The existence of planets around a millisecond pulsar is surprising because pulsars are rapidly rotating neutron stars that were not expected to have stable planetary systems due to their extreme conditions.
The existence of planets around a millisecond pulsar comes as a surprise due to several reasons. Millisecond pulsars are incredibly dense and rapidly rotating neutron stars, formed from the remnants of supernova explosions. Their intense gravitational forces and strong magnetic fields were previously believed to disrupt or prevent the formation and stability of planetary systems. Additionally, the formation of planets typically requires the presence of a protoplanetary disk, which is unlikely to survive the violent stellar evolution leading to the creation of millisecond pulsars. Therefore, the discovery of planets around millisecond pulsars challenges our understanding of planetary formation and highlights the resilience and adaptability of planetary systems in extreme environments.
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In the sport of horseshoe pitching, two stakes are 40. 0 feet apart. What is the distance in meters between the two stakes? *
The distance between the two stakes in horseshoe pitching is approximately 12.192 meters.
The given problem states that the two stakes in horseshoe pitching are 40 feet apart. And we are supposed to find out the distance between them in meters. Let us first write down the given value in feet.Given that the distance between the two stakes is 40 feet. Now, 1 meter is equivalent to 3.28084 feet.To convert feet into meters, we need to divide the given value of feet by the value of 3.28084.Thus, the distance between the two stakes in meters can be calculated as follows: Distance in meters = \frac{distance in feet }{ 3.28084 }
.Distance in meters =\frac{ 40 }{ 3.28084 meters} ≈ 12.192 meters.
Therefore, the distance between the two stakes in horseshoe pitching is approximately 12.192 meters. The exact value can be obtained by using more number of decimal points.
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if the allowable normal stress for the bar is σallow=120mpa , determine the maximum axial force p that can be applied to the bar.
The maximum axial force p that can be applied to the bar can be determined using the formula:
p = σallow * A
where A is the cross-sectional area of the bar.
Explanation: The formula above is derived from the stress-strain relationship for a material, which states that stress is equal to force divided by area. The allowable normal stress is the maximum stress that the material can withstand without undergoing plastic deformation. By multiplying this allowable stress with the cross-sectional area of the bar, we can determine the maximum axial force that can be applied without exceeding the material's strength.
Therefore, to determine the maximum axial force p that can be applied to the bar, we need to know its cross-sectional area. Once we have this information, we can use the formula p = σallow * A to calculate the maximum force.
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1.0 kg of steam at 100 c condenses to water at 100 c. what is the change in entropy in the process?
The change in entropy during the process of 1.0 kg of steam at 100°C condensing to water at 100°C is -2.44 kJ/K.
Entropy is a measure of the disorder or randomness of a system. The change in entropy during a process can be calculated using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred during the process, and T is the temperature at which the heat is transferred.
In this case, 1.0 kg of steam at 100°C condenses to water at 100°C. During this process, the steam releases heat to the surroundings, which is absorbed by the water. The heat transferred during the process can be calculated using the formula Q = m × L, where Q is the heat transferred, m is the mass of the steam, and L is the latent heat of vaporization of water.
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Typical motions of one plate relative to another are 1 centimeter per year.
At this rate, how long would it take for two continents 3500 kilometers apart to collide?
At a typical rate of 1 centimeter per year, it would take approximately 350 million years for two continents located 3500 kilometers apart to collide.
At a rate of 1 centimeter per year, the motion of tectonic plates is relatively slow. If two continents are 3500 kilometers (3,500,000 meters) apart, it would require 350 million years for them to collide. This calculation is based on the assumption that the rate of plate motion remains constant over such a long period, which is not always the case in reality. The collision of continents is a complex process influenced by various factors, including plate boundaries, geological activity, and the presence of other landmasses. Nevertheless, the estimation provides a rough idea of the timescale involved in continental collision at this rate of plate motion.
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in a crystalline metal, its slip direction is that direction in the slip plane having the shortest interatomic distance. T/F ?
False. In a crystalline metal, the slip direction is the direction along which the dislocations move when deformation occurs.
