The electrical charge on an atom is zero due to presence of same number of protons and electrons.
An atom is the smallest entity comprising of three components. These are protons, neutrons and electrons. Protons and neutrons are centrally located forming the nucleus while electrons revolve around the nucleus. Protons are positively charged while electrons are negatively charged. Neutrons are neutral due to lack of charge.
The number of protons and electrons are same in an atom owing to balancing the overall charge in an atom. This makes the atom electrical neutral and hence the charge on an atom is zero.
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Consider two plates, uniformly charged, one having a positive charge and the other having a negative charge. These two plates are parallel to each other and are a distance d from each other. Each plate has an electric field of E=2ϵ0σ a) Draw a well-labeled diagram of these two plates horizontally with the top plate having a positive charge and the bottom plate having a negative charge. Also, draw the electric field lines representing the electric field (E1) above the top plate, the electric field (E2) in the middle of the two plates, and the electric field (E3) below the bottom plate. b) With the aid of your drawing in question (a) explain where the electric field will be zero? c) Determine the electric field where it is not zero?
The electric field between two parallel plates with uniform charges will be zero in the region between the plates. Above and below the plates, the electric field will have a non-zero value.
a) Here is a well-labeled diagram of the two charged plates:
+Q
------------------------------------ Top Plate
↑ E1
-------------------|---------------- E2
↓ E3
------------------------------------ Bottom Plate
-Q
In the diagram, the top plate has a positive charge (+Q), and the bottom plate has a negative charge (-Q). The distance between the plates is labeled as 'd'.
Above the top plate, the electric field is represented by E1. In the middle of the two plates, the electric field is represented by E2. Below the bottom plate, the electric field is represented by E3.
b) The electric field will be zero at the points where the electric field lines from the positive plate (E1) and the negative plate (E3) cancel each other out. These points are equidistant between the plates and lie on a line perpendicular to the plates.
Therefore, the electric field will be zero in the region between the plates, precisely in the middle (where E2 is).
c) The electric field where it is not zero is above the top plate (E1) and below the bottom plate (E3). These electric fields have the same magnitude and point away from their respective plates.
The electric field E1 points away from the positive plate, and E3 points away from the negative plate. The magnitude of these electric fields is given by E = 2ε0σ, where ε0 is the permittivity of free space and σ is the surface charge density of the plates.
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An experiment is performed using a spring to determine its
constant. The mass is 0.0300kg and the period is 0.241s, what is
T^2?
T^2 is 0.05808099. The period of a spring-mass system is given by: T = 2*pi*sqrt(m/k). k is the spring constant.
The period of a spring-mass system is given by:
T = 2*pi*sqrt(m/k)
where:
m is the mass of the object
k is the spring constant
In this case, the mass is 0.0300 kg and the period is 0.241 s, so:
T^2 = (0.241 s)^2 = 0.05808099
Therefore, T^2 is 0.05808099.
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______ is (are) partially blocked by ozone in the stratosphere.
a. Ultraviolet radiation
b. Microwaves
c. X-rays
Ultraviolet radiation is partially blocked by ozone in the stratosphere.
Hence, the correct option is A.
Ultraviolet (UV) radiation is partially blocked by ozone in the stratosphere. The ozone layer in the Earth's stratosphere acts as a protective shield by absorbing much of the incoming UV radiation from the Sun. This absorption process helps to prevent a significant amount of harmful UV radiation from reaching the Earth's surface.
Ozone molecules are particularly effective at absorbing UV-B (shortwave) and UV-C (even shorter wavelength) radiation. The absorption of UV radiation by ozone in the stratosphere plays a crucial role in protecting living organisms from the damaging effects of excessive UV exposure, which can lead to sunburn, skin cancer, and other harmful health effects.
Therefore, Ultraviolet radiation is partially blocked by ozone in the stratosphere.
Hence, the correct option is A.
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The potential difference between two spherical shells, one with charge Q and outer radius r and the other with charge −Q and inner radius 2r, have a potential difference Δϕ=
4πϵ
0
Q
(
r
1
−
2r
1
) If the capacitance of this system is C
0
, what is the capacitance of this system if the charges are the same but the inner radius of the larger shell goes from 2r to 4r ? A. 1/2C
0
B. 2/3C
0
C. 3/4C
0
D. 4/5C
0
E. C
0
The capacitance of the system, if the inner radius of the larger shell goes from 2r to 4r, is C = 1/(2πϵ0). Potential difference between two spherical shells, one with charge Q and outer radius r and the other with charge −Q and inner radius 2r.Δϕ = 4πϵ0Q (1/r − 1/2r)Capacitance of this system is C0.
Now, we need to find the capacitance of the system if the charges are the same but the inner radius of the larger shell goes from 2r to 4r.
Capacitance of the system is given by,C = Q/Δϕ.
Put the value of Δϕ in above formula,C = Q/ (4πϵ0Q (1/r − 1/2r))C = 1/(4πϵ0) (1/(1/r − 1/2r))C = 1/(4πϵ0) (2r/(2r − r))C = 1/(4πϵ0) (2r/r)C = 1/(2πϵ0).
Now, capacitance of the system if the inner radius of the larger shell goes from 2r to 4r is C = 1/(2πϵ0).
Hence, option E is the correct answer.
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Name the three methods that we can use to determine the amount
of dark matter in the universe.
There are several methods used to estimate the amount of dark matter in the universe are Galaxy Rotation Curves, Gravitational Lensing, Cosmic Microwave Background (CMB) Anisotropies.
There are several methods used to estimate the amount of dark matter in the universe. Here are three commonly employed methods:
1. Galaxy Rotation Curves: This method involves studying the rotation curves of galaxies. By measuring the speeds at which stars or gas clouds orbit within a galaxy, scientists can infer the distribution of mass within the galaxy. If the observed rotation speeds cannot be explained by the visible matter alone, it suggests the presence of additional mass in the form of dark matter.
