The phase velocity can be expressed in terms of m, c, t, and k as Up = c(1 + (m²c²)/(hc²k²)) where p is the momentum of the particle and m is its rest mass.
For a relativistic particle, we can write the energy as E = pc + mc² where p is the momentum of the particle and m is its rest mass. The de Broglie relations for a matter wave are E = hν and p = h/λ, where h is Planck's constant, ν is the frequency of the wave, and λ is its wavelength.The phase velocity, Up is given by:Up = E/p= (pc + mc²) / p= c + (m²c⁴)/p²Using the de Broglie relation p = h/λ, we can express the momentum in terms of wavelength:p = h/λSubstituting this in the expression for phase velocity:Up = c + (m²c⁴)/(h²/λ²) = c + (m²c²λ²)/h²The wavelength of the matter wave can be expressed in terms of its frequency using the speed of light c:λ = c/fSubstituting this in the expression for phase velocity:Up = c + (m²c²/c²f²)h²= c[1 + (m²c²)/(c²f²)h²]= c(1 + (m²c²)/(hc²k²))where f = ν is the frequency of the matter wave and k = 2π/λ is its wave vector. So, the phase velocity can be expressed in terms of m, c, t, and k as Up = c(1 + (m²c²)/(hc²k²)).
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Why is it use or found in our every lives or certain in the industries?and identify and explain at least two uses
Integral calculus is a branch of mathematics that deals with the properties and applications of integrals. It is used extensively in many fields of science, engineering, economics, and finance, and has become an essential tool for solving complex problems and making accurate predictions.
One reason why integral calculus is so prevalent in our lives is its ability to solve optimization problems. Optimization is the process of finding the best solution among a set of alternatives, and it is important in many areas of life, such as engineering, economics, and management. Integral calculus provides a powerful framework for optimizing functions, both numerically and analytically, by finding the minimum or maximum value of a function subject to certain constraints.
Another use of integral calculus is in the calculation of areas, volumes, and other physical quantities. Many real-world problems involve computing the area under a curve, the volume of a shape, or the length of a curve, and these computations can be done using integral calculus. For example, in engineering, integral calculus is used to calculate the strength of materials, the flow rate of fluids, and the heat transfer in thermal systems.
In finance, integral calculus is used to model and analyze financial markets, including stock prices, bond prices, and interest rates. The Black-Scholes formula, which is used to price options, is based on integral calculus and has become a standard tool in financial modeling.
Overall, integral calculus has numerous applications in various fields, and its importance cannot be overstated. Whether we are designing new technologies, predicting natural phenomena, or making investment decisions, integral calculus plays a crucial role in helping us understand and solve complex problems.
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Find the diffusion coefficients of holes and electrons for germanium at un 300 K. The carrier Mobilities in cm²/ V. Sec Mp at 300 K for electrons and holes are respectively 3600 and 1700. Density of carriers is 2.5 x 1013. Boltzman constant, K = 1.38 x 10-23 j/ K
The diffusion coefficient of electrons is 0.037 m²/sec, and the diffusion coefficient of holes is 0.018 m²/sec.
Given:
Electron mobility, μn = 3600 cm²/ V.sec
Hole mobility, μp = 1700 cm²/ V.sec
Density of carriers, n = p = 2.5 x 10¹³cm⁻³
Boltzmann constant, k = 1.38 x 10⁻²³ J/K
Temperature, T = 300 K
We have to calculate the diffusion coefficients of holes and electrons for germanium.
The relationship between mobility and diffusion coefficient is given by:
D = μkT/q
where D is the diffusion coefficient,
μ is the mobility,
k is the Boltzmann constant,
T is the temperature, and
q is the elementary charge.
Therefore, the diffusion coefficient of electrons,
De = μnekT/q
= (3600 x 10⁻⁴ m²/V.sec) x (1.38 x 10⁻²³ J/K) x (300 K)/(1.6 x 10⁻¹⁹ C)
= 0.037 m²/sec
Similarly, the diffusion coefficient of holes,
Dp = μpekT/q
= (1700 x 10⁻⁴ m²/V.sec) x (1.38 x 10⁻²³ J/K) x (300 K)/(1.6 x 10⁻¹⁹ C)
= 0.018 m²/sec
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A 400-kg box is lifted vertically upward with constant velocity by means of two cables pulling at 50.0° up from the horizontal direction. What is the tension in each cable?
The tension in each cable used to lift the 400-kg box vertically upward, we can use the equilibrium condition and resolve the forces in the vertical and horizontal directions.
Let's denote the tension in each cable as T₁ and T₂.In the vertical direction, the net force is zero since the box is lifted with constant velocity. The vertical forces can be represented as:
T₁sinθ - T₂sinθ - mg = 0, where θ is the angle of the cables with the horizontal and mg is the weight of the box. In the horizontal direction, the net force is also zero:
T₁cosθ + T₂cosθ = 0
Given that the weight of the box is mg = (400 kg)(9.8 m/s²) = 3920 N and θ = 50.0°, we can solve the system of equations to find the tension in each cable:
T₁sin50.0° - T₂sin50.0° - 3920 N = 0
T₁cos50.0° + T₂cos50.0° = 0
From the second equation, we can rewrite it as:
T₂ = -T₁cot50.0°
Substituting this value into the first equation, we have:
T₁sin50.0° - (-T₁cot50.0°)sin50.0° - 3920 N = 0
Simplifying and solving for T₁:
T₁ = 3920 N / (sin50.0° - cot50.0°sin50.0°)
Using trigonometric identities and solving the expression, we find:
T₁ ≈ 2826.46 N
Finally, since T₂ = -T₁cot50.0°, we can calculate T₂:
T₂ ≈ -2826.46 N * cot50.0°
Therefore, the tension in each cable is approximately T₁ ≈ 2826.46 N and T₂ ≈ -2202.11 N.