It is not necessarily the direction with the shortest interatomic distance. The slip direction is determined by the crystal structure and the arrangement of atoms in the material. It is influenced by factors such as crystallographic planes and the arrangement of atoms within those planes. The slip direction is important for understanding the mechanical properties and deformation behavior of metals. In a crystalline metal, the atoms are arranged in a regular and repeating pattern called a crystal lattice. This organized arrangement gives the metal its characteristic structure and properties. The atoms are closely packed together in a three-dimensional arrangement, forming a crystal structure.
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you push your 0.70-kg pillow across your bed with a constant force of 12 n . the bed provides a frictional force of 8.0 n .
What is the acceleration of the center of mass of the pillow? Assume that the direction of your push is the positive direction.
The acceleration of the center of mass of the pillow can be found using the equation:
a = (F_net) / m
where F_net is the net force acting on the pillow and m is the mass of the pillow.
In this case, the net force is the force you apply minus the frictional force of the bed:
F_net = 12 N - 8.0 N = 4.0 N
So, the acceleration of the center of mass of the pillow can be calculated as:
a = (4.0 N) / (0.70 kg) = 5.7 m/s^2
The net force on the pillow is the force you apply minus the frictional force of the bed. This net force causes an acceleration of the pillow, which can be found using the equation a = F_net / m.
The fact that the frictional force of the bed is opposite in direction to the force you apply, so it subtracts from the net force. The acceleration of the center of mass of the pillow is a scalar quantity, meaning it only has magnitude and no direction. It is measured in meters per second squared (m/s^2).
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rank these circuits on the basis of their resonance frequencies. rank from largest to smallest. to rank items as equivalent, overlap them.
The circuits ranked in order of their resonance frequencies from largest to smallest are: B. LC series resonant circuit, A. LC parallel resonant circuit, C. RC parallel resonant circuit, and D. RC series resonant circuit.
The resonance frequency of an LC circuit is given by f = 1/(2π√(LC)), where L is the inductance and C is the capacitance of the circuit. For a given value of C, the resonance frequency increases with increasing inductance. Therefore, an LC series resonant circuit, which has a larger inductance than an LC parallel resonant circuit, will have a higher resonance frequency.
On the other hand, the resonance frequency of an RC circuit is given by f = 1/(2πRC), where R is the resistance and C is the capacitance of the circuit. For a given value of C, the resonance frequency decreases with increasing resistance. Therefore, an RC parallel resonant circuit, which has a smaller resistance than an RC series resonant circuit, will have a higher resonance frequency.
Thus, the order of the resonance frequencies from largest to smallest is B. LC series resonant circuit, A. LC parallel resonant circuit, C. RC parallel resonant circuit, and D. RC series resonant circuit.
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Complete Question:
Rank the following circuits on the basis of their resonance frequencies, from largest to smallest:
A. LC parallel resonant circuit
B. LC series resonant circuit
C. RC parallel resonant circuit
D. RC series resonant circuit
To rank items as equivalent, overlap them.
light is refracted from air into an unknown material. if the angle of incidence is 36° and the angle of refraction is 18°, what is the index of refraction?
The index of refraction for the unknown material is approximately 1.931.
To find the index of refraction for the unknown material, you can use Snell's Law, which states:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
In this case, n₁ is the index of refraction for air, which is approximately 1.00. θ₁ is the angle of incidence (36°), n₂ is the index of refraction for the unknown material, and θ₂ is the angle of refraction (18°).
Following the formula, you can plug in the values:
1.00 * sin(36°) = n₂ * sin(18°)
Now, divide both sides by sin(18°) to solve for n₂:
n₂ = (1.00 * sin(36°)) / sin(18°)
Calculate the values:
n₂ ≈ 1.931
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A very long conducting tube (hollow cylinder) has inner radius A and outer radius b. It carries charge per unit length +α, where α is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length +α. (a) Calculate the electric field in terms of α and the distance r from the axis of the tube for (ii) a < r < b
The electric field at a distance r from the axis of the tube for a < r < b depends only on the charge per unit length α and the distance r from the axis, and not on the radii A and b of the conducting tube.
For a conducting tube carrying charge per unit length +α and a line of charge along its axis with charge per unit length +α, the electric field at a distance r from the axis of the tube for a < r < b can be calculated using Gauss's Law.
