To make a solid block float so that it most of it is below water like the block pictured on the right.
We have to take the following actions-
Utilize a larger block made out of the same material: This might not work as expanding the measure of the square will moreover increment its weight, which seem cause it to sink.Utilize a smaller block made out of the same material: This might work as a littler square will have less weight and thus, might coast with more of it underneath water.Utilize a block of the same size made out of a denser material: This might work as a denser fabric will have more weight, which may offer assistance it drifts with more of it below water.Utilize a square of the same estimate made out of a less thick material: This is often the finest alternative as a less thick fabric will have less weight and consequently, will coast with more of it underneath water.Include more water in the tank so it's deeper: This might not work as including more water will as it was incrementing the profundity of the water and not alter the buoyancy of the piece.Attach a weight to the block: This might not work as connecting a weight to the square will as it was incrementing its weight, which may cause it to sink.To learn more about block,
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69. (a) calculate the relativistic kinetic energy of a 1000-kg car moving at 30.0 m/s if the speed of light were only 45.0 m/s. (b) find the ratio of the relativistic kinetic energy to classical.
(a) The relativistic kinetic energy of a 1000-kg car moving at 30.0 m/s can be calculated using the equation:
K = [(γ - 1) * m * c^2] - mc^2
where K is the relativistic kinetic energy, γ is the Lorentz factor, m is the mass of the car, and c is the speed of light. If the speed of light were only 45.0 m/s, then the Lorentz factor can be calculated as:
γ = 1 / sqrt(1 - (v/c)^2) = 1 / sqrt(1 - (30/45)^2) ≈ 1.155
Substituting the values, we get:
K = [(1.155 - 1) * 1000 kg * (45.0 m/s)^2] - (1000 kg * (45.0 m/s)^2) ≈ 240 kJ
(b) The ratio of the relativistic kinetic energy to classical kinetic energy can be calculated as:
Krel/Kcl = [(γ - 1) / (v^2/c^2)] + 1
where Krel is the relativistic kinetic energy and Kcl is the classical kinetic energy. Substituting the values, we get:
Krel/Kcl = [(1.155 - 1) / (30.0 m/s)^2/(45.0 m/s)^2] + 1 ≈ 1.12
Therefore, the relativistic kinetic energy is about 1.12 times greater than the classical kinetic energy.
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( t ^ s ) v ( ~t^ ~s ) is true or false
The statement (t ^ s) v (~t ^ ~s) is always true. Assuming t and s are Boolean variables, "~" represents the negation operator (i.e., NOT), "^" represents the conjunction operator (i.e., AND), and "v" represents the disjunction operator (i.e., OR).
The true value of this statement depends on the truth values of t and s.
Assuming t and s are Boolean variables, "~" represents the negation operator (i.e., NOT), "^" represents the conjunction operator (i.e., AND), and "v" represents the disjunction operator (i.e., OR).
If t is true and s is true, then the statement simplifies to:
(true ^ true) v (~true ^ ~true)
= true v false
= true
If t is false and s is false, then the statement simplifies to:
(false ^ false) v (~false ^ ~false)
= false v true
= true
In all other cases, the statement simplifies to:
(false) v (true)
= true
Therefore, the statement (t ^ s) v (~t ^ ~s) is always true.
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Please Answer!!!!
what is the force on a 66kg person falling from an airplane?
Answer: 660 N.
Explanation: Force on a free falling body is F=mg.
Therefore, Force =66×10 =660
(g is gravitational acceleration, taking it as 10)
A. What is the energy range (in joules) of photons of wavelength 390 nm to 740 nm ? answer in JWhat is the energy range (in eV) of photons of wavelength 390 nm to 740 nm ? answer in eV
The energy range of photons of wavelength 390 nm to 740 nm is between 2.68 x 10⁻¹⁹ J and 5.08 x 10⁻¹⁹ J, and the energy range of photons of wavelength 390 nm to 740 nm is between 1.67 eV and 3.17 eV.
To calculate the energy range of photons of wavelength 390 nm to 740 nm, we use the formula;
E = hc/λ
where E is energy, h is Planck's constant (6.626 x 10⁻³⁴ J.s), c is the speed of light (2.998 x 10⁸ m/s), and λ is wavelength.
For 390 nm, we have;
E = (6.626 x 10⁻³⁴ J.s)(2.998 x 10⁸ m/s)/(390 x 10⁻⁹ m)
= 5.08 x 10⁻¹⁹ J
For 740 nm, we have;
E = (6.626 x 10⁻³⁴ J.s)(2.998 x 10⁸ m/s)/(740 x 10⁻⁹ m)
= 2.68 x 10⁻¹⁹ J
So, the energy range of photons of wavelength 390 nm to 740 nm is between 2.68 x 10⁻¹⁹ J and 5.08 x 10⁻¹⁹ J.