2. Gravitational Lensing: Gravitational lensing occurs when the gravitational field of a massive object, such as a galaxy cluster, bends the path of light from more distant objects behind it. By observing the distortion of light caused by gravitational lensing, astronomers can deduce the distribution of mass, including dark matter, within the lensing object. This method provides indirect evidence of dark matter by revealing its gravitational effects on visible light.
3. Cosmic Microwave Background (CMB) Anisotropies: The cosmic microwave background is the radiation left over from the early universe. Tiny temperature fluctuations, known as anisotropies, in the CMB can be analyzed to provide insights into the composition of the universe. By studying the patterns of these anisotropies, scientists can estimate the total amount of matter in the universe, including both visible matter and dark matter.
These methods, along with other observational and theoretical approaches, help researchers refine their understanding of the amount and distribution of dark matter in the universe.
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A block of mass m is initially at rest at the origin x = 0. A one-dimension force given by F = Fo e-x, where Fo & λ are positive constants, is appl block. a. What are the units of Fo & λ? (2pts) b. Argue that the force is conservative. (1pt) c. Find the potential energy associated with the force. (2pts) Find the total energy of the block. (1pt) d. e. f. Find the velocity of the block as a function of position x. (4pts) What is the terminal speed of the block as x→ [infinity]o? Justify the valu (2pts)
a. Units of Fo = Newton (N).Units of λ = Inverse of distance, for example, 1/m.
b. A force is said to be conservative if the work done by the force to move an object from point A to point B depends on the initial and final position of the object and not on the path it follows. A force is also said to be conservative if its work done is path-independent. The given force F = Fo e^(-x) is conservative because it is derived from the potential energy, and its work done depends only on initial and final positions, and not on the path followed.
c. The potential energy associated with the force F is given by - Fo e^(-x) + C, where C is an arbitrary constant. Because the force is conservative, it is derived from a potential function, which is the opposite of the potential energy. Therefore, the potential function is U(x) = Fo e^(-x) + C. Total energy E of the block is the sum of kinetic energy and potential energy. E = 1/2 mv^2 + U(x)
d. Work done by the force to move the block from position x1 to x2 is given by W(x1 to x2) = U(x1) - U(x2). By the work-energy theorem, the work done is equal to the change in kinetic energy. Therefore, 1/2 mv^2 = Fo (e^(-x1) - e^(-x2)), where m is the mass of the block. Using this equation, we can find the velocity v of the block as a function of position x. v(x) = {2Fo/m} (e^(-x) - e^(-x2))^(1/2)e. As x → ∞, v(x) → 0. Therefore, the terminal speed of the block as x → ∞ is 0. It means that the block will stop moving as it approaches infinity.f. Terminal speed is the maximum speed attained by the object when the force of resistance is equal and opposite to the applied force. In this case, there is no force of resistance, and hence, the terminal speed of the block as x → ∞ is 0.
About Potential energyPotential energy is energy that affects objects because of the position of the object, which tends to go to infinity with the direction of the force generated from the potential energy. The SI unit for measuring work and energy is the Joule. What are the benefits of potential energy? Potential energy is energy that is widely used to generate electricity. Even so, there are two objects that are used to store potential energy. These objects, namely water and fuel, are used to store potential energy. In general, water can store potential energy like a waterfall.
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A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 6.97 g coins stacked over the 29.2 cm mark, the stick is found to balance at the 47.1 cm mark. What is the mass of the meter stick?
The principle of moments is a fundamental concept in physics, that refers to the statement. For an object to be in rotational equilibrium, the sum of the moments acting on that object must be zero.
”Let's find out the mass of the meter stick:
Let the mass of the meter stick be m1 grams and its center of gravity be at a distance of x from the left end.
Since the stick balances horizontally on a knife edge at the 50 cm mark, the distance of its center of gravity from the left end is 50 cm.
M1 × 50 = 2 × 6.97 × (50 - 29.2) + m2 × (50 - 47.1)
Where M1 = mass of the meter stick,
M2 = mass of coins stacked over 29.2 cm markm1 = (2 × 6.97 × (50 - 29.2) + m2 × (50 - 47.1))/50
Since M2 = 2 × 6.97 g and the stick balances at the 47.1 cm mark,
Distance of center of gravity of meter stick from left end = 47.1 cm
Thus, m1 = (2 × 6.97 × (50 - 29.2) + 2 × 6.97 × (50 - 47.1))/50= (2 × 6.97 × 20.8 + 2 × 6.97 × 2.9)/50= (2 × 6.97 × 23.7)/50= 3.1 g
Therefore, the mass of the meter stick is 3.1 grams .A solution is a process of balancing the moments that will be helpful for students to know.
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6. A wheel spins counterclockwise through three revolutions for 2 seconds. What is the average angular velocity of the wheel? 7. The fan blades of a jet engine in an airplane rotate counterclockwise with an initial angular velocity of 100rad/s. As the airplane takes off, the angular velocity of the blades reaches 400rad/s in 10 seconds. Calculate the average angular acceleration. 8. A new car takes 10 seconds to accelerate from rest to 30 m/s. Its mass is 1500 kg. What is the net average force that acts on the car? 9. A 2 kg ball, moving to the right at a velocity of 2 m/s on a frictionless table, has an elastic head-on collision with a stationary 5 kg ball. What is the total kinetic energy before the collision? What is the total kinetic energy after the collision? 10. Starting from rest, Amy and Jane push off against each other on the smooth frictionless ice rink. The mass of Amy is 50 kg and that of Jane is 60 kg. Amy moves to the right (positive direction) with a velocity of 3 m/s. What is the recoil velocity of Jane?
The average angular velocity of the wheel is 3π rad/s. The average angular acceleration is 30 rad/s². The net average force that acts on the car is 4500 N. The total kinetic energy before the collision is 4 J. The total kinetic energy after the collision is 10 J.The recoil velocity of Jane is 15 m/s.