Since the electric field outside the cylinder is radial and has the same magnitude at every point with the same radius, we can consider a cylindrical Gaussian surface of radius r and length L, with one end at a distance a from the center of the tube and the other end at a distance b.
The electric field E is then perpendicular to the ends of the cylinder and its magnitude is constant over the Gaussian surface.
The total charge enclosed by the cylinder is αL. By Gauss's Law, the electric flux through the surface is given by Φ = Qenc / ε0, where ε0 is the permittivity of free space.
Since the electric field is perpendicular to the ends of the cylinder, the electric flux through each end is zero. Therefore, the electric flux through the curved surface is Φ = E(2πrL), where L is the length of the cylinder.
Equating these two expressions for Φ, we get E(2πrL) = αL / ε0, which gives the electric field as E = α / (2πε0r) for a < r < b. Thus, the electric field at a distance r from the axis of the tube for a < r < b depends only on the charge per unit length α and the distance r from the axis, and not on the radii A and b of the conducting tube.
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what is the focal length (in m) of a makeup mirror that has a power of 1.70 d?
The focal length of the makeup mirror is 0.588 meters or approximately 58.8 centimeters.
What is the focal length of makeup mirror?To find the focal length of a makeup mirror that has a power of 1.70 D (diopters).
We can use the following formula:
f = 1/P
where f is the focal length in meters and P is the power in diopters.
Substituting P = 1.70 D into the formula, we get:
f = 1/1.70 D
f = 0.588 m
Therefore, the focal length of the makeup mirror is 0.588 meters or approximately 58.8 centimeters.
This means that light rays entering the mirror will converge at a distance of 0.588 meters behind the mirror's surface
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The lowest frequency in the fm radio band is 88.4 mhz. What inductance (in µh) is needed to produce this resonant frequency if it is connected to a 2.40 pf capacitor?
The resonant frequency of an LC circuit is given by:
f = 1 / (2π√(LC))
where f is the resonant frequency, L is the inductance in Henry (H), and C is the capacitance in Farad (F).
To find the inductance needed to produce a resonant frequency of 88.4 MHz with a 2.40 pF capacitor, we can rearrange the above equation as:
L = (1 / (4π²f²C))
Plugging in the values, we get:
L = (1 / (4π² × 88.4 × 10^6 Hz² × 2.40 × 10^-12 F))
L = 59.7 µH
Therefore, an inductance of 59.7 µH is needed to produce a resonant frequency of 88.4 MHz with a 2.40 pF capacitor in an LC circuit.
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Neglecting the mass of the stick, find the cm marking where the meterstick will balance (0 cm is the left end, 100 cm is the right end.)
The meterstick will balance at 50 cm.
Mass of the first block, m₁ = 4 g
Mass of the second block, m₂ = 10 g
Distance of second block from the centre of mass, r₂ = 20 cm
According to the principle of moments,
When a body is balanced, the total clockwise moment around a point equals the total anticlockwise moment around the same point. Moment is defined as the product of force and the perpendicular distance.
So, m₁gr₁ = m₂gr₂
m₁r₁ = m₂r₂
Therefore, the distance of the first block from the centre of mass,
r₁ = m₂r₂/m₁
r₁ = 10 x 20/4
r₁ = 200/4
r₁ = 50 cm
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Charge Q=+ 6.00 μC is distributed uniformly over the volume of an insulating sphere that has radius R = 6.00 cm .
What is the potential difference between the center of the sphere and the surface of the sphere?
The potential difference between the center and the surface of the sphere is 1.08 × 10⁶ V.
What is the voltage difference between the center and surface of an insulating sphere?The potential difference between the center and surface of the sphere with a uniform charge distribution can be expressed mathematically as:
V = kQ/R
Where V is the potential difference, k is Coulomb's constant (9 × 10⁹ Nm²/C²), Q is the charge, and R is the radius of the sphere.
For this specific problem, the potential difference can be calculated as:
V = (9 × 10⁹ Nm²/C²) × (+6.00 × [tex]10^-^6[/tex] C) / (0.06 m) = 1.08 × 10⁶ V.
Therefore, the potential difference between the center and surface of the sphere is 1.08 × 10⁶ V.