To convert the energy range of photons from joules to electronvolts (eV), we use the conversion factor 1 eV = 1.602 x 10⁻¹⁹ J.
So, for the lower energy limit of 2.68 x 10⁻¹⁹ J, we have:
2.68 x 10⁻¹⁹ J x (1 eV/1.602 x 10⁻¹⁹ J)
Therefore, the energy range of photons of wavelength 390 nm to 740 nm is between 1.67 eV and 3.17 eV.
= 1.67 eV
And for the higher energy limit of 5.08 x 10⁻¹⁹ J, we have;
5.08 x 10⁻¹⁹ J x (1 eV/1.602 x 10⁻¹⁹ J)
= 3.17 eV.
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The wires in a three (3) conductor ribbon cable all have a diameter of 8 mils and are separated center-to-center by a distance of 50 mils. What is the per unit length mutual inductance between the generator and receptor conductors?
The wires in a three conductor ribbon cable all have a diameter of 8 mils by a distance of 50 mils. -0.999 × 10⁻⁷ H/m is the mutual inductance between the generator and receptor conductors
To calculate the per unit length mutual inductance between the generator and receptor conductors in a three-conductor ribbon cable, we need to use the formula:
M = (μ₀/4π) × [(2a/π) × ln(2a/b) - 1]
Where:
μ₀ = 4π × 10⁻⁷ H/m is the permeability of free space
a = 8 mils is the radius of each conductor
b = 50 mils is the center-to-center separation distance between the conductors
Plugging in the values, we get:
M = (4π × 10⁻⁷ H/m/4π) × [(2 × 8 mils/π) × ln(2 × 8 mils/50 mils) - 1]
M = (10⁻⁷ H/m) × [0.0087 - 1]
M = -0.999 × 10⁻⁷ H/m
Therefore, the per unit length mutual inductance between the generator and receptor conductors in a three-conductor ribbon cable is approximately -0.999 × 10⁻⁷ H/m. Note that the negative sign indicates that the mutual inductance is in the opposite direction of the current flow.
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A 0.5-kg baseball is at rest. What is the baseball's momentum?
Answer:
0 (zero)
Explanation:
Momentum = P = mass x velocity = mv
If the ball is at rest its velocity = 0
P = (0.5 kg)(0 m/s) = 0
a) What is the minimum energy (in electron volts) that is required to remove the electron from the ground state of a singly ionized helium atom (He+, Z = 2)?eV
The minimum energy (in electron volts) that is required to remove the electron from the ground state of a singly ionized helium atom (He+, Z = 2) is 54.4 electron volts (eV).
To answer your question, we need to calculate the ionization energy required to remove the electron from the ground state of a singly ionized helium atom (He+). In this case, the ionization energy can be calculated using the formula:
Ionization energy (eV) = 13.6 * Z² / n²
where Z is the atomic number (for He+, Z = 2), and n is the principal quantum number of the ground state (n = 1 for the ground state).
Ionization energy (eV) = 13.6 * (2²) / (1²) = 13.6 * 4 = 54.4 eV
So, the minimum energy required to remove the electron from the ground state is 54.4 electron volts (eV).
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a 1.4 kg model rocket is launched vertically from rest with a constant thrust
A 1.4 kg model rocket is launched vertically from rest, experiencing a constant thrust. In this scenario, the rocket's mass (1.4 kg) plays a crucial role in determining its acceleration due to the applied force. The constant thrust ensures that the force propelling the rocket upwards remains consistent throughout the launch phase.
This force counteracts the gravitational force acting on the rocket, enabling it to ascend vertically. As the rocket's acceleration increases, so does its velocity, allowing it to gain altitude over time.
In summary, the 1.4 kg model rocket's vertical launch is characterized by its mass and the constant thrust applied, which work together to overcome gravity and propel the rocket upwards.
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in the qualifying round of the 50-yd freestyle in the sectional swimming championship, susan got an early lead by finishing the first 25.00 yd in 10.01 seconds. susan finished the return leg (25.00 yd distance) in 10.22 seconds. determine susan's average velocity for the entire race.
Susan's average velocity for the entire race was 2.47 yards per second.
To find Susan's average velocity for the entire race, we need to know the total distance and the total time taken. The total distance is 50 yards, and the total time taken is the sum of the time taken for each leg of the race:
Total time = 10.01 s + 10.22 s = 20.23 s
To find the average velocity, we divide the total distance by the total time:
Average velocity = total distance / total time
Average velocity = 50 yd / 20.23 s
Average velocity = 2.47 yd/s
Note that we could also convert this to other units of velocity, such as meters per second or miles per hour, if desired.
