pressure amplitude, ρ is the density of seawater at 13°C, and v is the speed of sound in seawater at 13°C. Solving for p:
p = sqrt(2ρvI)
We can find the sound intensity (I) using the measured SPL at 0.5 m from the wall, since the incident wave should have the same intensity everywhere in the absence of any reflections or absorption:
SPL0.5m = 177 dB
I = 10^((SPL0.5m - 120) / 10) = 2.51 x 10^-4 W/m²
Using the formula for p above:
p = sqrt(2(1025 kg/m³)(1532 m/s)(2.51 x 10^-4 W/m²)) = 0.00785 Pa
Finally, we can convert the sound pressure amplitude to an SPL using the formula:
SPL = 20 log10(p / p_ref)
where p_ref is the reference sound pressure (1 μPa):
SPL = 20 log10(0.00785 Pa / 1 μPa) = 147 dB re 1 μPa
Therefore, the SPL of the incident wave is approximately 147 dB re 1 μPa.
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A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen.
a) half of the image will disappear
b) no part of image will disappear .
c) Intensity of the image will increased
d. Intensity of the image will decrease
When the upper half of a converging lens is covered by an opaque screen, only the lower half of the image will be visible on the screen.
This is because a converging lens refracts light rays that pass through it, causing them to converge at a focal point.
When the upper half of the lens is covered, only the lower half of the light rays can pass through the lens and converge at the focal point. As a result, only the lower half of the image will be visible on the screen.
The intensity of the image will not be affected by covering the upper half of the lens. The intensity of the image is determined by the amount of light that reaches the screen, which is not affected by covering part of the lens.
However, the overall brightness of the image may be affected if the amount of light entering the lens is reduced by the opaque screen.
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in a wire with a 2.11 mm2 cross-sectional area, 6.09×10^20 electrons flow past any point during 2.45 s. what is the current in the wire?
The current in the wire is calculated by dividing the total charge passing through any point (9.77×10^1 C) by the time taken for it to pass (2.45 s), resulting in a current of 3.74 A.
The formula for current (I) is I = Q/t, where Q is the charge passing through a point in the wire and t is the time taken for the charge to pass through. In this case, we need to calculate the charge (Q) first. The number of electrons passing through any point in the wire is given as 6.09×10^20. Since each electron has a charge of 1.602×10^-19 C, the total charge passing through the point in the wire is:
Q = 6.09×10^20 × 1.602×10^-19 = 9.77×10^1 C
Now, we can use the formula for current:
I = Q/t = 9.77×10^1 C / 2.45 s = 3.74 A
Therefore, the current in the wire is 3.74 A.
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Suppose we have a simple AC circuit with just an inductor in it, and we want a 22 kΩ reactance at a frequency of 510 Hz.
What value of inductance should be used in H?
The inductance should be approximately 0.068 H (68 mH) to achieve a reactance of 22 kΩ at a frequency of 510 Hz.
The reactance (X) of an inductor in an AC circuit is given by the formula:
X = 2πfL
where f is the frequency in Hz, L is the inductance in Henrys, and π is the constant pi (approximately 3.14159).
In this case, we want a reactance of 22 kΩ at a frequency of 510 Hz. So we can rearrange the formula to solve for L:
L = X / (2πf)
L = (22,000 Ω) / (2π × 510 Hz)
L ≈ 0.068 H
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An electron travels with speed 1.3�107m/s between the two parallel charged plates shown in the figure(Figure 1) . The plates are separated by 1.0 cm and are charged by a 200 V battery.
What magnetic field strength will allow the electron to pass between the plates without being deflected?
What is the direction of the magnetic field?
The required magnetic field strength is 1.54×10^-3 T and it must be perpendicular to the plane of the paper and into the paper.
When a charged particle moves through a magnetic field, it experiences a force given by the equation F = qvBsinθ, where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this problem, the electron is moving horizontally between the plates, and we want it to pass through the plates without being deflected. This means that the force on the electron due to the electric field between the plates must be equal and opposite to the force on the electron due to the magnetic field.
The force on the electron due to the electric field is given by F = Eq, where E is the electric field strength and q is the charge of the electron. The electric field strength is given by E = V/d, where V is the voltage of the battery and d is the distance between the plates. Substituting the values given in the problem, we get E = 200 V/0.01 m = 2.0×10^4 V/m.
The force on the electron due to the magnetic field is given by F = qvB, where q is the charge of the electron, v is the velocity of the electron, and B is the magnetic field strength. Substituting the values given in the problem, we get F = 1.6×10^-19 C × 1.3×10^7 m/s × B = 2.08×10^-12 B N.
Since the electron is not deflected by the magnetic field, we know that the magnetic force is equal and opposite to the electric force. Therefore, we can equate the two forces and solve for B:
F = Eq
1.6×10^-19 C × 1.3×10^7 m/s × B = 2.0×10^4 V/m × 1.6×10^-19 C
B = (2.0×10^4 V/m × 1.6×10^-19 C)/(1.3×10^7 m/s)
B ≈ 1.54×10^-3 T
The direction of the magnetic field can be found using the right-hand rule. If you point your right thumb in the direction of the electron's velocity (to the right in this case), and your fingers in the direction of the electric field (from positive to negative), then the magnetic field must be directed into the paper (towards you) in order to produce a force that opposes the electric force and allows the electron to pass through the plates without being deflected.