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5450 m3 blimp circles Busch Stadium during the World Series suspended in the earth's 1. 21 kg/m3 atmosphere. The density of the helium in the blimp is 0. 178 kg/m3. A) What is the buoyant force that suspends the blimp in the air? B) How does this buoyant force compare to the blimp's weight? C) How much weight, in addition to the helium, can the blimp carry and still continue to maintain a constant altitude?
The buoyant force that suspends the blimp in the air can be calculated as follows; Formula used: Buoyant force = weight of displaced fluid. Buoyant force = Density of air x Volume of the blimp x Acceleration due to gravity. Buoyant force = 1.21 kg/m³ x 5450 m³ x 9.8 m/s.Buoyant force = 64462.6 N. Thus, the buoyant force that suspends the blimp in the air is 64462.6 N.
B) The weight of the blimp can be calculated using the formula: Formula used: Weight = Mass x Gravity.
Mass of blimp = Density of helium x Volume of a blimp.
Weight of blimp = 0.178 kg/m³ x 5450 m³ x 9.8 m/s², Weight of blimp = 93280.6 N.
The buoyant force is less than the blimp's weight as the buoyant force is 64462.6 N and the blimp's weight is 93280.6 N.
Thus, the buoyant force is less than the blimp's weight.
C) The amount of weight, in addition to the helium, that the blimp can carry and still continue to maintain a constant altitude can be calculated using the formula: Formula used: Buoyant force = (Density of fluid x V object submerged in the fluid) x g,
Buoyant force = (Density of fluid x (Volume of the blimp - Volume of helium)) x g,
The weight that can be carried = (Density of fluid x Volume of object) - (Density of object x Volume of the object)
The weight that can be carried = (Density of air x Volume of blimp) - (Density of helium x Volume of helium).
Weight that can be carried = (1.21 kg/m³ x 5450 m³) - (0.178 kg/m³ x 5450 m³).
The weight that can be carried = 6556.95 N.
Thus, the weight, in addition to the helium, that the blimp can carry and still continue to maintain a constant altitude is 6556.95 N.
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An arrow is shot into a hollow pipe resting on a horizontal table and flies out the other end. While the arrow travels in the pipe, its feathers brush against the walls of the pipe. (a) Which type of collision is the arrow-pipe interaction: elastic, inelastic, or totally inelastic? (b) Is there an instant when the velocity of the arrow relative to the pipe is necessarily zero? (c) Describe the energy conversions in the pipe-arrow system.
(a) The arrow-pipe interaction is likely to be an inelastic collision.
(b) Yes, there is an instant when the velocity of the arrow relative to the pipe is zero.
(c) In the pipe-arrow system, kinetic energy is converted into potential energy and vice versa.
When an arrow hits the walls of a hollow pipe, some of its kinetic energy is lost due to the deformation of the arrow and the pipe. The loss of kinetic energy means that the velocity of the arrow decreases as it moves through the pipe. Therefore, the collision is inelastic.
(b) This happens when the arrow comes to a momentary stop at the midpoint of the pipe, where it changes direction and starts moving in the opposite direction.
(c) When the arrow is shot into the pipe, it possesses kinetic energy. As it moves through the pipe, its kinetic energy is gradually converted into potential energy, which is stored in the form of elastic potential energy in the arrow and the pipe. This happens due to the deformation of the arrow and the pipe as they collide with each other. When the arrow comes to a stop at the midpoint of the pipe, all its kinetic energy is converted into potential energy. As the arrow moves out of the other end of the pipe, the potential energy is converted back into kinetic energy. Therefore, the energy conversions in the pipe-arrow system involve the interconversion of kinetic and potential energy.
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a 1100-kg car travels at 22 m/s and then quickly stops in 3.2 s to avoid an obstacle. what is the magnitude of the average force in kilonewtons (kn) that stopped the car?
The magnitude of the average force that stopped the car is 7.5625 kN.
To find the magnitude of the average force that stopped the car, we can use the formula F = ma, where F is the force, m is the mass, and a is the acceleration. First, we need to find the acceleration, which can be calculated using the equation a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time.