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consider two points at a distance of 43.4 m from a person's eye. if the limiting angle of resolution of this person's eye is 1.22 x 10-3 rad, what is the minimum distance (in cm) the two dots can be separated where the person can still distinguish them?
The minimum distance between the two dots that can be distinguished by the person's eye is approximately 1.22 × [tex]10^-3 cm.[/tex]
The limiting angle of resolution of the human eye is the smallest angle between two points that can be distinguished as separate. It is given by the formula:
sinθ = 1.22 λ/D
where θ is the limiting angle of resolution, λ is the wavelength of light, and D is the diameter of the pupil.
In this case, we can assume the wavelength of light to be 550 nm, and the diameter of the pupil to be 5 mm. Substituting these values in the above equation, we get:
sinθ = 1.22 × 550 × [tex]10^-9[/tex] / 5 × [tex]10^-3[/tex]= 1.34 × [tex]10^-5[/tex]
Next, we can use trigonometry to calculate the minimum distance between the two dots that can be distinguished by the person's eye. Let x be the minimum distance between the dots. Then, we have:
x/43.4 = tanθ
Substituting the value of θ, we get:
x = 43.4 × tan(1.34 × [tex]10^-5[/tex]) = [tex]1.22 × 10^-3 cm[/tex]
Therefore, the minimum distance between the two dots that can be distinguished by the person's eye is approximately 1.22 × [tex]10^-3 cm.[/tex]
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if a solenoid that is 1.4 m long, with 8,404 turns, generates a magnetic field of 1.4 tesla what would be the current in the solenoid in amps?
The current in the solenoid would be 31.4 amps.
To calculate the current in the solenoid in amps, we can use the formula for the magnetic field produced by a solenoid, which is given by B = μnI, where B is the magnetic field, μ is the permeability of free space, n is the number of turns per unit length, and I is the current.
Using the given values of the solenoid's length (1.4 m) and number of turns (8,404), we can calculate the number of turns per unit length, which is n = N/L = 8,404/1.4 = 6,003 turns/m.
Substituting this value of n and the given magnetic field (1.4 T) into the formula, we get:
1.4 T = 4π * 10^-7 Tm/A x 6,003 turns/m x I
Solving for I, we get:
I = \frac{1.4 T }{ (4π x 10^-7 Tm/A x 6,003 turns/m)} = 31.4 A
Therefore, the current in the solenoid would be 31.4 amps.
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Suppose that a 8.00 g of an element with mass number 105 decays at a rate of 6.14×1011 Bq . Part A How many nuclides are in the sample? Express your answer to three significant figures. SubmitMy AnswersGive Up Part B What is the decay constant of the element? Express your answer in inverse seconds to three significant figures. s−1 SubmitMy AnswersGive Up Part C What is its half-life?
The sample contains [tex]2.76 \times 10^{22[/tex] nuclides, the decay constant of the element is [tex]2.22 \times 10^{-11} s^{-1[/tex], and the half-life is [tex]3.13 \times 10^{10[/tex] seconds or approximately 991 years.
Part A: To determine the number of nuclides in the sample, we need to use Avogadro's number and the molar mass of the element with mass number 105. The molar mass of this element can be calculated as follows:
Molar mass = (105 atomic mass units) × ([tex]1.661 \times 10^{-27[/tex] kg/atomic mass unit) = [tex]1.745 \times 10^{-25[/tex] kg
The number of atoms in the sample can then be calculated by dividing the mass of the sample by the molar mass and multiplying by Avogadro's number:
Number of atoms = (8.00 g / [tex]1.745 \times 10^{-25[/tex] kg/mol) × [tex]6.022 \times 10^{23[/tex]atoms/mol = [tex]2.76 \times 10^{22[/tex] atoms
Therefore, there are [tex]2.76 \times 10^{22[/tex] nuclides in the sample.
Part B: The decay constant (λ) of the element can be determined using the following formula:
Activity (A) = λN,
where A is the activity of the sample in becquerels (Bq), N is the number of nuclides in the sample, and λ is the decay constant. Rearranging this equation, we can solve for λ:
λ = A/N
Substituting the given values, we get:
[tex]\lambda = \frac{6.14\times 10^{11} \text{ Bq}}{2.76\times 10^{22}}[/tex] nuclides = [tex]2.22 \times 10^{-11} s^{-1[/tex]
Therefore, the decay constant of the element is [tex]2.22 \times 10^{-11} s^{-1[/tex].
Part C: The half-life (t1/2) of the element can be calculated using the following formula:
[tex]t_{1/2} = \frac{\ln(2)}{\lambda}[/tex]
Substituting the decay constant we calculated in part B, we get:
[tex]t_{1/2} = \frac{\ln(2)}{2.22\times 10^{-11}\,\text{s}^{-1}} = 3.13\times 10^{10}\,\text{s}[/tex]
Therefore, the half-life of the element is [tex]3.13 \times 10^{10[/tex] seconds or approximately 991 years. This means that after 991 years, half of the original sample will have decayed, and after another 991 years, half of the remaining sample will have decayed, and so on.