The required magnetic field strength is 1.54×10^-3 T, and it must be perpendicular to the plane of the paper and into the paper.
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Fred (mass 65 Kg) is running with the football at a speed of 5.7 m/s when he is met head-on by Brutus (mass 120 kg), who is moving at 4.3m/s. Brutus grabs Fred in a tight grip, and they fall to the ground. How far do they slide? The coefficient of kinetic friction between football uniforms and Astroturf is 0.30.
Fred (mass 65 Kg) is running with the football at a speed of 5.7 m/s when he is met head-on by Brutus (mass 120 kg), who is moving at 4.3m/s. They will slide about 5.30 meters.
To solve this problem, we need to use conservation of momentum to find the velocity of the combined system after the collision. Then, we can use the coefficient of kinetic friction and the laws of motion to find how far they will slide.
Let's first find the velocity of the combined system using conservation of momentum
m1v1 + m2v2 = (m1 + m2)v
Where m1 and v1 are the mass and velocity of Fred, m2 and v2 are the mass and velocity of Brutus, and v is the velocity of the combined system after the collision.
Substituting the given values, we get
(65 kg)(5.7 m/s) + (120 kg)(4.3 m/s) = (65 kg + 120 kg)v
v = [(65 kg)(5.7 m/s) + (120 kg)(4.3 m/s)] / (65 kg + 120 kg)
v = 4.58 m/s
So the velocity of the combined system after the collision is 4.58 m/s.
Now, let's use the coefficient of kinetic friction and the laws of motion to find how far they will slide. We can use the equation
Ffriction = μk * Fnorm
Where Ffriction is the force of friction, μk is the coefficient of kinetic friction, and Fnorm is the normal force, which is equal to the weight of the combined system
Fnorm = (m1 + m2)g
Where g is the acceleration due to gravity, which is approximately 9.81 m/[tex]s^{2}[/tex].
Substituting the given values, we get
Fnorm = (65 kg + 120 kg)(9.81 m/[tex]s^{2}[/tex]) = 1764.45 N
Ffriction = μk * Fnorm = 0.30 * 1764.45 N = 529.34 N
The force of friction is acting in the opposite direction of the motion, so we can use the equation
Ffriction = mtotal * a
Where m_total is the total mass of the system and a is the acceleration of the system.
Substituting the given values, we get
529.34 N = (65 kg + 120 kg) * a
a = 1.97 m/[tex]s^{2}[/tex]
Now, we can use the equation
d = [tex]v^{2}[/tex] / (2 * a)
Where d is the distance they slide.
Substituting the given values, we get
d = [tex](4.58m)^{2}[/tex] / (2 * 1.97 m/[tex]s^{2}[/tex]) = 5.30 m
So they will slide about 5.30 meters.
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estimate the total mass of the earth's atmosphere? (the radius of the earth is 6.37 106 m, and atmospheric pressure at the surface is 1.01 105 n/m2.)
The estimated total mass of the Earth's atmosphere is approximately 5.14 x 10¹⁸ kg, based on the given information about the Earth's radius and surface atmospheric pressure.
To estimate the total mass of the Earth's atmosphere, we need to use the following formula
mass = density x volume
We know that the atmospheric pressure at the surface of the Earth is 1.01 x 10⁵ N/m². We can use this value to estimate the density of air using the ideal gas law
PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature.
Assuming the temperature is 293 K (20°C), we can calculate the number of moles of air per unit volume
n/V = P/RT = (1.01 x 10⁵ N/m²) / (8.31 J/K/mol x 293 K) = 1.293 kg/m³
This is the density of air at sea level.
Now we need to calculate the volume of the Earth's atmosphere. To do this, we can assume that the atmosphere extends up to a height of about 100 km. The radius of the Earth is 6.37 x 10⁶ m, so the volume of the atmosphere is
V = 4/3 x pi x (6.47 x 10⁶)³ - 4/3 x pi x (6.37 x 10⁶)³ = 1.09 x 10^21 m^3
Therefore, the total mass of the Earth's atmosphere is
mass = density x volume = 1.293 kg/m³ x 1.09 x 10²¹ m³ = 5.14 x 10¹⁸ kg
So the estimated total mass of the Earth's atmosphere is about 5.14 x 10¹⁸ kg.
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the mean time period of a simple pendulum is 1.92 s. mean absolute error in the time period is 0.05 s. to express the maximum estimate of error, the time period should be written
To express the maximum estimate of the error in the time period of a simple pendulum, we can use the formula:
Maximum Estimate of Error = Mean Absolute Error + (Standard Deviation / √n)
Here, we are given that the mean time period of the simple pendulum is 1.92 s and the mean absolute error is 0.05 s. However, we do not have information about the standard deviation or the sample size (n).