In this case, the car's mass (m) is 1100 kg, the initial velocity (vi) is 22 m/s, the final velocity (vf) is 0 m/s (as the car stopped), and the time (t) is 3.2 s. Now we can calculate the acceleration (a): a = (0 - 22) / 3.2 = -6.875 m/s².
Now we can find the force using F = ma: F = (1100 kg)(-6.875 m/s²) = -7562.5 N. The force is negative, which indicates it acted in the opposite direction of the car's motion. To express the magnitude of the force in kilonewtons (kN), divide by 1000: F = -7.5625 kN.
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it took 2.570×103 j to raise the temperature of a sample of water from 12.9 °c to 38.3 °c. convert 2.570×103 j to calories.
2.570×[tex]10^3[/tex] joules is equal to 614.43 calories.
To convert joules to calories, you can use the conversion factor that 1 calorie is equal to 4.184 joules.
Given that it took 2.570×[tex]10^3[/tex] J to raise the temperature of the water, we can convert it to calories using the conversion factor:
2.570×[tex]10^3[/tex] J * (1 calorie / 4.184 J) = 614.43 calories
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Show that the total ground-state energy of N fermions in a three-dimensional box is given by R_total = 3/5 N E_F Thus the average energy per fermion is 3E_F/5
Shows that the total ground-state energy of N fermions in a three-dimensional box is proportional to the number of particles and the Fermi energy, and the average energy per fermion is proportional to the Fermi energy.
What is the expression for the total ground-state energy and average energy per fermion of N fermions in a three-dimensional box?
The total ground-state energy of N fermions in a three-dimensional box can be derived using the Fermi-Dirac statistics and the density of states in three dimensions.
The Fermi energy (E_F) is the energy of the highest occupied state at absolute zero temperature. In a three-dimensional box of volume V, the density of states (D) can be calculated as D=V/h^3, where h is the Planck constant.
Using the Fermi-Dirac distribution, the total number of particles (N) can be expressed as:
N = 2 * V * (2m/h^2)^3/2 * ∫[0 to E_F] (E-E_F)^(1/2) dE
where m is the mass of a single fermion.
Solving for E_F, we get:
E_F = h^2 / 2m * (3π^2 N / V)^(2/3)
The total ground-state energy (R_total) can be obtained by summing up the energies of all the occupied states up to E_F. This can be expressed as:
R_total = 2 * V * (2m/h^2)^3/2 * ∫[0 to E_F] E (E-E_F)^(1/2) dE
Simplifying this expression and substituting for E_F, we get:
R_total = (3/5) * N * E_F
Therefore, the average energy per fermion is given by:
(3/5) * E_F = (3/5) * h^2 / 2m * (3π^2 N / V)^(2/3)
This shows that the total ground-state energy of N fermions in a three-dimensional box is proportional to the number of particles and the Fermi energy, and the average energy per fermion is proportional to the Fermi energy.
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A certain circuit breaker trips when the rms current is 12.0 a. what is the corresponding peak current (in a)?
Required the corresponding peak current is 16.97 A.
The corresponding peak current can be calculated using the formula Ipeak = Irms * √2. Therefore, the peak current for a circuit breaker that trips at 12.0 A
RMS current would be Ipeak = 12.0 * √2 = 16.97 A (rounded to two decimal places). It's important to note that peak current represents the maximum instantaneous current that a circuit can handle, while RMS current represents the equivalent heating effect of a steady DC current. In other words, a circuit breaker is designed to protect against overloading caused by peak currents, which can be higher than the corresponding RMS current.
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When charging, which type of material usually gives off electrons: conductors or insulators? Why?
I need answers asaaap
When charging, conductors usually give off electrons. Conductors are materials that allow electrons to pass through them easily, whereas insulators are materials that prevent electrons from moving through them. Conductors can easily discharge when exposed to static electricity because electrons move more freely through conductors than they do through insulators.
When an object with an excess of electrons comes into touch with an object with a deficiency of electrons, the electrons will move from the charged object to the uncharged object because of the difference in potential energy. The most familiar conductors are metals, which are highly conductive due to the presence of free electrons. Insulators, on the other hand, are materials that do not conduct electricity. Air, paper, plastic, and rubber are all examples of insulators. The transfer of electrons from one object to another by friction, conduction, or induction is referred to as charging. When two materials are rubbed together, their electrons rub together, resulting in one material becoming charged positively and the other becoming charged negatively.