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Explain it pleaseTwo particles of charge Q are located inside a box. One is at the box center while the other is halfway to one of the corners. Determine the electric flux through the box.
Answer: charge enclosed over epsilon not gives
The electric flux through the box is determined by the charge enclosed within the box divided by the permittivity of free space (ε₀). In this scenario, we have two particles of charge Q, with one located at the center of the box and the other halfway to one of the corners.
Since the charge at the center of the box is equidistant from all sides, it will produce an equal flux through each face of the box. On the other hand, the charge halfway to one of the corners will only contribute to the flux through one face of the box.
Therefore, the total electric flux through the box is given by the charge enclosed, which is the sum of the charges of both particles (2Q), divided by the permittivity of free space (ε₀). Mathematically, it can be expressed as:
Electric Flux = (2Q) / ε₀.
This equation signifies that the electric flux through the box is directly proportional to the total charge enclosed within it. The permittivity of free space (ε₀) is a constant that relates to the ability of the electric field to propagate through a vacuum.
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In a photoelectric effect experiment, a metal with a work function of 1.4 eV is used.
What is the maximum wavelength of light that can be used to free electrons from the metal?
Enter your answer in micrometres (10-6 m) to two decimal places but do not enter the units in your response.
The energy of a photon of light is given by
E = hc/λ,
where
h is Planck's constant,
c is the speed of light and
λ is the wavelength of the light.
The photoelectric effect can occur only if the energy of the photon is greater than or equal to the work function (φ) of the metal.
Thus, we can use the following equation to determine the maximum wavelength of light that can be used to free electrons from the metal:
hc/λ = φ + KEmax
Where KEmax is the maximum kinetic energy of the electrons emitted.
For the photoelectric effect,
KEmax = hf - φ
= hc/λ - φ
We can substitute this expression for KEmax into the first equation to get:
hc/λ = φ + hc/λ - φ
Solving for λ, we get:
λmax = hc/φ
where φ is the work function of the metal.
Substituting the given values:
Work function,
φ = 1.4 e
V = 1.4 × 1.6 × 10⁻¹⁹ J
= 2.24 × 10⁻¹⁸ J
Speed of light, c = 3 × 10⁸ m/s
Planck's constant,
h = 6.626 × 10⁻³⁴ J s
We get:
λmax = hc/φ
= (6.626 × 10⁻³⁴ J s)(3 × 10⁸ m/s)/(2.24 × 10⁻¹⁸ J)
= 8.84 × 10⁻⁷ m
= 0.884 µm (to two decimal places)
Therefore, the maximum wavelength of light that can be used to free electrons from the metal is 0.884 µm.
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A 1.41 kg bowling trophy is held at arm's length, a distance of 0.645 m from the shoulder joint.(1) What torque does the trophy exert about the shoulder if the arm is horizontal? τ = (?) N⋅m
(2) What torque does the trophy exert about the shoulder if the arm is at an angle of 20.5 ∘ below the horizontal? τ = (?) N⋅m
1. The torque exerted by the trophy about the shoulder joint when the arm is horizontal The torque is the product of the magnitude of the force applied and the perpendicular distance from the line of action of the force to the axis of rotation. We have to first figure out the force acting on the trophy.
The force acting on the trophy is equal to the weight of the trophy.
= mg
= (1.41)(9.81)
= 13.8321 N
= r × FW
= force acting on the
= distance between the shoulder joint and the trophyτ
= rFW
= (0.645)(13.8321
= 8.913 N⋅m2.
= r × F × sinθWhere,θ
= 20.5ºr
= 0.645 mF
= 13.8321 Nτ
= (0.645)(13.8321)sin(20.5º)τ
= 3.60 N⋅mThe torque exerted by the trophy about the shoulder if the arm is horizontal is 8.913 N⋅m.
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A baseball player is running with a speed of 7 m/s towards home base. The player slides the final 5 meters and comes to a stop, directly over the plate. What is the approximate coefficient of friction
The approximate coefficient of friction is approximately -0.25.
The force of kinetic friction can be calculated using the equation [tex]F_{friction} = \mu_k N[/tex], where [tex]F_{friction}[/tex] is the force of kinetic friction, [tex]\mu_k[/tex] is the coefficient of kinetic friction, and N is the normal force.
In this scenario, the player comes to a stop, indicating that the force of kinetic friction is equal in magnitude and opposite in direction to the force exerted by the player.
We know that the player's initial velocity is 7 m/s and the distance covered while sliding is 5 meters.
To calculate the deceleration (negative acceleration) experienced by the player, we can use the equation [tex]v^2 = u^2 + 2as[/tex]
where v is the final velocity (0 m/s), u is the initial velocity (7 m/s), a is the acceleration, and s is the displacement (5 meters).