6. The average angular velocity can be calculated by dividing the total angle rotated by the time it took to rotate that angle.A wheel spins counterclockwise through three revolutions, so it rotates 3 × 2π = 6π radians.
The time it took to do this is 2 seconds. Average angular velocity (ωav) = θ ÷ tωav = 6π ÷ 2ωav = 3π rad/s
7. The formula for average angular acceleration is given byω = ω0 + αt where ω0 is the initial angular velocity, ω is the final angular velocity, t is the time interval, and α is the angular acceleration.
The initial angular velocity is 100 rad/s.The final angular velocity is 400 rad/s.The time interval is 10 s.
The average angular acceleration is:αav = (ω - ω0) ÷ tαav = (400 - 100) ÷ 10αav = 30 rad/s²
8. Force = Mass × AccelerationNet Average Force = Change in Momentum ÷ Time taken to change momentumInitial Velocity (u) = 0m/s Final Velocity (v) = 30 m/s, Time taken (t) = 10 s, Mass (m) = 1500 kg.
Using the formula,v = u + at30 m/s = 0 + a × 10sa = 3 m/s².
Using the formula,Net Average Force = Change in Momentum ÷ Time taken to change momentum Change in momentum = Mass × (Final Velocity - Initial Velocity) Change in momentum = 1500 × (30 - 0) Change in momentum = 45000 Ns.
Net Average Force = 45000 ÷ 10Net Average Force = 4500 N
9. Kinetic energy (KE) can be calculated using the formula, KE = ½mv².
KE of the 2 kg ball before the collision:Initial Velocity (u) = 2 m/sMass (m) = 2 kg.
Using the formula,KE = ½mv²KE = ½ × 2 × 2²KE = 4 JKE of the 5 kg ball before the collision:Mass (m) = 5 kg.
Using the formula,KE = ½mv²KE = ½ × 5 × 0²KE = 0 J.
Total Kinetic Energy before the collision = KE of the 2 kg ball + KE of the 5 kg ball.
Total Kinetic Energy before the collision = 4 J + 0 J.
Total Kinetic Energy before the collision = 4 JKE of the 2 kg ball after the collision:
Using the principle of conservation of energy, the total kinetic energy after the collision is equal to the total kinetic energy before the collision.
Initially, only the 2 kg ball had kinetic energy, so the total kinetic energy after the collision will be equal to the kinetic energy of the 2 kg ball.
KE = ½mv²KE = ½ × 2 × 2²KE = 4 JKE of the 5 kg ball after the collision:
Since the 5 kg ball was stationary before the collision, it will gain some of the kinetic energy of the 2 kg ball after the collision.
Using the principle of conservation of momentum,m1v1i + m2v2i = m1v1f + m2v2f0 + 2 × 2 = 2v1f + 5v2fv1f + v2f = 0
Since the collision was elastic, the relative velocity of the balls will remain the same after the collision.
Therefore, the velocity of the 2 kg ball after the collision is 0 m/s, since it hit the stationary 5 kg ball and stuck to it.
Using the formula,KE = ½mv²KE = ½ × 5 × 2²KE = 10 J.
Total Kinetic Energy after the collision = KE of the 2 kg ball + KE of the 5 kg ballTotal Kinetic Energy after the collision = 0 J + 10 JTotal Kinetic Energy after the collision = 10 J
10. Momentum is conserved in this scenario.
Using the principle of conservation of momentum,m1v1i + m2v2i = m1v1f + m2v2fAmy moves to the right (positive direction) with a velocity of 3 m/s.
Since the ice rink is frictionless, there are no external forces acting on the system.
Therefore, momentum is conserved.The initial momentum of the system is:Initial Momentum (p) = Mass of Amy × Velocity of Amy + Mass of Jane × Velocity of JaneInitial Momentum (p) = 50 × 3 + (-60) × 0 Initial Momentum (p) = 150 kg m/s.
The final momentum of the system is:Final Momentum (p) = Mass of Amy × Velocity of Amy + Mass of Jane × Velocity of Jane + Mass of Jane × Velocity of Jane (after the collision)Final Momentum (p) = 50 × v + (-60) × v + (-60) × (-v)Final Momentum (p) = -10v kg m/s.
Since momentum is conserved,Initial Momentum = Final Momentum 150 = -10vv = -15 m/s.
Since Jane moves to the left (negative direction), her velocity is -15 m/s.
Therefore, the recoil velocity of Jane is 15 m/s.
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the brakes are applied, skopping the vehdel in a uniform manner in an addaionai 5.90 s. (a) Haw iang is the three-wheeled car in motian (in s)? (b) What is the averege velocty of the theee-wheeied car for the metian describedi (Enter the magnituste in mis.)
The three-wheeled car comes to a stop in 5.90 seconds. Its average velocity during this time is X m/s.
To determine the answer, we need to consider the given information. The brakes are applied, causing the car to skid uniformly for an additional 5.90 seconds.
In this case, the car is experiencing uniform deceleration as it comes to a stop. The time taken for the car to stop, as given, is 5.90 seconds. This time can be considered as the total time for the car's motion.
To calculate the average velocity, we need to determine the magnitude of the displacement of the car during this time. Since the car comes to a stop, its displacement is equal to zero. Therefore, the average velocity during this time period is also zero.
Hence, the main answer is that the three-wheeled car comes to a stop in 5.90 seconds, and its average velocity during this time is zero m/s.
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What season is this in the northern hemisphere?
The season in the northern hemisphere depends on the current date. Without the specific date, it is not possible to determine the exact season.
The northern hemisphere experiences four distinct seasons: spring, summer, autumn (fall), and winter. The season depends on the tilt of the Earth's axis and its position in orbit around the Sun. However, the specific season at any given time varies throughout the year.
As the Earth orbits the Sun, the tilt of the northern hemisphere determines the amount of sunlight received. During summer, the northern hemisphere is tilted towards the Sun, resulting in longer days and warmer temperatures. Spring and autumn occur during the transitional periods as the tilt gradually changes.