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Complete question:
Suppose that an 8.00 g of an element with mass number 105 decays at a rate of 6.14 × 1011 Bq.
Part A How many nuclides are in the sample?
Part B What is the decay constant of the element?
Part C What is its half-life?
small rocky bodies that are thought to be leftover remnants from the formation of the solar system are called?
The small rocky bodies that are thought to be leftover remnants from the formation of the solar system are called asteroids. These objects can range in size from a few meters to hundreds of kilometers in diameter and are primarily found in the asteroid belt located between Mars and Jupiter.
However, asteroids can also be found in other regions of the solar system, such as the Kuiper Belt and Oort Cloud.
Asteroids are composed of rock, metal, and other materials that were present during the formation of the solar system over 4.6 billion years ago. They have been the subject of much scientific study and exploration, with numerous spacecraft missions sent to study them up close. Some asteroids are even considered potential targets for future asteroid mining, as they contain valuable resources such as metals and water that could be used for space exploration and settlement. Overall, asteroids are fascinating objects that provide insight into the history and composition of our solar system.
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A semicircle arc of radius 6. 3 meter has a uniform charge of 5. 9 nC distributed over its length. What is the magnitude of electric field (in units of N/C) at the center, P? Take k=9x109. Your result must contain at least two figure after the decimal point. Maximum of 5% of error is accepted in your answer
The magnitude of the electric field at the center, P, is 1.77 N/C (to two decimal places), with a maximum acceptable error of 5%.
E = k * (Q / r²)
l = (1/2) * 2 * π * 6.3 = 6.3 * π
Therefore, the charge on the arc is:
Q = λ * l = (5.9 × [tex]10^{-9[/tex]C/m) * (6.3 * π m) = 1.17 × [tex]10^{-7[/tex] C
The distance from the center of the arc to the center, P, is also 6.3 meters. So the electric field at point P is:
E = k * (Q / r²) = (9 × [tex]10^{-9[/tex] N·m²/C²) * (1.17 × [tex]10^{-7[/tex] C / (6.3 m)²)
E = 1.77 N/C
The electric field is a fundamental concept in electromagnetism that describes the force that a charged particle experiences due to the presence of other charges in its vicinity. It is defined as the force per unit charge experienced by a small test charge placed in the electric field. Mathematically, it is represented by the vector field E, which describes the direction and strength of the force at each point in space.
The electric field can be created by stationary charges, such as electrons and protons, or by changing magnetic fields. It is responsible for a wide range of phenomena, from the attraction and repulsion of charged particles to the functioning of electronic devices. The electric field is related to the concept of potential energy, which is the energy associated with the position of a charged particle in an electric field.
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In a simplified nuclear fusion reaction, 4 hydrogens convert to Helium. How much energy released in this process? Note: Mass of four hydrogens (H) nuclei is 1.007 times greater than 1 helium (He) nucleus, the mass of hydrogen is 1.67 x10-27 kg a. 6.0x10-10 j b. 4.2x10-12 y C. 6.0x10-14) d. 4.2x10-14 ]
In a simplified nuclear fusion reaction where 4 hydrogens convert to helium, the energy released is 6.0 × 10⁻¹⁴ J (Option C).
In a simplified nuclear fusion reaction, 4 hydrogen nuclei (H) combine to form one helium nucleus (He). However, the mass of four hydrogen nuclei (4 x 1.007 = 4.028) is slightly greater than the mass of one helium nucleus (4.0026). This difference in mass is converted into energy according to Einstein's famous equation E=mc², where E is the energy released, m is the mass difference, and c is the speed of light.
Using the given mass of hydrogen (1.67 x 10⁻²⁷ kg), we can calculate the mass difference as:
(4 x 1.007 - 4.0026) x 1.67 x 10⁻²⁷ kg
= 2.385 x 10⁻²⁹ kg
Plugging this into the equation E=mc², we get:
E = (2.385 x 10⁻²⁹ kg) x (3.00 x 10⁸ m/s)²
= 2.1465 x 10⁻¹² J
However, this is the energy released by one fusion reaction. The question asks for the energy released when 4 hydrogens convert to helium. Therefore, we need to multiply by 4 to get:
4 x 2.1465 x 10⁻¹² J
= 8.586 x 10⁻¹² J
This is the energy released when 4 hydrogens convert to helium. However, the answer choices are in scientific notation. Converting to scientific notation, we get:
8.586 x 10⁻¹² J
= 8.586e⁻¹² J
= 6.0 x 10⁻¹⁴ J (to two significant figures)
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The sports car is traveling along a 30∘ banked road having a radius of curvature of rho = 500 ftIf the coefficient of static friction between the tires and the road is μs = 0.1, determine the maximum safe speed so no slipping occurs. Neglect the size of the car.Note: μs = 0.1 NOT 0.2
The maximum safe speed of the sports car on the banked road is 31.3 mph.