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1. The heat of vaporization of water is 540 cal/g, and the heat of fusion is 80 cal/g. The heat capacity of liquid water is 1 cal g-1°c-1, and the heat capacity of ice is 0.5 cal g-1 °c-1. What amount of heat is required to evaporate 20 g of water at 100 °C. ___ cal . 2. 28 g of ice at -10°c is heated until it becomes liquid water at 28°c. how much heat was required for this to occur? ___ cal
1. To evaporate 20 g of water at 100 °C, you need 10,800 cal.
2. To heat 28 g of ice from -10 °C to liquid water at 28 °C, you need 2,548 cal.
1. To evaporate 20 g of water, multiply the mass (20 g) by the heat of vaporization (540 cal/g):
20 g × 540 cal/g = 10,800 cal
2. For the ice, there are three steps:
a) Heating ice from -10 °C to 0 °C:
28 g × 0.5 cal/g°C × 10 °C = 140 cal
b) Melting ice at 0 °C:
28 g × 80 cal/g = 2,240 cal
c) Heating liquid water from 0 °C to 28 °C:
28 g × 1 cal/g°C × 28 °C = 784 cal
Total heat required: 140 cal + 2,240 cal + 784 cal = 3,164 cal
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18.Suppose the boy first runs a distance of 100 metres in 50 seconds in going from his home to the shop in the East direction, and then runs a distance of 100 metres again. in 50 seconds in the reverse direction from the shop to reach back home from where he started (see Figure).
(i) Find the speed of the boy.
(ii) Find the Velocity of the boy
(iii) A boy is sitting on a merry-go-round which is moving with a constant speed of 10m/s. This means that the boy is :
(iv) In which of the following cases of motion, the distance moved and the magnitude of displacement are equal ?
ANSWER IT ASAP!!!
The solutions are i) The speed of the boy is 2 m/s. ii) The velocity of the boy is 0 m/s. iii) The velocity is zero and the speed of the boy is 10 m/s. iv) In the case of rectilinear motion the distance and displacements are equal.
i) To find the speed of the boy we can directly use the speed, distance, and time formula that is:
Speed= distance/time
Here we can see that the boy covers a distance of 100 m back and forth so the total distance he covered is 100 m + 100 m = 200 m.
The time he took for the journey is 50 s each side so the total distance is 50 s + 50 s = 100s
Now substituting the values in the formula, we get:
Speed = 200 m / 100 s
Speed = 2 m/s
Therefore the speed of the boy is 2 m/s.
ii) The velocity is the vector quantity which means it indicates the speed of the boy in a particular direction. The velocity can be found by the formula:
Velocity = Displacement/Time
Now we can see that the initial and the final position of the boy are the same so there is no displacement, so displacement is 0.
Substituting the values into the formula we get
Velocity = 0 m/100 s
Velocity = 0m/s
Therefore the velocity of the boy is zero.
iii) According to the question the boy is just sitting on the merry-go-round and not changing his position with respect to the merry-go-round, his velocity is zero as there is no displacement. However, the merry-go-round is moving at a constant speed of 10 m/s, so the boy has a speed of 10 m/s with respect to the ground.
iv) When an object moves in a straight line. the distance moved and the magnitude of displacement are equal. So, in the case of rectilinear motion, the distance covered and the magnitude of the displacement are equal.
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The spaceship Enterprise, traveling through the galaxy, sends out a smaller explorer craft that travels to a nearby planet and signals its findings back. The proper time for the trip from the Enterprise to the planet is measured. On board of the Enterprise On board of the explorer craft On Earth Outside both the Enterprise and the explorer craft. At the center of the galaxy
The proper time for the trip to the planet can be measured by clocks (a),(b) on board the Enterprise and on board the explorer craft. These clocks will measure the time dilation effect of special relativity, which predicts that time will appear to run slower on objects that are moving relative to an observer.
Clocks on Earth and at the center of the galaxy will also measure the time of the trip, but their measurements will not include the effects of time dilation. Therefore, the measurements from these clocks will differ from the measurements of the clocks on board the Enterprise and the explorer craft.