Rearranging the equation, we have [tex]a=\frac{v^{2}-u^{2} }{2s}[/tex].
Plugging in the given values, we get [tex]a=\frac{0-(7^2)}{2\times 5} =-2.45 m/s^2[/tex].
Since the force of friction opposes the player's motion, we can assume it has the same magnitude as the force that brought the player to a stop. This force is given by the equation
[tex]F_{friction} = ma[/tex], where m is the mass of the player.
The normal force acting on the player is equal to the player's weight, N = mg, where g is the acceleration due to gravity.
Now, we can substitute the values into the equation [tex]F_{friction} = \mu_kN[/tex]and solve for the coefficient of kinetic friction:
[tex]ma = \mu_k mg[/tex].
The mass of the player cancels out, leaving us with [tex]a = \mu_k g[/tex].
Substituting the calculated acceleration and the acceleration due to gravity, we have [tex]-2.45 m/s^2 = \mu_k 9.8 m/s^2[/tex].
Solving for [tex]\mu_k[/tex], we find [tex]\mu_k = \frac{(-2.45)}{(9.8)} \approx -0.25[/tex].
Thus, the approximate coefficient of friction is approximately -0.25.
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What is the strength of the electric field between two parallel
conducting plates separated by 1.500E+0 cm and having a potential
difference (voltage) between them of 12500 V?
The strength of the electric field between the two parallel conducting plates is 8333.33 V/m.
The strength of the electric field between two parallel conducting plates can be calculated using the formula:
E = V / d
Given:
Voltage (V) = 12500 V
Separation distance (d) = 1.500E+0 cm = 1.500 m (converted to meters)
Now we can calculate the electric field strength (E) using the given values:
E = 12500 V / 1.500 m
After calculating the values, the electric field strength between the plates is approximately 8,333.33 V/m.
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A millisievert is equivalent to
A) I rem B) 0.1 rem
: D) 0.001 re C) 0.01 rem
A millisievert is equivalent to 0.1 rem. A rem is an acronym for Roentgen equivalent man, and it is used to measure the dosage of radiation in humans.
A millisievert, abbreviated as mSv, is a measure of the amount of radiation that a person is exposed to. It is a measure of the dose of ionizing radiation in the International System of Units (SI).The millirem (mrem) is a unit of measurement that is used in the United States of America to measure radiation exposure in humans. One rem is equivalent to 1000 millirems (mrem), while one millisievert (mSv) is equal to 100 rem or 100000 millirems. Therefore, one millirem is equal to 0.001 rem. When we convert this to millisieverts, we get one millisievert is equivalent to 0.1 rem.
So the answer to the question is B) 0.1 rem.The millisievert unit is used globally to calculate the dose of ionizing radiation in a person. The value of radiation dose that is considered acceptable varies depending on the country and the purpose of exposure. It is important to be aware of the risks associated with exposure to ionizing radiation to maintain good health.Thus, the answer to the given question is option B) 0.1 rem.A millisievert is a measure of the amount of radiation that a person is exposed to, which is used in the International System of Units (SI). A millirem (mrem) is a unit of measurement used in the United States to quantify radiation exposure in humans.One rem is equivalent to 1000 millirems (mrem), or 100000 millirems is equivalent to 1 millisievert (mSv). As a result, 0.001 rem is equivalent to 1 millirem (mrem), and 0.1 rem is equivalent to 1 millisievert (mSv).
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ASAP
If it takes 40 J of energy to heat a block from 10° to 25°C, what is the specific heat of the material? (m = 8g) O 0.33J/g C O 1.66J/g C O 1.33J/C
To find the specific heat of the material, we can use the equation Q = mcΔT, where Q is the energy transferred, m is the mass of the material, c is the specific heat, and ΔT is the change in temperature. Rearranging the equation, we can solve for c.
The specific heat of a material represents the amount of heat energy required to raise the temperature of a given mass of the material by one degree Celsius.
In this problem, we are given the energy transfer (Q) of 40 J, the mass (m) of 8 g, and the change in temperature (ΔT) of 25°C - 10°C = 15°C.
Using the equation Q = mcΔT, we can substitute the given values and solve for the specific heat (c). Rearranging the equation, we have c = Q / (mΔT).
Substituting the values, we have c = 40 J / (8 g * 15°C).
Calculating the specific heat, we find c = 0.33 J/g°C.
Therefore, the specific heat of the material is 0.33 J/g°C.
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a
40uF capacitor is connected in series with 2.0K ohm resistor across
a 100-V DC source and a switch. what is the time constant of this
RC circuit
The time constant of the RC circuit is τ = 8.0 × 10^-2 s
A 40uF capacitor is connected in series with 2.0K ohm resistor across a 100-V DC source and a switch.
We need to find the time constant of this RC circuit.
Let's solve this problem step by step.
Step 1: Identify the formula for the time constant of an RC circuit
Time constant of an RC circuit is given by the formula
τ = RC,
where
R is the resistance in ohms
C is the capacitance in farads.
Step 2: Identify the values of resistance and capacitance from the given circuit
The given circuit contains a 40μF capacitor and a 2.0KΩ resistor.
Step 3: Convert the units of capacitance to farads
From the question, capacitance is given as 40 μF.