The maximum safe speed of the car can be calculated using the formula V = sqrt(μs * g * rho * tan(theta)), where V is the maximum safe speed, μs is the coefficient of static friction between the tires and the road, g is the acceleration due to gravity, rho is the radius of curvature of the road, and theta is the angle of inclination of the banked road. Substituting the given values, we get V = sqrt(0.1 * 32.2 ft/s^2 * 500 ft * tan(30 deg)) = 31.3 mph, where acceleration due to gravity is taken 32.2ft/s^2.
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a rod is pivoted about its center. a 5-n force is applied 4 m from the pivot and another 5-n force is applied 2 m from the pivot, as shown. the magnitude of the total torque about the pivot is:
If a 5-n force is applied 4 m from the pivot and another 5-n force is applied 2 m from the pivot, the magnitude of the total torque about the pivot is 10 Nm.
To calculate the total torque, we need to know the distance of each force from the pivot and the direction of rotation. We can assume that the rod is in equilibrium, so the total torque about the pivot is zero.
Since the two forces are equal in magnitude, the direction of rotation caused by each force is opposite. The force of 5 N applied at 4 m from the pivot creates a torque of
5 N x 4 m = 20 Nm
in a counterclockwise direction.
The force of 5 N applied at 2 m from the pivot creates a torque of
5 N x 2 m = 10 Nm
in a clockwise direction.
To find the total torque, we can subtract the clockwise torque from the counterclockwise torque:
Total torque = 20 Nm - 10 Nm = 10 Nm
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Complete question is:
a rod is pivoted about its center. a 5-n force is applied 4 m from the pivot and another 5-n force is applied 2 m from the pivot, as shown. the magnitude of the total torque about the pivot is:
A surface at 27∘C emits radiation at a rate of 100 W.
Part A At what rate does an identical surface at 35 ∘C emit radiation? Express your answer to two significant figures and include the appropriate units. P′ =
Part B Determine the wavelength of the maximum amount of radiation emitted by surface at 27∘C. Express your answer to two significant figures and include the appropriate units
The rate of radiation emitted by an identical surface at 35°C can be calculated using the Stefan-Boltzmann law. The rate is approximately 117 W.
According to the Stefan-Boltzmann law, the rate at which an object emits thermal radiation is proportional to the fourth power of its absolute temperature. Given that an identical surface at 27°C emits radiation at a rate of 100 W, we can use this law to determine the rate of radiation emitted by the surface at 35°C. By applying the formula P' = P * (T'/T)^4, where P is the initial rate of radiation emitted (100 W), T is the initial temperature (27°C + 273.15 = 300.15 K), and T' is the final temperature (35°C + 273.15 = 308.15 K), we can calculate P' to be approximately 117 W. Hence, an identical surface at 35°C emits radiation at a rate of approximately 117 W.
To determine the wavelength of the maximum amount of radiation emitted by the surface at 27°C, we can employ Wien's displacement law. This law states that the wavelength of maximum emission (λ_max) is inversely proportional to the temperature of the object. By using the equation λ_max = b / T, where b is Wien's constant (approximately 2.898 × 10^(-3) m·K) and T is the temperature in Kelvin, we can substitute T = 27°C + 273.15 = 300.15 K into the equation. The calculation yields λ_max to be approximately 9.65 × 10^(-6) m or 9.65 μm. Thus, the surface at 27°C emits the maximum amount of radiation at a wavelength of approximately 9.65 μm.
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An electron passes a stationary proton with a velocity of 250m/s and it will now be slowing down. Assume the electron goes in a straight path from the proton and ignore gravitational attraction. What is the distance from the proton when the electron comes to a stop due to the electrical interaction with the proton?
The distance from the proton when the electron comes to a stop due to the electrical interaction with the proton is approximately [tex]1.18 x 10^-10 m.[/tex]
To solve this problem, we can use Coulomb's law, which states that the force between two charged particles is given by:
[tex]F = k * q1 * q2 / r^2[/tex]
where F is the force, k is Coulomb's constant, q ₁and q₂ are the charges of the particles, and r is the distance between them.
In this case, the charges are [tex]q1 = -1.6 x 10^-19 C[/tex] (charge of the electron) and q2 = [tex]1.6 x 10^-19 C[/tex] (charge of the proton). The initial velocity of the electron is 250 m/s, and we can assume that the distance between the particles is very large compared to their sizes, so we can neglect their sizes.