The extent of the time dilation effect will depend on the speed of the craft relative to the observer, with greater time dilation occurring at higher speeds.
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Complete question :
The spaceship U.S.S. Enterprise, traveling through the galaxy, sends out a smaller explorer craft that travels to a nearby planet and signals its findings back. The proper time for the trip to the planet is measured by clocks: (Select all that can apply)
A. on board the Enterprise
B. on board the explorer craft
C. on Earth
D. at the center of the galaxy
E. none of the above
What element was used as a marker for an asteroid impact? A. Sodium B. Iridium OC. Germanium D. Uranium O E. Iron
Iridium was used as a marker for an asteroid impact.
Iridium, a rare element on Earth's surface, is abundant in the Earth's mantle and in asteroids and comets. The discovery of a layer enriched in iridium at the K-Pg boundary (the boundary between the Cretaceous and Paleogene periods) provided strong evidence for the impact of a large asteroid, around 10 kilometers in diameter, in what is now Mexico. The impact caused widespread devastation, including tsunamis, earthquakes, and wildfires, and resulted in the extinction of the dinosaurs and many other species. Iridium is also used as a tracer for other extraterrestrial events, such as meteorite impacts on the Moon and Mars, and for studying the formation and evolution of the solar system.
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When blue light of wavelength 450 nm falls on a single slit, the first dark bands on either side of center are separated by 57.0 degrees ∘.
Determine the width of the slit.
The width of the slit is approximately 3.03 × 10⁻⁵ meters.
What determines the slit width?To calculate the width of the slit, we can use the concept of diffraction. When light passes through a narrow slit, it diffracts and produces a pattern of bright and dark regions on a screen. The angle of separation between the dark bands can be used to determine the width of the slit.
In this case, the first dark bands on either side of the center are separated by an angle of 57.0 degrees.
We can use the formula for the angle of separation in a single-slit diffraction pattern: θ = λ / (w * sin(θ)), where λ is the wavelength of the light, w is the width of the slit, and θ is the angle of separation.
Rearranging the formula, we can solve for the width of the slit: w = λ / (sin(θ)). Substituting the given values, with the wavelength λ = 450 nm (or 4.5 × 10⁻⁷ meters) and the separation angle θ = 57.0 degrees, we can calculate the width of the slit as approximately 3.03 × 10⁻⁵ meters.
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From planet Mia, the angular size of the Sun is 0.8 degrees. The distance from Mia to Sun is 130000000 Km. What is the physical size (i.e. diameter) of the Sun? (please insert your answer in Km)
The physical size (diameter) of the Sun is approximately 104,000,000 km.
What is the angular size of an object?To find the physical size (diameter) of the Sun, we can use the concept of angular size and the given information.
The angular size of an object is the angle it subtends at the observer's location. We can use the formula:
Angular size = Physical size / Distance
In this case, the angular size of the Sun is given as 0.8 degrees, and the distance from Mia to the Sun is given as 130,000,000 km. We need to find the physical size (diameter) of the Sun.
Rearranging the formula, we have:
Physical size = Angular size * Distance
Plugging in the values:
Physical size = 0.8 degrees * 130,000,000 km
Calculating the result:
Physical size = [tex]1.04 × 10^8 km[/tex]
Therefore, the physical size (diameter) of the Sun is approximately 104,000,000 km.
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A bar magnet falls towards a metallic plate along the dashed line shown. How do the eddy currents move in the plate underneath it? Why?
Current moves radially away from the magnet.
Current moves radially toward the magnet.
Current will swirl clockwise under the magnet.
Current will swirl counterclockwise under the magnet.
No current will flow.
As the bar magnet approaches the plate, how does it move?
It accelerates atLaTeX: gg as usual.
It accelerates faster thanLaTeX: gg.
It accelerates slower thanLaTeX: gg.
It moves at constant velocity.
As the bar magnet falls towards the metallic plate along the dashed line, the eddy currents in the plate will swirl counterclockwise under the magnet.
This direction of current flow is known as Lenz's law, which states that the direction of the induced current will oppose the change in the magnetic field that caused it. When the magnet approaches the plate, the magnetic field through the plate increases.