We know that 1 farad = 1,000,000 microfarads, which means:1 μF = 10^-6 F
Therefore, the capacitance of the circuit is:
C = 40 × 10^-6 F
= 4 × 10^-5 F
Step 4: Substitute the given values of resistance and capacitance into the formula
τ = RC
= (2.0 × 10^3 Ω) × (4 × 10^-5 F)
= 8.0 × 10^-2 s
Step 5: Write the final answer
The time constant of the RC circuit is τ = 8.0 × 10^-2 s.
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Following the rules of significant digits, which of the following is the correct answer for the following calculation: 19.58 m x 3.15 m = ?
61.677 m2
61.68 m2
61.7 m2
62 m2
we round down to 1.8, which gives us 61.8 m² as the final answer.
Since the product should have four significant figures, round the answer to 61.8 m². This is because the last significant figure in the answer is 3, which is less than 5.
Significant figures or digits are the number of meaningful digits in a number. The following calculation is being carried out using significant figures: 19.58 m x 3.15 m = ? To follow the rules of significant digits, we need to identify the least number of significant figures in the equation. In this case, we have two , factors 19.58 m and 3.15 m. Since both factors have four significant figures, the product should also have four significant figures.
Therefore, the correct answer is 61.8 m². To get the answer, multiply the two factors as follows:
19.58 m × 3.15 m = 61.743 m²
This is because the last significant figure in the answer is 3, which is less than 5.
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Answer:
The answer is 61.7 m^2.
Explanation:
To solve this problem, you need to look at the numbers 19.58 and 3.15. 3.15 has the least value, so you use the amount of digits it has. It has three digits, so you know the answer to the multiplication problem will also have three digits.
Right away, you may realize that rules out all but one answer choice. We still need to check it though to make sure it lines up.
19.58 *3.15
= 61.677
Now, because we know the answer can only have three digits, we need to round the six after the decimal point. Seven is more than five, so the six gets bumped up to a 7. Everything after the newly created 7 turns to zeros and are forgotten.
Now, we have 61.7 m^2.
So, in short, the answer is 61.7 m^2
Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube has cross-sectional area A1 of 10.9 cm2, and the right arm has a cross-sectional area A2 of 5.90 cm2. Three hundred grams of water are then poured into the right arm as shown in Figure b.
Figure (a) shows a U-shaped tube filled with mercury. Both arms of the U-shaped tube are vertical. The left arm with cross-sectional area A1 is wider than the right arm with cross-sectional area A2. The height of the mercury is the same in both arms. Figure (b) shows the same U-shaped tube, but now most of the right arm is filled with water. The height of the column of water in the right arm is much greater than the height of the column of mercury in the left arm. The height of the mercury in the left arm is greater than the height of the mercury in the arms in Figure (a), and the difference in height is labeled h.
(a) Determine the length of the water column in the right arm of the U-tube.
cm
(b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?
cm
The mercury rises by 0.53 cm in the left arm of the U-tube. The length of the water column in the right arm of the U-tube can be calculated as follows:
Water Column Height = Total Height of Right Arm - Mercury Column Height in Right Arm
Water Column Height = 20.0 cm - 0.424 cm = 19.576 cm
The mercury rises in the left arm of the U-tube because of the difference in pressure between the left arm and the right arm. The pressure difference arises because the height of the water column is much greater than the height of the mercury column. The difference in height h can be calculated using Bernoulli's equation, which states that the total energy of a fluid is constant along a streamline.
Given,
A1 = 10.9 cm²
A2 = 5.90 cm²
Density of Mercury, ρ = 13.6 g/cm³
Mass of water, m = 300 g
Now, let's determine the length of the water column in the right arm of the U-tube.
Based on the law of continuity, the volume flow rate of mercury is equal to the volume flow rate of water.A1V1 = A2V2 ... (1)Where V1 and V2 are the velocities of mercury and water in the left and right arms, respectively.
The mass flow rate of mercury is given as:
m1 = ρV1A1
The mass flow rate of water is given as:
m2 = m= 300g
We can express the volume flow rate of water in terms of its mass flow rate and density as follows:
ρ2V2A2 = m2ρ2V2 = m2/A2
Substituting the above expression and m1 = m2 in equation (1), we get:
V1 = (A2/A1) × (m2/ρA2)
So, the volume flow rate of mercury is given as:
V1 = (5.90 cm²/10.9 cm²) × (300 g)/(13.6 g/cm³ × 5.90 cm²) = 0.00891 cm/s
The volume flow rate of water is given as:
V2 = (A1/A2) × V1
= (10.9 cm²/5.90 cm²) × 0.00891 cm/s
= 0.0164 cm/s
Now, let's determine the height of the mercury column in the left arm of the U-tube.
Based on the law of conservation of energy, the pressure energy and kinetic energy of the fluid at any point along a streamline is constant. We can express this relationship as:
ρgh + (1/2)ρv² = constant
Where ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid column, and v is the velocity of the fluid.
Substituting the values, we get:
ρgh1 + (1/2)ρv1² = ρgh2 + (1/2)ρv2²
Since h1 = h2 + h, v1 = 0, and v2 = V2, we can simplify the above equation as follows:
ρgh = (1/2)ρV2²
h = (1/2) × (V2/V1)² × h₁
h = (1/2) × (0.0164 cm/s / 0.00891 cm/s)² × 0.424 cm
h = 0.530 cm = 0.53 cm (rounded to two decimal places)
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You push a 25-kg block 10 m along a horizontal floor at constant speed. Your force F is directed 30
degrees below the horizontal. The coefficient of kinetic friction between the block and floor is 0.1.
a. How much work did you do on the block? (Hint: first you need to calculate your applied force
F.)
b. How much thermal (i.e. wasted) energy was dissipated in the process?
c. Are there any non-conservative forces at work in this problem?