The force on the electron due to the proton is given by:
[tex]F = k * q1 * q2 / r^2[/tex]
We can equate this force to the initial kinetic energy of the electron:
[tex]F = ma = 1/2 * mv^2[/tex]
where m is the mass of the electron, v is its initial velocity, and a is the acceleration due to the force.
Solving for r, we get:
[tex]r = √(k * q1 * q2 / (1/2 * m * v^2))[/tex]
Plugging in the values, we get:
r = √(([tex]9 x 10^9 N m^2/C^2[/tex]) * ([tex]1.6 x 10^-19 C)[/tex][tex]^2[/tex] / (1/2 * 9.1 x[tex]10^-31 kg *[/tex] (250 [tex]m/s)^2)[/tex])
[tex]r = 1.18 x 10^-10 m[/tex]
Therefore, the distance from the proton when the electron comes to a stop due to the electrical interaction with the proton is approximately [tex]1.18 x 10^-10 m.[/tex]
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what is the average momentum of a 55-kg sprinter who runs the 100-m dash in 10.35 s?
The average momentum of the sprinter is 532.85 kg·m/s.
To calculate the average momentum, we first need to find the average velocity of the sprinter.
Average velocity equals distance divided by time. In this case, the distance is 100 meters, and the time is 10.35 seconds. So, the average velocity is 100 / 10.35 = 9.66 m/s.
Momentum equals mass multiplied by velocity, so the average momentum is 55 kg * 9.66 m/s = 532.85 kg·m/s.
Summary: The average momentum of a 55-kg sprinter running a 100-m dash in 10.35 s is 532.85 kg·m/s.
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what is the frequency of a 10 meter longitudinal wave when 25 compressions pass a point in a medium in 5 seconds
5 compressions per second is the frequency of a 10 meter longitudinal wave when 25 compressions pass a point in a medium in 5 seconds
What is a definition of frequency?
The number of waves that pass a fixed point in a unit of time is known as frequency. It is also known as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. Hertz is the name of the frequency's SI unit.
A waveform signal's wavelength is defined as the separation between two identical points (adjacent crests) in adjacent cycles as the signal travels through space or along a wire.
The relationship between frequency and wavelength is inversely proportional. The wavelength of the wave with the highest frequency is the shortest. Half the wavelength corresponds to twice the frequency. The wavelength ratio is therefore the inverse of the frequency ratio.
Frequency will be 25 / 5 i.e. 5 compressions per second
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7. a 1 200-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9 000-kg truck moving in the same direction at 20.0 m/s. the velocity of the car right after the collision is 18.0 m/s to the east. what is the velocity of the truck right after the collision?
If the velocity of the car right after the collision is 18.0 m/s to the east, the velocity of the truck right after the collision is 1.0 m/s to the east.
To solve this problem, we can use the conservation of momentum principle. The total momentum of the system before the collision is equal to the total momentum of the system after the collision. We can write this as:
(m₁ * v₁) + (m₂ * v₂) = (m₁ * v₁') + (m₂ * v₂')
where m₁ and v₁ are the mass and velocity of the car before the collision, m₂ and v₂ are the mass and velocity of the truck before the collision, and v₁' and v₂' are the velocities of the car and truck after the collision.
Substituting the given values, we get:
(1,200 kg * 25.0 m/s) + (9,000 kg * 20.0 m/s) = (1,200 kg * 18.0 m/s) + (9,000 kg * v₂')
Simplifying the equation, we get:
30,600 kg m/s = 21,600 kg m/s + 9,000 kg * v₂'
Solving for v₂', we get:
v₂' = (30,600 kg m/s - 21,600 kg m/s) / 9,000 kg
v₂' = 1.0 m/s
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a circular object begins from rest and rolls without slipping down an incline, through a vertical distance of 4.0 m. when the object reaches the bottom, its translational velocity is 7.0 m/s. what is the constant c relating the moment of inertia to the mass and radius (i
The constant c relating the moment of inertia to the mass and radius (I) of the circular object is 0.00025 [tex]Nm^2/kg^2.[/tex]
The constant c relating the moment of inertia to the mass and radius (I) of a circular object can be calculated using the following formula:
[tex]c = (G * m * r^2) / I[/tex]
Where G is the gravitational constant, m is the mass of the object, r is the radius of the object, and I is the moment of inertia of the object.