Without knowing the current date, it is not possible to determine the exact season in the northern hemisphere. The transition between seasons occurs gradually, and the date determines the tilt of the Earth and its position in orbit. To determine the season, one must refer to the current date and consider the specific time of the year.
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the moon must be on the ________ for a solar eclipse to occur.
The moon must be on the line between the Earth and the Sun for a solar eclipse to occur.
This line is called the ecliptic and it is the path that the Sun, Moon, and planets follow across the sky. When the Moon passes in front of the Sun along this line, it blocks out the light of the Sun and creates a solar eclipse. Solar eclipses can only occur during a new moon when the Moon is between the Sun and the Earth. There are different types of solar eclipses such as total solar eclipses, partial solar eclipses, and annular solar eclipses.
Total solar eclipses occur when the Moon completely covers the disk of the Sun, while partial solar eclipses occur when only a portion of the Sun is covered by the Moon. Annular solar eclipses occur when the Moon is too far away from the Earth to completely cover the Sun, creating a ring of fire effect. So therefore the moon must be on the line between the Earth and the Sun for a solar eclipse to occur.
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A helim-neon laser beam has a wavelength in air of 633 nm. It takes 1.42 ns for the light to travel through 26.0 cm of an unknown liquid. What is the wavelength of the laser beam in the liquid? Express your answer with the appropriate units.
The wavelength of the laser beam in the unknown liquid is 474 nm.
Determine the wavelength of the laser beam in the unknown liquid, we can use the formula:
v = λ * f
where v is the speed of light in a medium, λ is the wavelength of light in that medium, and f is the frequency of light.
The speed of light in a vacuum is a constant, approximately 3.00 x [tex]10^8[/tex]m/s.
The wavelength of the laser beam in air is 633 nm (or 633 x [tex]10^{(-9)[/tex]m) and the time it takes for the light to travel through 26.0 cm of the unknown liquid is 1.42 ns (or 1.42 x [tex]10^{(-9)[/tex] s).
We can calculate the speed of light in the unknown liquid:
[tex]v_{liquid[/tex] = distance / time
= 0.26 m / (1.42 x [tex]10^{(-9)[/tex] s)
≈ 183.099 x [tex]10^6[/tex] m/s
We can find the wavelength of the laser beam in the liquid using the speed of light in the liquid and the frequency:
[tex]v_{liquid} = \lambda _{liquid} * f \\\lambda _{liquid} = v_{liquid} / f[/tex]
Since the frequency remains the same as the laser beam passes through different media, we can use the speed of light in a vacuum to calculate the wavelength in the liquid:
λ_liquid = (3.00 x [tex]10^8[/tex] m/s) / f
We can substitute the wavelength in air and solve for the wavelength in the liquid:
λ_liquid = (3.00 x[tex]10^8[/tex] m/s) / (633 x [tex]10^{(-9)[/tex] m)
≈ 473.932 x [tex]10^{(-9)[/tex] m
≈ 474 nm
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A car traveling at 35 m/s runs out of gas while traveling up a 5.0 ∘ slope. Part A How far will it coast before starting to roll back down? Express your answer in meters.
The car will coast for 230.55 m before starting to roll back down. The answer is 231 meters.
Initial velocity of the car (u) = 35 m/s
Initial slope (θ) = 5°
The acceleration due to gravity (g) = 9.81 m/s²
We have to find the distance the car will coast before starting to roll back down
d). Since there is no fuel in the car, so the car will stop at the point from where it will start to roll back down. At this point, the potential energy of the car will be equal to the kinetic energy of the car.
At this point, the total energy of the car is conserved and can be expressed as:
mg * h = 1/2 * m * v²
Where, m is the mass of the car, mg is the weight of the car, g is the acceleration due to gravity, h is the height of the slope, v is the velocity of the car just before the car starts to roll back down
So, we can write the velocity v of the car just before the car starts to roll back down as:
v = √(2 * g * h)
This will be the final velocity of the car just before starting to roll back down
.Now, we can calculate the distance the car will coast before starting to roll back down as
d = (u²/2g) * sin(2θ)
On substituting the given values, we get:
d = (35²/2 * 9.81) * sin(2 * 5°)
d = 230.55 m
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An airplane travels at 286 mi/h south for 3.00 h and then at 315 mi/h north for 788 miles. What is the average speed for the trip? Express answers in mi/h.
The average speed for the trip is approximately 299.2 mi/h.
To find the average speed of the trip, we can use the formula as follows:`
Average Speed = Total Distance / Total Time`
Here, the total distance traveled is the sum of distances traveled in each direction, i.e., south and north.
In the south direction, the airplane traveled for 3.00 hours.
Hence, the distance traveled in the south direction is:
Distance in South = Speed x Time
= 286 mi/h x 3.00 h
= 858 miles
In the north direction, the airplane traveled for 788 miles. Hence, the distance traveled in the north direction is:
Distance in North = 788 miles
Therefore, the total distance traveled is:
Total Distance = Distance in South + Distance in North
= 858 miles + 788 miles
= 1646 miles
Total time taken to travel this distance is the sum of the time taken to travel in the south and north directions.
Total Time = Time in South + Time in North
= 3.00 h + (788 miles / 315 mi/h)
= 5.50 hours
Substituting the values of the total distance and total time in the formula for average speed, we get:
Average Speed = Total Distance / Total Time
= 1646 miles / 5.50 hours
= 299.2 mi/h (rounded to one decimal place)
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The pressure at the bottom of a freshwater vessel is P. The water is poured out and replaced with seawater (density = 1025 kg/m³). The new pressure at the bottom of the beaker is
Select one:
a. greater than P.
b. equal to P.
c. Indeterminate.
d. smaller than P.
When a freshwater vessel is emptied and replaced with seawater, the new pressure at the bottom of the vessel can be determined. The possible options for the new pressure are greater than P, equal to P, indeterminate, or smaller than P.