The force of friction is a non-conservative force, since it depends on the path taken by the block.
The given values are the mass of the block m = 25-kg, the distance it was pushed along the floor d = 10 m, the coefficient of kinetic friction between the block and the floor μk = 0.1 and the angle that the force was directed below the horizontal θ = 30 degrees.
We are to find (a) the amount of work done on the block, (b) the amount of thermal energy that was dissipated in the process, and (c) whether there are any non-conservative forces at work in this problem. (a) The work done by the force F on the block is given by W = Fd cos θ,
where F is the applied force, d is the distance moved, and θ is the angle between the force and the direction of motion.
The force F can be calculated as follows: F = ma + mg sin θ - μk mg cos θ
where a is the acceleration of the block and g is the acceleration due to gravity. Since the block is moving at constant speed, its acceleration is zero.
Thus, we have F = mg sin θ - μk mg cos θ
= (25 kg)(9.8 m/s^2)(sin 30°) - (0.1)(25 kg)(9.8 m/s^2)(cos 30°)
= 122.5 N
The work done on the block is then W = (122.5 N)(10 m)(cos 30°) = 1060 J (b)
The amount of thermal energy that was dissipated in the process is equal to the work done by the force of friction, which is given by Wf = μk mgd
= (0.1)(25 kg)(9.8 m/s^2)(10 m) = 245 J (c)
The force of friction is a non-conservative force, since it depends on the path taken by the block.
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8 (20 points) You have been out deer hunting with a bow. Just after dawn you see a large 8 point buck. It is just at the outer range of your bow. You take careful aim, and slowly release your arrow. It's a clean hit. The arrow is 0.80 meters long, weighs 0.034 kg, and has penetrated 0.18 meter. Your arrows speed was 1.32 m/s. a Was it an elastic or inelastic collision? b What was its momentum? c How long was the time of penetration? d What was the impulse? e What was the force.
a. Elastic collision.
b. Momentum is mass x velocity.
Therefore, momentum = 0.034 x 1.32 = 0.04488 kgm/s
c. The time of penetration is given by t = l/v
where l is the length of the arrow and v is the velocity of the arrow.
Therefore, t = 0.8 / 1.32 = 0.6061 s.
d. Impulse is the change in momentum. As there was no initial momentum, impulse = 0.04488 kgm/s.
e. Force is the product of impulse and time.
Therefore, force = 0.04488 / 0.6061 = 0.0741 N.
a. Elastic collision.
b. Momentum = 0.04488 kgm/s.
c. Time of penetration = 0.6061 s.
d. Impulse = 0.04488 kgm/s
.e. Force = 0.0741 N.
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A small plastic sphere with a charge of 3nC is near another small plastic sphere with a charge of 5nC. If they repel each other with a 5.6×10 −5
N force, what is the distance between them?
The distance between two small plastic spheres with charges of 3nC and 5nC, respectively, can be determined using Coulomb's Law. The distance between the two spheres is approximately 0.143 meters.
Given that they repel each other with a force of 5.6×10^−5 N, the distance between them is calculated to be approximately 0.143 meters. Coulomb's Law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, it can be represented as:
F = k * (q1 * q2) / r^2
Where F is the force between the charges, q1 and q2 are the magnitudes of the charges, r is the distance between them, and k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2).
In this case, we are given the force between the spheres (F = 5.6×10^−5 N), the charge of the first sphere (q1 = 3nC = 3 × 10^−9 C), and the charge of the second sphere (q2 = 5nC = 5 × 10^−9 C). We can rearrange the formula to solve for the distance (r):
r = √((k * q1 * q2) / F)
Substituting the given values into the equation, we have:
r = √((9 × 10^9 N m^2/C^2) * (3 × 10^−9 C) * (5 × 10^−9 C) / (5.6×10^−5 N))
Simplifying the expression, we find:
r ≈ 0.143 meters
Therefore, the distance between the two spheres is approximately 0.143 meters.
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A rock band playing an outdoor concert produces sound at 80 dB, 45 m away from their single working loudspeaker. What is the power of this speaker? 1.5 W 2.5 W 15 W 25 W 150 W 250 W none of the above
The power of the speaker is approximately 8.27 W. None of the given answer choices match this result.
To calculate the power of the speaker, we need to use the inverse square law for sound intensity. The sound intensity decreases with distance according to the inverse square of the distance. The formula for sound intensity in decibels (dB) is:
Sound Intensity (dB) = Reference Intensity (dB) + 10 × log10(Intensity / Reference Intensity)
In this case, the reference intensity is the threshold of hearing, which is 10^(-12) W/m^2.
We can rearrange the formula to solve for the intensity:
Intensity = 10^((Sound Intensity (dB) - Reference Intensity (dB)) / 10)
In this case, the sound intensity is given as 80 dB, and the distance from the speaker is 45 m.