Without knowing the standard deviation and sample size, we cannot accurately calculate the maximum estimate of error. The standard deviation indicates how much variation there is in the data, while the sample size indicates how representative the data is of the entire population.
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When novices close their eyes when trying a somersault, _____.
a. their performance typically improves
b. there is no impact on their performance
c. their performance is not degraded as much as is the performance of experts
d. their performance seems to change randomly
Closed eyes improve performance as they reduce distractions and focus on the task at hand, but may decrease performance due to reliance on visual cues and muscle memory.
Novices may find it easier to concentrate on their body's motion and internal feelings when their eyes are closed. Additionally, it can lessen anxiety and distracting visuals, which will make the performer feel more at ease and assured when performing the exercise. Experts, however, rely significantly on visual feedback to make corrections and keep their balance in midair, thus doing so while performing a somersault could be detrimental to their performance.
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Ayren purchased a 1000W (120V) hair dryer in the USA, and she took it to Europe where the standard voltage is 220V. Assuming the hair dryer can operate safely at the higher voltage, can you actually use it if the outlet is rated at 15A, or will it draw too much current (How much current)?
Yes, the hair dryer can be used in Europe if the outlet is rated for 15A. The difference in voltage is accounted for by a reduction in the current draw.
At 220V, the hair dryer will draw just 4.5A to provide the same 1000W of power, which is easily within the 15A rating of the outlet. To calculate the current draw, we use Ohm's Law (I=P/V). In this case, I is the current draw, P is the power (1000W), and V is the voltage (220V).
By solving the equation, we get I=4.5A. This means that the hair dryer will not draw too much current and can be used in Europe with no safety issues.
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the reversing valve on most air-source residential heat pumps is energized:
The reversing valve on most air-source residential heat pumps is energized in order to change the direction of refrigerant flow.
The reversing valve is a key component of the heat pump system, as it allows the unit to switch between heating and cooling modes. When the valve is energized, it redirects the flow of refrigerant so that it can either absorb heat from the air outside and transfer it inside (during heating mode), or absorb heat from inside and transfer it outside (during cooling mode).
The reversing valve is typically controlled by the thermostat in the home, which signals the heat pump to switch modes as needed. When the thermostat calls for heat, the heat pump will energize the reversing valve to switch from cooling mode to heating mode, and vice versa when cooling is required.
It's important to note that not all heat pumps have reversing valves. Some units are designed to only provide either heating or cooling, and do not have the ability to switch modes. However, for those that do, the reversing valve is a crucial component that allows for efficient and effective temperature control in residential buildings.
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the maximum clock frequency, given a critical path of 10 ns, is f = _____ mhz ?
The maximum clock frequency, given a critical path of 10 ns, is f = 100,000 MHz.
To calculate the maximum clock frequency (f) given a critical path of 10 ns, we need to invert the critical path time to find the frequency. The formula to calculate frequency is:
f = 1 / t
where f is the frequency and t is the time period.
In this case, the time period (t) is given as 10 ns. However, to convert the time period to seconds, we need to divide it by 1 billion (1 ns = 1/1,000,000,000 seconds):
t = 10 ns / 1,000,000,000 seconds
Now, we can calculate the frequency:
f = 1 / (10 ns / 1,000,000,000 seconds)
Simplifying the equation:
f = 1,000,000,000 seconds / 10 ns
f = 100,000,000,000 Hz
To convert the frequency to megahertz (MHz), we divide it by 1 million:
f = 100,000,000,000 Hz / 1,000,000
f = 100,000 MHz
Therefore, the maximum clock frequency is 100,000 MHz (or 100 GHz).
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A block initially at rest on a horizontal, frictionless surface is accelerated by a constant horizontal force of 5.0 newtons. If 15 joules of work is done on the block by this force while accelerating it, the kinetic energy of the block increases by a. 3.0 J b. 15 J c. 20 J d. 75 J
The correct answer is b. 15 J.
The work done on the block is equal to the change in kinetic energy of the block, as given by the work-energy principle:
Work = Change in kinetic energy
15 J = 1/2 * m * v^2 - 0
where m is the mass of the block and v is its final velocity.
Since the block starts from rest, its initial kinetic energy is zero. Therefore, the final kinetic energy of the block is equal to the work done on it by the force:
Final kinetic energy = Work = 15 J
So the correct answer is b. 15 J.
The kinetic energy of the block increases by 15 J (option b). This is because the work done on the block (15 J) is equal to the change in its kinetic energy, as stated by the work-energy theorem. Since the block is initially at rest and the force applied is constant, the entire amount of work done will be converted into kinetic energy.
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what is the gravitational force between two identical bodies with a mass of 2.0x104 kg each if they are 2.0 m apart?
The gravitational force between two identical bodies with a mass of 2.0x104 kg each, if they are 2.0 m apart is given by Newton's Law of Gravitation.
According to this law, the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Therefore, the force of gravity between the two bodies is given by: F=G*((m1*m2)/r^2). Here, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.
Therefore, the force of gravity between the two bodies is given by: F=6.673*((2.0x104*2.0x104)/2.0^2). This gives a force of gravity of 1.337x106 N.
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