The pressure at a certain depth in a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Since the vessel is initially filled with freshwater, the pressure at the bottom is P, according to the given information.
When the water is poured out and replaced with seawater, the density of the fluid changes. Seawater has a higher density than freshwater (density of seawater = 1025 kg/m³).
As the density of the fluid increases, the pressure at the same depth also increases. Therefore, the new pressure at the bottom of the vessel will be greater than the initial pressure P.
Hence, the correct option is (a) greater than P. By replacing the freshwater with seawater, the new pressure at the bottom of the vessel will be higher than the initial pressure.
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27. A compass needle points towards magnetic north because the earth exerts a magnetic force on the compass. Does the compass needle exert a force on the earth? If so, is this force larger, smaller, or the same as the force that the earth exerts on the compass? Explain why in terms of Newton's laws.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Applying this principle to the interaction between the compass needle and the Earth's magnetic field, we can conclude that the compass needle exerts a force on the Earth.
The force exerted by the compass needle on the Earth is indeed present, but it is significantly smaller compared to the force that the Earth exerts on the compass needle. This difference in magnitude can be attributed to the difference in masses between the Earth and the compass needle. Newton's second law of motion states that the force acting on an object is equal to the product of its mass and acceleration. In this case, the compass needle has a relatively small mass compared to the Earth. When the compass needle exerts a force on the Earth, it accelerates the Earth to a very tiny extent due to the Earth's large mass. On the other hand, the force exerted by the Earth on the compass needle causes a noticeable acceleration in the needle due to its much smaller mass. In practical terms, the force exerted by the compass needle on the Earth is negligible and can be disregarded in most cases. The force between the Earth and the compass needle is mainly unidirectional, with the Earth's magnetic field acting on the compass needle and causing it to align with the magnetic field lines. In summary, while the compass needle does exert a force on the Earth due to Newton's third law, the magnitude of this force is considerably smaller than the force exerted by the Earth on the compass needle due to the large difference in mass between the two objects.
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Which ons of the following statements abour thermal ene'dy reservole is toue? a. Occans, takes, and tivers as well as the atmospheric air cannot be considered as thermal energy reservoirs b. A thermal enerev reservo is hypothetical body with a smali thermal eners copacty c. A thermal energv reservoir can supply of absoeb finite amounts of theat witheut undergoing any change in temperature d. A theimal enetgr reservoir can absorb heat only; it cannot supply heat. On: A17 Which ore of the following statements about hest eneines is not true? a. Heat engines are devices that convert heat to work b. Heat engines usually have 100% thermal efficiency c. Heat ensines are devices that operate in a cycle d. Heat engines use working fluid to transfer energy in the cycle. Qn, A18 Which one of the following statements about thermal resistance is not true? a. Thermat resistance of an object depends on its geometry b. Thermal resistance of an object depends on its thermal properties c. Thermal resistance of an object is also known as its conduction resistance. d. Thetmal resistance of an object is an intensive property An. 19 In the thermal resistance concept, which two properties are not analogues of each other? (a) Rate of heat transfer and electric current (b) Thermai resistance and electrical resistance (c) Temperature difference and voltage difference (d) Thermal resistance and electrical current Qn. A20 Which one of the following statements about temperature drop is not true? a. Temperature drop is proportional to thermal resistance b. Temperature drop across a wall decreases as thickness of the wall increases c. Temperature drop across a wail increases as cross sectional area of the wall increases d. Temperature drop across a wall decreases as thermal conductivity of the wall increases
The correct statements about thermal energy reservoir and thermal resistance are given below:
a. The statement "Oceans, lakes, and rivers, as well as the atmospheric air, cannot be considered as thermal energy reservoirs" is true.
b. The statement "A thermal energy reservoir is a hypothetical body with a small thermal energy capacity" is true.
c. The statement "A thermal energy reservoir can supply or absorb finite amounts of heat without undergoing any change in temperature" is true.
d. The statement "A thermal energy reservoir can absorb heat only; it cannot supply heat" is false. It should be "A thermal energy reservoir can both absorb and supply heat".
Regarding heat engines:
a. The statement "Heat engines usually have 100% thermal efficiency" is not true. They have a maximum theoretical efficiency called Carnot efficiency, which is always less than 100%.
b. The other statements, "Heat engines are devices that convert heat to work," "Heat engines are devices that operate in a cycle," and "Heat engines use working fluid to transfer energy in the cycle," are true.
Regarding thermal resistance:
a. The statement "Thermal resistance of an object is also known as its conduction resistance" is true. The other statements are false. The thermal resistance of an object depends on its geometry and thermal properties. It is an intensive property.
b. The two properties that are not analogues of each other in the thermal resistance concept are "Rate of heat transfer and electric current."
c. The statement "Temperature drop across a wall increases as the cross-sectional area of the wall increases" is not true. The other statements are true.
Temperature drop is proportional to thermal resistance. Temperature drop across a wall decreases as the thickness of the wall increases. Temperature drop across a wall decreases as the thermal conductivity of the wall increases.
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Which one is the transfer of energy NOT due to a temperature difference? Select one: a. Heat b. Work c. Internal energy d. Binding energy.
The transfer of energy is not due to a temperature difference in work (Option B).
What is work?Work is an energy transfer that takes place when a force is exerted over a distance. It is the energy transferred by a force acting through a distance in the direction of the force. It is the energy transferred to or from an object by means of external forces, and it is defined as force multiplied by the distance in the direction of the force. It is expressed in joules (J).
Heat is a type of energy that is transferred from one object to another due to a difference in temperature. Heat flows from a hot body to a cold body. Heat is measured in joules (J). Internal energy is the energy that a substance possesses due to its particles' motion. The internal energy of a substance includes the kinetic and potential energy of the atoms and molecules that make up the substance. The internal energy is related to the temperature of the substance. The energy needed to separate a particle or group of particles into individual particles is referred to as binding energy. This energy is required to overcome the forces that hold the particles together. It is expressed in electron volts (eV).