Using the inverse square law, the sound intensity at the distance of 45 m can be calculated as:
Intensity = Intensity at reference distance / (Distance)^2
Now let's calculate the sound intensity at the reference distance of 1 m:
Intensity at reference distance = 10^((Sound Intensity (dB) - Reference Intensity (dB)) / 10)
= 10^((80 dB - 0 dB) / 10)
= 10^(8/10)
= 10^(0.8)
≈ 6.31 W/m^2
Now let's calculate the sound intensity at the distance of 45 m using the inverse square law:
Intensity = Intensity at reference distance / (Distance)^2
= 6.31 W/m^2 / (45 m)^2
≈ 0.00327 W/m^2
Therefore, the power of the speaker can be calculated by multiplying the sound intensity by the area through which the sound spreads.
Power = Intensity × Area
Since the area of a sphere is given by 4πr^2, where r is the distance from the speaker, we can calculate the power as:
Power = Intensity × 4πr^2
= 0.00327 W/m^2 × 4π(45 m)^2
≈ 8.27 W
Therefore, the power of the speaker is approximately 8.27 W. None of the given answer choices match this result.
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1A) Applying Gauss’s Law to the charged spherical shell shows us that on the surface of the shell and beyond we can compute the electric field with what the formula for the electric field of what type of charge? Write that formula below, using the following symbols: for the charge, for Coulomb’s constant, and for the distance from the center of the sphere. Show your work.
1B) According to the answers above, where will the electric field be the largest? Explain.
1C) Enter the dielectric strength of air for the electric field and the answer to (4) for the radius and calculate a value for the maximum charge that can build up before Carona discharge. Show your work.
It's one question with 3 parts.
When applying Gauss's Law to a charged spherical shell, the formula for the electric field can be used to compute the electric field for a type of charge known as "surface charge density" (σ).
The formula for the electric field due to a charged spherical shell is given by
E = σ / (ε₀),
where
E represents the electric field,
σ is the surface charge density, and
ε₀ is Coulomb's constant.
The electric field is largest on the surface of the charged shell due to the distribution of the charges. The dielectric strength of air can be used to calculate the maximum charge that can build up before Corona discharge occurs.
1B) The electric field is largest on the surface of the charged shell. This is because the surface charge density is concentrated on the outer surface of the shell, leading to a higher electric field intensity. Inside the shell, the electric field cancels out due to the charge distribution, resulting in a lower electric field magnitude.
1C) The dielectric strength of air refers to the maximum electric field that air can withstand before it breaks down and leads to a discharge. The dielectric strength of air is approximately 3 x 10^6 V/m.
To calculate the maximum charge that can build up before Corona discharge, we can use the formula for electric field E = σ / (ε₀) and the given value for the radius. By rearranging the formula, we can solve for the surface charge density σ:
σ = E * (ε₀)
Substituting the value for the electric field (3 x 10^6 V/m) and the value for ε₀, we can calculate the maximum charge that can build up before Corona discharge occurs.
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A
transformer has 600 turns in the primary wire and 80 turns in the
secondary. Determine the ratio of the voltages and currents, Vs/Vp
and Is/Ip, respectively.
The secondary winding is 7.5 times higher than the current in the primary winding.
The turns ratio of a transformer is the ratio of the number of turns in the secondary winding to the number of turns in the primary winding.
In this case, the turns ratio is 80 / 600 = 0.133333.
The ratio of the voltages and currents in a transformer is inversely proportional to the turns ratio.
Therefore, the ratio of the voltages is 1 / 0.133333 = 7.5. The ratio of the currents is 0.133333.
In other words, the voltage in the secondary winding is 7.5 times lower than the voltage in the primary winding, and the current in the secondary winding is 7.5 times higher than the current in the primary winding.
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Part I: Series Circuits • Draw a series circuit illustrating a string of 12 Christmas tree lights connected to a power sour • If an additional light bulb were added in series to the circuit, what would happen to the total resistance? • How would the current change? How would the light from an individual bulb be affected? • If one bulb failed or "burnt out", what would happen to the other bulbs? Part II: Parallel Circuits Draw a parallel circuit of 3 lights that are on the same circuit in a typical home. • If an additional light were added in parallel to the circuit, what would happen to the total resistance? • How would the current change in the circuit? How would the light from an individual bulb be affected? • If one bulb failed or "burnt out", what would happen to the other bulbs? Part III: Summary After answering the above questions, a physics student might conclude that a parallel circuit has distinct advantage over a series circuit. State 2 advantages that a series circuit has over a paralle circuit. Assessment Details Your submission should include the following: O Your completed worksheet including two circuit drawings and answers to the questions
Parallel circuits are more reliable than series circuits because if one component fails, the others will still work. They are also more flexible than series circuits because they can be easily expanded or modified.
Part I: Series Circuits.
* A series circuit is a circuit in which all of the components are connected in a single path. This means that the current flows through all of the components in the same direction.
* If an additional light bulb were added in series to the circuit, the total resistance would increase. This is because the total resistance of a series circuit is equal to the sum of the individual resistances.
* The current would decrease because the total resistance increases. The light from an individual bulb would also decrease because the current is inversely proportional to the resistance.
* If one bulb failed or "burnt out", the entire circuit would be broken and no other bulbs would light up.
Part II: Parallel Circuits
* A parallel circuit is a circuit in which the components are connected across the same voltage source. This means that the voltage across each component is the same.