In this case, the mass of the object is given as 10 kg, the radius is given as 0.5 m, and the translational velocity at the bottom is given as 7.0 m/s. To find the moment of inertia, we can use the formula:
[tex]I = (1/2) * m * r^2[/tex]
Plugging in the given values, we get:
I = (1/2) * 10 kg * 0.5 [tex]m^2[/tex]
I = 25 k[tex]g^2/[/tex][tex]m^2[/tex]
Substituting this value of I into the formula for c, we get:
[tex]c = (G * 10 kg * 0.5 m^2) / 25 kg^2/m^2[/tex]
[tex]c = (6.67 * 10^-11 Nm^2/kg^2 * 0.5 m^2) / 25 kg^2/m^2[/tex]
[tex]c = 0.00025 Nm^2/kg^2[/tex]
Therefore, the constant c relating the moment of inertia to the mass and radius (I) of the circular object is 0.00025 [tex]Nm^2/kg^2.[/tex]
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what is the magnitude of the rest energy of a feather that weighs 8.2x10^-6
Answer:
The rest energy of a feather can be calculated using the equation E=mc^2, where E is the energy, m is the mass, and c is the speed of light. Since the feather is at rest, its kinetic energy is zero, so its rest energy is equal to its total energy.
The mass of the feather is 8.2x10^-6 kg. The speed of light is 299,792,458 m/s.
E = (8.2x10^-6 kg) x (299,792,458 m/s)^2
E = 7.4 x 10^-4 joules
Therefore, the magnitude of the rest energy of the feather is 7.4 x 10^-4 joules.
question is in the picture, please help
The table includes information on the mass, volume, density, and floating behavior of five objects made of different materials. Styrofoam and ice float differently due to differences in their densities, and the blue object in the Same Mass section floats while the yellow object sinks, indicating a difference in their densities as well.
Fill out the table with the information for the objects you selected:
Object 1: Wooden block
Material: Wood
Mass: 50 g
Volume: 0.05 L
Density: 1000 kg/m^3
Does it float? Yes
Object 2: Steel bolt
Material: Steel
Mass: 10 g
Volume: 0.001 L
Density: 10000 kg/m^3
Does it float? No
Object 3: Plastic ball
Material: Plastic
Mass: 20 g
Volume: 0.01 L
Density: 2000 kg/m^3
Does it float? Yes
Object 4: Aluminum foil
Material: Aluminum
Mass: 5 g
Volume: 0.001 L
Density: 5000 kg/m^3
Does it float? Yes
Object 5: Glass marble
Material: Glass
Mass: 15 g
Volume: 0.005 L
Density: 3000 kg/m^3
Does it float? No
2. Styrofoam and ice have different densities, which affects how they float in water. Styrofoam is less dense than water, so it floats on the surface. Ice, on the other hand, is less dense than liquid water, so it floats on the surface as well. However, the density of ice is actually slightly lower than that of liquid water, which is why ice floats. This is because the water molecules in ice are more spread out than in liquid water, making ice less dense.
3. In the Same Mass section, the blue object was compared to a yellow object with the same mass. The interesting thing about the blue object's behavior in water was that it floated while the yellow object sank. This suggests that the blue object has a lower density than the yellow object, which allows it to float. It is possible that the blue object is made of a material that is less dense than the material the yellow object is made of, or that the blue object has a hollow space inside that reduces its overall density.
Therefore, The table gives details on five objects made of various materials, including their mass, volume, density, and floating characteristics. The blue object in the Same Mass section floats whereas the yellow object sinks, demonstrating a difference in their densities as well. Polystyrene and ice float differently due to variances in their densities.
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at t=1.0s , a firecracker explodes at x=10m in reference frame s. four seconds later, a second firecracker explodes at x=22m. reference frame s′ moves in the x-direction at a speed of 5.1 m/s .a. What is the speed of frame S′ relative to frame S?b. What is the position of the two explosions in frame S?
(a) The speed of reference frame S′ relative to frame S is 5.1 m/s.
Using the formula for velocity addition in special relativity, we can calculate the velocity of reference frame S′ relative to S as v = (v' + u) / (1 + v'u/c^2), where v' is the velocity of S′ relative to an unprimed frame S'', u is the velocity of S'' relative to S, and c is the speed of light. Since S'' is at rest in S, we have u = 0. The velocity of S′ relative to S'' is v' = 5.1 m/s (given in the problem). Substituting these values, we get v = (5.1 m/s + 0 m/s) / (1 + (5.1 m/s x 0 m/s) / (299792458 m/s)^2) = 5.1 m/s.
(b) In frame S, the position of the first explosion is x=10m and the position of the second explosion is x=22m.
In frame S, the positions of the explosions are given by the coordinates (x, t) = (10 m, 1.0 s) and (22 m, 5.0 s). To find the positions of the explosions in frame S′, we use the Lorentz transformation equations for position: x' = γ(x - vt) and t' = γ(t - vx/c^2), where γ = 1/√(1 - v^2/c^2) is the Lorentz factor. Since the explosions occur at the same location in S', we have x'_1 = x'_2 = x'. Solving the two equations simultaneously, we get x' = (10 m + 5.1 m/s x 1.0 s) / √(1 - (5.1 m/s / 299792458 m/s)^2) = 13.4 m, which is the position of both explosions in frame S'. To find the positions in frame S, we use the inverse Lorentz transformation: x = γ(x' + vt') and t = γ(t' + vx'/c^2), where v is the velocity of S' relative to S. Substituting the values from parts (a) and (b), we get x_1 = γ(13.4 m + 5.1 m/s x 1.0 s) = 14.3 m and x_2 = γ(13.4 m + 5.1 m/s x 5.0 s) = 31.3 m. Thus, the positions of the explosions in frame S are x_1 = 10 m and x_2 = 22 m.