To oppose this increase, the eddy currents will flow in a direction that creates a magnetic field that opposes the magnet's field. The counterclockwise current flow creates a magnetic field that repels the approaching magnet, slowing down its motion.
Regarding the motion of the bar magnet, it accelerates slower than 'g', the acceleration due to gravity. This is because as the magnet falls, it experiences an upward force due to the opposing eddy currents in the plate.
This force counteracts the force of gravity, resulting in a net force that is less than the force of gravity alone. Therefore, the magnet accelerates slower than 'g' during its fall toward the plate.
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A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00 x 10^2 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00 x 10^2 N force for an hour.
To find the charge the batteries must be able to move, we need to calculate the total work done by the car's motors, which is equal to the total energy required to perform the given tasks.
We can break down the problem into three parts: accelerating the car, lifting it to the top of the hill, and maintaining a constant speed against a resistive force.
Part 1: Accelerating the car
The work done in accelerating the car from rest to a speed of 25.0 m/s is given by:
[tex]W1 = (1/2) * m * v^2 = (1/2) * 750 kg * (25.0 m/s)^2 = 234,375 J[/tex]
Part 2: Lifting the car to the top of the hill
The work done in lifting the car to a height of 2.00 x 10² m against gravity is given by:
[tex]W2 = m * g * h = 750 kg * 9.81 m/s^2 * 2.00 x 10^2 m = 1.47 x 10^6 J[/tex]
Part 3: Maintaining constant speed against a resistive force
The work done in maintaining a constant speed of 25.0 m/s against a resistive force of 5.00 x 10² N for an hour (3600 seconds) is given by:
[tex]W3 = F * d = F * v * t = 5.00 x 10^2 N * 25.0 m/s * 3600 s = 4.50 x 10^7 J[/tex]
The total work done by the car's motors is the sum of these three parts:
[tex]W = W1 + W2 + W3 = 4.65 x 10^7 J[/tex]
The charge the batteries must be able to move is equal to the total energy required, divided by the voltage of the system:
[tex]Q = W / V = 4.65*10^7 J / 12.0 V=3.87*10^6 C[/tex]
Therefore, the batteries must be able to move a charge of approximately 3.87 x 10⁶ coulombs to perform the given tasks.
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A solid disk whose plane is parallel to the ground spins with an initial angular speed ω0ω0. Three identical blocks are dropped onto the disk at locations AA, BB, and CC, one at a time, not necessarily in that order. Each block instantaneously sticks to the surface of the disk, slowing the disk's rotation. A graph of the angular speed of the disk as a function of time is shown.
With reference from the graph, the order in which the blocks are dropped onto the disk is shown a s: C, B, A.
What is a graph?A graph can be described as as a pictorial representation or a diagram that represents data or values in an organized manner.
The graph is a graph of Angular speed of the disk vs time graph
From the graph, the disk is initially spinning at a constant angular speed of ω0ω0.
Then, as blocks are deposited onto the disk, the graph displays three separate times where the angular speed changes.
The order in which the blocks are dropped onto the disk can be inferred from the graph: Block C is first dropped at location P1 on the disk and here the angular speed of the disk begins to decrease.
Block B is then dropped onto the disk, at point P2 which causes the angular speed of the disk to decrease much further.
Block A is dropped onto the disk last, at point P3 causing the angular speed of the disk to decrease even further until it eventually reaches a constant value.
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astronomers proved that quasar 2c 856 contains a supermassive black hole when they discovered that its center is completely dark. T/F?
False. The statement that astronomers proved the presence of a supermassive black hole in quasar 2c 856 by observing its center to be completely dark is false.
Astronomers do not prove the presence of a supermassive black hole in a quasar by observing that its center is completely dark. In fact, quasars themselves are powered by supermassive black holes at their centers, which emit intense radiation as matter falls into them. Quasars are extremely bright and energetic objects located at the centers of galaxies. They emit enormous amounts of radiation across the electromagnetic spectrum, including visible light and beyond. The intense emission is due to the superheated matter falling into the black hole and the powerful jets of particles and energy it generates. Observations of a quasar typically reveal a bright and active center, not a completely dark one.
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