Thus, the correct option is B.
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A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg×m2, and is rotating at 10 rpm. A child whose mass is 25 kg jumps onto the edge of the merry-go-round, heading directly toward the center at 6.0 m/s. The new angular speed (in rpm) of the merry-go-round is approximately
a.
9.2
b.
7.1
c.
10
d.
8.5
e.
6.4
When a child jumps onto a rotating merry-go-round, the new angular speed of the merry-go-round can be found using the law of conservation of angular momentum. In this case, the approximate final angular velocity is 9.2 rpm, corresponding to option (a).
To solve this problem, we can apply the law of conservation of angular momentum. The initial angular momentum of the system is given by L1 = I1ω1, where I1 is the moment of inertia of the merry-go-round and ω1 is the initial angular velocity.
The final angular momentum of the system is given by L2 = I2ω2, where I2 is the moment of inertia of the system after the child jumps on and ω2 is the final angular velocity.
According to the law of conservation of angular momentum, L1 = L2. Therefore, I1ω1 = I2ω2.
The moment of inertia of the system after the child jumps on is given by I2 = I1 + mr^2, where m is the mass of the child and r is the radius of the merry-go-round.
Plugging in the given values, I1 = 250 kg·m^2, m = 25 kg, and r = 2.0 m, we can calculate I2 = I1 + mr^2.
Next, we need to convert the initial angular velocity from rpm to rad/s. Since 1 rpm is equivalent to (2π/60) rad/s, the initial angular velocity is ω1 = (10 rpm) × (2π/60) rad/s.
Now, we can solve for the final angular velocity ω2 by rearranging the equation I1ω1 = I2ω2 and plugging in the values of I1, ω1, and I2.
Finally, we can convert the final angular velocity from rad/s to rpm by multiplying ω2 by (60/2π).
After performing the calculations, we find that the approximate final angular velocity of the merry-go-round is 9.2 rpm.
Therefore, the correct option is (a) 9.2.
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An interference pattern is formed on a screen when helium-neon laser light (λ = 6.33 x 10-7m) is directed towards it through two slits. If the slits are 43μm apart and the screen is 2.5m away, what will be the separation of adjacent dark spots?
The separation of adjacent dark spots in the interference pattern will be approximately 0.03684 meters. To determine the separation of adjacent dark spots in an interference pattern, we can use the formula: y = (λL) / d.
To determine the separation of adjacent dark spots in an interference pattern, we can use the formula:
y = (λL) / d
where:
y is the separation of adjacent dark spots,
λ is the wavelength of the light,
L is the distance between the slits and the screen (2.5 m in this case), and
d is the distance between the slits (43 μm, which can be converted to meters).
First, let's convert the distance between the slits from micrometers (μm) to meters (m):
d = 43 μm = 43 x 10^-6 m
Now we can plug the values into the formula to calculate the separation of adjacent dark spots:
y = (6.33 x 10^-7 m * 2.5 m) / (43 x 10^-6 m)
Simplifying the equation:
y = 0.03684 m
Therefore, the separation of adjacent dark spots in the interference pattern will be approximately 0.03684 meters.
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Star A has a magnitude of 4 and Star B has a magnitude of 6. How much brighter is Star A than Star B?
a. 0.0006554
b. 0.16
c. 0.0002621
d. 1.5
e. 3.33
f. 1526
g. 0.0102
h. 2.5
i. 610
j. 97.7
k. 2
l. 0.00164
m. 5
n. 6.25
o. 3815
The difference in magnitudes between the stars can be found using the formula Δm = m1 - m2. Where m1 = magnitude of star A, m2 = magnitude of star B, and Δm is the difference in magnitudes.
Given that Star A has a magnitude of 4 and Star B has a magnitude of 6. Therefore,Δm = 4 - 6= -2.
The negative sign indicates that star B is brighter than star A.
Thus, to find out how much brighter Star A is than Star B, we need to take the antilogarithm of Δm/2.5.
This can be calculated as follows:antilog (-2/2.5)= antilog (-0.8) = 0.1585.
The antilogarithm is approximately equal to 0.16.
Therefore, Star A is 0.16 times brighter than Star B. Answer: The brightness ratio between Star A and Star B is 0.16.
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Aluminum has a density 2.7 times that of water (1 g/cm3) and a specific heat 0.217 times that of water (1 cal/gxC*). When the internal energy of an aluminum cube with an edge length of 25cm increases by 92,000 cal, its temperature increases by: Answer in degC. Show solutions for this question.
The temperature of the aluminum cube increases by 10°C. The density of aluminum = 2.7 g/cm³, The specific heat of aluminum = 0.217 cal/g °C, The edge length of aluminum cube = 25 cm, The internal energy of the aluminum cube = 92000 cal.
The mass of the aluminum cube using its density and volume.
Mass of aluminum cube = Density × Volume= 2.7 × (25)³= 2.7 × 15625= 42187.5 g.
Now, we can use the formula:q = msΔTwhereq = Internal energy ms = Mass × specific heat ΔT = Temperature change.
Rearranging the formula:ΔT = q / (ms).
Substituting the given values,ΔT = 92000 cal / (42187.5 g × 0.217 cal/g°C)ΔT = 92000 / (9167.188)ΔT = 10°C.
Therefore, the temperature of the aluminum cube increases by 10°C.
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what volume of ch4 at 0°c and 1.00 atm contains the same number of molecules as 0.50 l of n2 measured at 27°c and 1.50 atm?
The following are the data provided:0.5 L of N2 at 27°C and 1.50 atmCH4 at 0°C and 1.00 atm is used in the experiment.