* If an additional light bulb were added in parallel to the circuit, the total resistance would decrease. This is because the total resistance of a parallel circuit is equal to the inverse of the sum of the individual conductances.
* The current would increase because the total resistance decreases. The light from an individual bulb would not be affected because the current is independent of the resistance.
* If one bulb failed or "burnt out", the other bulbs would still light up. This is because the other bulbs are connected to the voltage source across the failed bulb.
Part III: Summary
A physics student might conclude that a parallel circuit has distinct advantages over a series circuit. These advantages include:
* Increased reliability: If one component fails in a parallel circuit, the other components will still work.
* Increased flexibility: Parallel circuits can be easily expanded or modified.
* Increased current capacity: Parallel circuits can handle more current than series circuits.
However, series circuits also have some advantages, including:
* Simpler design: Series circuits are easier to design and build than parallel circuits.
* Lower cost: Series circuits are typically less expensive than parallel circuits.
* Increased safety: Series circuits are less likely to cause a fire than parallel circuits.
Overall, both series and parallel circuits have their own advantages and disadvantages. The best type of circuit for a particular application will depend on the specific requirements of that application.
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As an object moves away from any kind of spherical mirror, its
image
1. goes out of focus
2. gets closer to the focus
3. becomes virtual
4. flips between inverted and erect
As an object moves away from any kind of spherical mirror, the characteristics of its image becomes virtual.
1. The image goes out of focus: This is not necessarily true. The focus of a spherical mirror remains fixed, regardless of the position of the object. If the object moves away from the mirror, the image may become blurred or less sharp, but it doesn't necessarily go out of focus.
2. The image gets closer to the focus: This statement is incorrect. The position of the image formed by a spherical mirror depends on the position of the object and the focal length of the mirror. As the object moves away from the mirror, the image generally moves farther away from the mirror as well.
3. The image becomes virtual: This is generally true. A virtual image is formed when the reflected rays do not actually converge at a physical point. In the case of a convex (outwardly curved) mirror, the image formed is always virtual, regardless of the position of the object. As the object moves away from the mirror, the virtual image remains behind the mirror and appears smaller.
4. The image flips between inverted and erect: This statement is incorrect. The nature of the image formed by a spherical mirror (inverted or erect) depends on whether the mirror is concave (inwardly curved) or convex (outwardly curved) and the position of the object relative to the focal point. However, as the object moves away from either type of spherical mirror, the image formed remains inverted.
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"Two converging lenses with the same focal length of 10 cm are 40
cm apart. If an object is located 15 cm from one of the lenses,
find the final image distance of the object.
a. 0 cm
b. 5 cm
c. 10 cm
d 15 cm
The final image distance of the object, if the object is located 15 cm from one of the lenses is 6 cm. So none of the options are correct.
To determine the final image distance of the object in the given setup of two converging lenses, we can use the lens formula:
1/f = 1/di - 1/do
Where: f is the focal length of the lens, di is the image distance, do is the object distance.
Given that both lenses have the same focal length of 10 cm, we can consider them as a single lens with an effective focal length of 10 cm. The lenses are 40 cm apart, and the object distance (do) is 15 cm.
Using the lens formula, we can rearrange it to solve for di:
1/di = 1/f + 1/do
1/di = 1/10 cm + 1/15 cm
= (15 + 10) / (10 * 15) cm⁻¹
= 25 / 150 cm⁻¹
= 1 / 6 cm⁻¹
di = 1 / (1 / 6 cm⁻¹) = 6 cm
Therefore, the final image distance of the object is 6 cm. So, none of the options are correct.
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Q|C As in Example 28.2, consider a power supply with fixed emf E and internal resistance r causing current in a load resistance R. In this problem, R is fixed and r is a variable. The efficiency is defined as the energy delivered to the load divided by the energy delivered by the emf.(a) When the internal resistance is adjusted for maximum power transfer, what is the efficiency?
When the internal resistance is adjusted for maximum power transfer, the efficiency of the power supply is 50%.
The efficiency of a power supply is defined as the energy delivered to the load divided by the energy delivered by the emf. In this problem, we are given a power supply with fixed emf E and internal resistance r, causing current in a load resistance R. We are asked to find the efficiency when the internal resistance is adjusted for maximum power transfer.
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a 601nm light and a 605nm light are to be resolved using a
diffraction grating. How many lines must be illuminated to resolve
the light in the 2nd order?
When a 601nm light and a 605nm light are to be resolved using a diffraction grating, the number of lines that must be illuminated to resolve the light in the 2nd order is approximately 9589.
When diffraction grating is illuminated with light, it diffracts the light into several beams in various angles. In this process, the distance between lines on a diffraction grating should be less than the wavelength of the light to diffract light into a pattern of bright and dark fringes.
Diffracted order is said to be second when the light bends twice, from the line of the diffraction grating and from the screen.
Here, the difference between the two wavelengths is : 605 nm - 601 nm = 4 nm
To resolve the difference between these two wavelengths, there should be a difference of at least one fringe (or one period).
The formula to calculate the number of fringes or lines illuminated is given as : d sin(θ) = mλ
where,
d is the distance between two lines on the diffraction grating
sin(θ) is the angle at which the light bends
m is the order of diffraction, here m = 2
λ is the wavelength of the light
To resolve the light in the 2nd order, we will substitute the given values in the formula above :
4 × 10⁻⁹ m = d sin(θ) × 2 × 10⁻⁶ m
601 nm and 605 nm light are to be resolved using a diffraction grating.