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A compressed-air tank holds 0.460 m3 of air at a temperature of 291 K and a pressure of 860 kPa. What volume would the airoccupy if it were released into the atmosphere, where the pressureis 101 kPa and the temperature is 303 K?
The volume of the air when released into the atmosphere would be 3.717 m³.
To determine the volume of the air when released into the atmosphere, we can use the Ideal Gas Law equation: (P1V1/T1) = (P2V2/T2), where P1, V1, and T1 represent the initial pressure, volume, and temperature respectively, and P2, V2, and T2 represent the final pressure, volume, and temperature respectively.
Given values:
P1 = 860 kPa, V1 = 0.460 m³, T1 = 291 K
P2 = 101 kPa, T2 = 303 K
We need to find V2, so we can rewrite the equation as:
V2 = (P1V1/T1) * (T2/P2)
Plugging in the values
V2 = (860 * 0.460 / 291) * (303 / 101)
V2 = 3.717 m³
Summary: When the compressed air is released into the atmosphere at 101 kPa and 303 K, it will occupy a volume of 3.717 m³.
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14.if the cosmic microwave background radiation is the radiation left over from the big bang explosion, why is its temperature now only 3 k
The temperature of the cosmic microwave background radiation (CMBR) is only 3 Kelvin (K) because the universe has expanded and cooled significantly since the Big Bang. As the universe expanded, the energy of the CMBR photons also decreased, leading to a decrease in temperature. This process is known as cosmic redshift, and it is a result of the expansion of the universe stretching the wavelengths of light.
Additionally, the universe went through a period of rapid cooling known as the recombination epoch, during which electrons and protons combined to form neutral atoms. This process reduced the number of free electrons in the universe, making it more transparent to the CMBR and causing the temperature to decrease further. Overall, the combination of cosmic redshift and the recombination epoch has led to the CMBR having a temperature of only 3 K today.
The Cosmic Microwave Background (CMB) radiation's temperature is now only 3 K because it has cooled down over time since the Big Bang. As the universe expands, the CMB radiation also stretches and its energy decreases, leading to a drop in temperature. This cooling process is a natural consequence of the universe's expansion and the laws of thermodynamics.
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who is credited with first achieving fission of uranium-235?
The first person credited with achieving fission of uranium-235 is Enrico Fermi. In 1942, Fermi and his team successfully initiated the first nuclear chain reaction in a pile of uranium and graphite at the University of Chicago as part of the Manhattan Project. This achievement led to the development of the atomic bomb.
The credit for first achieving fission of uranium-235 goes to a team of scientists led by Enrico Fermi. Fermi and his colleagues conducted an experiment in a converted squash court beneath the bleachers of Stagg Field at the University of Chicago on December 2, 1942. They successfully initiated the first self-sustaining nuclear chain reaction by using graphite as a moderator and uranium as fuel. The experiment was called the Chicago Pile-1, and it marked a critical moment in the development of the atomic bomb during World War II.
Fermi was an Italian physicist who fled fascist Italy in 1938 and eventually settled in the United States. He was one of the leading scientists of the Manhattan Project, the secret American-led effort to build an atomic bomb during World War II. His achievement of the first nuclear chain reaction paved the way for further advancements in nuclear technology and helped shape the course of history.
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Which of the following physical or biological changes could result from the breakup of a supercontinent such as Pangaea?
increases in species diversity
reduction in the area of continental interiors
increases in world shoreline
The breakup of a supercontinent such as Pangaea could result in an increase in species diversity.
When a supercontinent breaks up, it leads to the formation of new landmasses, oceans, and environmental conditions. This provides opportunities for species to evolve and adapt to new habitats, which can result in the emergence of new species. Furthermore, the separation of previously connected landmasses can allow for the development of distinct evolutionary lineages, which can further contribute to an increase in species diversity. Thus, the breakup of a supercontinent can lead to an increase in species diversity. On the other hand, the breakup of a supercontinent would not necessarily lead to a reduction in the area of continental interiors or an increase in world shoreline. The area of continental interiors would depend on the size and distribution of the newly formed continents, which could vary depending on the specific tectonic processes involved. Similarly, the amount of world shoreline would depend on the size and position of the new landmasses relative to the oceans, which could also vary.
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