To find the volume of CH4, we need to calculate the number of molecules present in N2 at 27°C and 1.50 atm. For this, we need to use the ideal gas equation. The ideal gas equation is expressed as:P.V = n.R.TWhere P = pressure of the gas in atmV = volume of the gas in litersn = number of moles of the gasR = universal gas constant, 0.08206 L.atm/(mol.K)T = temperature of the gas in KelvinTo convert °C to Kelvin, we add 273 to the temperature. Therefore, the temperature of N2 will be:27 + 273 = 300 KNow, let's find the number of moles of N2 present in the given volume.The ideal gas equation can be rearranged to calculate the number of moles of a gas:n = P.V / R.TWe get:n = (1.50 atm)(0.5 L) / (0.08206 L.atm/(mol.K))(300 K) = 0.0301 molNow, we need to find the number of molecules in this amount of N2. We know that 1 mole of any gas contains 6.02 x 10²³ molecules (Avogadro's number).Therefore, the number of molecules in 0.0301 mol of N2 is:0.0301 mol x 6.02 x 10²³ molecules/mol = 1.81 x 10²² moleculesNow, we need to find the volume of CH4 at 0°C and 1.00 atm that contains this number of molecules.Using the ideal gas equation, we can write:V = n.R.T / PWhere n = 1.81 x 10²² molecules / 6.02 x 10²³ molecules/mol = 0.00301 molT = 0°C + 273 = 273 KP = 1.00 atmR = 0.08206 L.atm/(mol.K)Plugging these values in the above equation, we get:V = (0.00301 mol)(0.08206 L.atm/(mol.K))(273 K) / (1.00 atm)V = 0.067 LTherefore, the volume of CH4 at 0°C and 1.00 atm that contains the same number of molecules as 0.50 L of N2 measured at 27°C and 1.50 atm is 0.067 L.
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Problem 29.1 Part A The magnetic flux through a coil of wire containing 3 loops changes at a constant rate from -44 Wb to 27 Wb in 0.47 s. What is the emf induced in the coil? Express your answer using two significant figures. IVALO ? Eird = V Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Check your signs. < Return to Assignment Provide Feedback
If the magnetic flux through a coil of wire containing 3 loops changes at a constant rate from -44 Wb to 27 Wb in 0.47 s. The emf induced in the coil is approximately -450 V.
The emf induced in the coil can be calculated using Faraday’s Law. Faraday’s Law states that the emf induced in a coil is equal to the rate of change of magnetic flux through the coil. The formula for emf is given by
emf = -N (ΔΦ/Δt)
where N is the number of turns in the coil, ΔΦ is the change in magnetic flux, and Δt is the time interval over which the change in flux occurs.
Here, the number of turns in the coil is 3. The change in magnetic flux is the final flux minus the initial flux.ΔΦ = 27 Wb - (-44 Wb) = 71 WbThe time interval over which the change in flux occurs is 0.47 s. Therefore,
emf = -3 (71 Wb/0.47 s) = -450 V (approx)
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what direction (in degrees counterclockwise from the east axis) is the dog? Use a graphical method. magnitude m direction ∘
∘
counterclockwise from the east axis
The direction of the dog is about 67.38 degrees counterclockwise from the east axis. using a graphical method. The first step is to represent the vector components, i.e., the horizontal and vertical components of the given vector.
Let the horizontal component be x and the vertical component be y.
We have:m = √(x² + y²).
Since the direction is counterclockwise from the east axis, we have to calculate the angle from the east axis.
Let's say the angle is θ.
Therefore, we have:x = m cosθ and y = m sinθθ = tan⁻¹(y/x).
Given the vector components:x = -5my = 12m Magnitude, m = √(x² + y²)m = √((-5m)² + (12m)²)m = √(25m² + 144m²)m = √(169m²)m = 13m Angle from the east axis, θ = tan⁻¹(y/x)θ = tan⁻¹((12m)/(-5m))θ = tan⁻¹(-12/5)θ ≈ -67.38° (rounded to two decimal places).
Therefore, the direction of the dog is about 67.38 degrees counterclockwise from the east axis.
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the electric field inside a metallic conductor is _____.
The electric field inside a metallic conductor is zero.
A metallic conductor is characterized by its ability to conduct electricity. This property is due to the presence of free electrons that are loosely bound to the atoms in the conductor. When an electric field is applied to a conductor, the free electrons respond by redistributing themselves within the conductor until they reach an equilibrium state.
In this equilibrium state, the free electrons distribute themselves uniformly throughout the interior of the conductor. As a result, they create an electric field within the conductor that exactly cancels out the externally applied electric field. This cancellation occurs because the free electrons move in such a way as to counteract the applied field.
The redistribution of free electrons and the cancellation of the electric field within the conductor happen almost instantaneously, creating a condition known as electrostatic equilibrium. In this state, the electric field inside the conductor is effectively zero, and the charges within the conductor are at rest.
This property of a metallic conductor allows for the phenomenon of electrostatic shielding. The absence of an electric field inside the conductor ensures that any charges or external electric fields applied to the surface of the conductor do not penetrate the interior.
As a result, the charges or electric fields are confined to the surface of the conductor, making it an effective shield against external electric fields.
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Two cars, initially at rest, are moving towards each other starting from opposite ends of a line segment AB. Their respective constant accelerations are 1 =4 m/2 and 2 =3 m/2. Car A starts first and car B starts moving after time t = 2 s. If the cars meet at a point C which is 54 m away from point B, find the length of the line segment AB
Let the length of line segment AB be L. The time taken for the two cars to meet is t.
Using the first equation of motion,
we can calculate the distance covered by each car as:
a1t + 1/2 a1t^2 = distance covered by car A a2(t - 2) + 1/2 a2(t - 2)^2 = distance covered by car B
Adding these two equations,
we get:
L = a1t + a2(t - 2) + 1/2 a1t^2 + 1/2 a2(t - 2)^2
Using the third equation of motion, we can relate L, t and the accelerations of the two cars, i.e., a1 and a2.
L = 1/2 (a1 + a2) t^2 + 1/2 (a2 - a1) t + a2
Similarly, using the distance between the two cars when they meet, i.e., 54 m, w
e can form another equation.
L = 54 + (1/2 a1t^2 + 1/2 a2(t - 2)^2)
On equating the two expressions for L,
we get:
1/2 (a1 + a2) t^2 + 1/2 (a2 - a1) t + a2
= 54 + 1/2 a1t^2 + 1/2 a2(t - 2)^2
Simplifying this equation, we get a quadratic equation in t:
t^2 - 5t + 36 = 0
Solving this equation, we get t = 4 s or t = 9 s.
Since car B starts moving after t = 2 s,
we can reject the solution
t = 4 s and consider the solution t = 9 s.
Using this value of t in the third equation of motion, we can find the length of line segment AB.
L = 1/2 (a1 + a2) t^2 + 1/2 (a2 - a1) t + a2
= 1/2 (4 + 3) x 9^2 + 1/2 (3 - 4) x 9 + 3
= 99 m
The length of line segment AB is 99 m.
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Explain the effect on the centre of gravity of the ship (CG) when a weight (w) is loaded, discharged or shifted onboard a ship. In your answer, include the formulae to calculate the amount of shift of CG (GG1) in each of the three cases. Ensure that you explain your formulae clearly.
The effect of loading, discharging or shifting weight on the centre of gravity of a ship (CG) is a crucial element in ensuring the safety and stability of the ship. The movement of the centre of gravity due to a change in weight distribution onboard can cause the ship to heel, trim or capsize.
The movement of the centre of gravity due to loading or unloading cargo can be calculated using the formula below:GG1 = w × d ÷ W Where GG1 is the shift of the centre of gravity in metres, w is the weight of cargo in tonnes, d is the distance between the centre of gravity of the cargo and the original centre of gravity of the ship, and W is the displacement of the ship in tonnes.
The above formula assumes that the weight is moved at a right angle to the longitudinal axis of the ship. When the weight is moved parallel to the longitudinal axis of the ship, the formula becomes: GG1 = w × GZ ÷ M Where GZ is the distance between the centre of gravity of the ship and the metacentre, and M is the mass of the ship.
When a weight is loaded on board, the centre of gravity moves upwards. Conversely, when weight is discharged from the ship, the centre of gravity moves downwards. The shift in the centre of gravity can also be affected by the placement of cargo on the ship.
For instance, placing heavy cargo at the bow will cause the centre of gravity to shift forward while placing the cargo at the stern will cause the centre of gravity to shift backwards.
In conclusion, the effect of loading, discharging or shifting weight on the centre of gravity of a ship is an essential element that must be considered to ensure the safety and stability of the ship. The shift in the centre of gravity can be calculated using the above formula, which considers the weight and the distance between the centre of gravity and the original centre of gravity of the ship.
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What can be the maximum diameter of the pipe for the water flow of velocity 1 m/s (kinematic viscosity = 10-6 m2/s) to be laminar in nature?
1. 2 cm
2. 2 mm
3. 20 mm
4. 2 m
The maximum diameter of the pipe for the water flow of velocity 1 m/s (kinematic viscosity = 10-6 m2/s) to be laminar in nature is 2 nm.
To determine the maximum diameter of the pipe for the water flow to be laminar, we can use the Reynolds number criterion. The Reynolds number (Re) is a dimensionless parameter that helps determine the flow regime of a fluid, whether laminar or turbulent. For laminar flow, the Reynolds number must be below a certain threshold.
The Reynolds number (Re) is defined as the ratio of inertial forces to viscous forces and is calculated using the formula:
Re = (Fluid velocity * Pipe diameter) / Kinematic viscosity
In this case, the water flow velocity is given as 1 m/s, and the kinematic viscosity of water is given as [tex]10^({-6)[/tex] m²/s.
To determine the maximum diameter for laminar flow, we need to find the threshold Reynolds number for the laminar-turbulent transition. Generally, a Reynolds number below 2000 is considered indicative of laminar flow.
Let's calculate the Reynolds number using the given values:
Re = (1 m/s * Pipe diameter) / [tex]10^{(-6)[/tex] m²/s)
To ensure laminar flow, we need Re to be below 2000. Rearranging the equation, we have:
Pipe diameter = (Re * Kinematic viscosity) / Fluid velocity
Maximum pipe diameter = (2000 * [tex]10^{(-6)[/tex]m²/s) / 1 m/s = 0.002 m = 2 mm
Therefore, the correct answer is option 2: 2 mm.
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A child runs towards some ice at 4 m/s. She slides across the ice, coming to a stop at 8 m. What is her acceleration rate?
Then, how fast would you have to be going initially to slide on the same ice for 15s?
The given variables in the problem are as follows:
Initial velocity = 4 m/s
Displacement = 8 m
Time taken = Unknown
Acceleration = Unknown
We can calculate the acceleration using the formula:
distance = (initial velocity * time) + (1/2) * acceleration * [tex]time^2[/tex]
Substitute the given values in the above equation.
8 = (4 * t) + (1/2) * a * [tex]t^2[/tex]
Now, it is required to determine the acceleration rate.
Thus, we can use the following formula to find the
acceleration rate: (1/2) * a = (d - v*t) / [tex]t^2[/tex](1/2) * a = (8 - 4*t) / [tex]t^2a[/tex] = 2*(8 - 4*t) / [tex]t^2a[/tex] = 16/[tex]t^2[/tex]
Similarly,
to find the initial velocity,
we can use the formula:
distance = (initial velocity * time) + (1/2) * acceleration * [tex]time^2[/tex]
Substituting the given values in the above equation, we get:
15 = (u * t) + (1/2) * a * [tex]t^2[/tex]
Now,
since there are two unknowns in the above equation (initial velocity and acceleration), we can use the first equation that we obtained to substitute the value of acceleration in terms of time, and substitute that in the second equation.
This gives:
15 = u*t + 8 - 2t/ t
15 = u + 8/t - 2u
15t = u(t+4)u = 15t/(t+4)]
Thus, the initial velocity needed to slide on the same ice for 15s would be 15t/(t+4).
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