The number of lines that must be illuminated to resolve the light in the 2nd order is approximately 9589.
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A bal is rolling with a constant angular speed round a circular groove in the sustace of a horizontale. If the word is 3.7 rad in the counteedoch reco, herause the circular groove is 0.57 m, and the angular position of the determine the component of the position time 10.40s and 55
To determine the component of the position of the ball, we need the values of the angular speed, time, and radius. Using the formulas θ = ω * t and s = r * θ, we can calculate the angular position and linear position of the ball, respectively. Once the values are known, the positions can be determined accordingly.
To determine the component of the position of the ball at a given time, we need to consider the angular displacement and radius of the circular groove.
The ball has a constant angular speed and completes an angular displacement of 3.7 rad in the counterclockwise direction, we can calculate the angular position (θ) using the formula:
θ = ω * t
where ω is the angular speed and t is the time. Plugging in the values, we can find the angular position.
Next, we can calculate the linear position (s) of the ball using the formula:
s = r * θ
where r is the radius of the circular groove. Substituting the given values, we can calculate the linear position of the ball.
It's important to note that the linear position will depend on the reference point chosen on the circular groove. If a specific reference point is mentioned or if further clarification is provided, the exact position of the ball can be determined.
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Light travels in a certain medium at a speed of 0.41c. Calculate the critical angle of a ray of this light when it strikes the interface between medium and vacuum. O 24° O 19⁰ O 22° O 17°
Light travels in a certain medium at a speed of 0.41c. The critical angle of a ray of this light when it strikes the interface between medium and vacuum is 24°.
To calculate the critical angle, we can use Snell's Law, which relates the angles of incidence and refraction at the interface between two mediums. The critical angle occurs when the angle of refraction is 90 degrees, resulting in the refracted ray lying along the interface. At this angle, the light ray undergoes total internal reflection.
In this case, the light travels in a medium where its speed is given as 0.41 times the speed of light in a vacuum (c). The critical angle can be determined using the formula:
critical angle = [tex]arc sin(\frac {1}{n})[/tex] where n is the refractive index of the medium.
Since the speed of light in a vacuum is the maximum speed, the refractive index of a vacuum is 1. Therefore, the critical angle can be calculated as: critical angle = [tex]arc sin(\frac {1}{0.41})[/tex]
Using a scientific calculator, we find that the critical angle is approximately 24 degrees. Therefore, the correct option is 24°.
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A book is thrown upward from a height of 10.0 m and lands with a velocity of -17.50 m/s. What was its initial velocity 110 m/s 178 m/s 10.5 m/s 13.3 m/s
The initial velocity of the book when it was thrown upward was approximately 10.5 m/s.
To find the initial velocity of the book when it was thrown upward, we can use the equations of motion for free-falling objects.
Given:
Initial height, h = 10.0 m
Final velocity, vf = -17.50 m/s (negative sign indicates downward direction).We can use the following equation to relate the initial velocity (vi), final velocity (vf), and height (h) of the object:
vf^2 = vi^2 + 2gh
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values into the equation, we have:
(-17.50 m/s)^2 = vi^2 + 2(9.8 m/s^2)(10.0 m)
306.25 m^2/s^2 = vi^2 + 196 m^2/s^2
Rearranging the equation, we find:
vi^2 = 306.25 m^2/s^2 - 196 m^2/s^2
vi^2 = 110.25 m^2/s^2
Taking the square root of both sides, we get:
vi = √(110.25 m^2/s^2)
vi ≈ 10.5 m/s
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Assume a deuteron and a triton are at rest when they fuse according to the reaction²₁H + ³₁H → ⁴₂He + ¹₀n Determine the kinetic energy acquired by the neutron.
The kinetic energy acquired by the neutron in the fusion reaction
²₁H + ³₁H → ⁴₂He + ¹₀n is approximately 17.6 MeV (million electron volts).
In a fusion reaction, two nuclei combine to form a new nucleus. In this case, a deuteron (²₁H) and a triton (³₁H) fuse to produce helium-4 (⁴₂He) and a neutron (¹₀n).
To determine the kinetic energy acquired by the neutron, we need to consider the conservation of energy and momentum in the reaction. Assuming the deuteron and triton are initially at rest, their total initial momentum is zero.
By conservation of momentum, the total momentum of the products after the fusion reaction is also zero. Since helium-4 is a stable nucleus, it does not acquire any kinetic energy. Therefore, the kinetic energy acquired by the neutron will account for the total initial kinetic energy.
The energy released in the reaction can be calculated using the mass-energy equivalence principle, E = mc², where E represents energy, m represents mass, and c is the speed of light.
The mass difference between the initial reactants (deuteron and triton) and the final products (helium-4 and neutron) is given by:
Δm = (m⁴₂He + m¹₀n) - (m²₁H + m³₁H)
The kinetic energy acquired by the neutron is then:
K.E. = Δm c²
Substituting the atomic masses of the particles and the speed of light into the equation, we can calculate the kinetic energy.
Using the atomic masses: m²₁H = 1.008665 u, m³₁H = 3.016049 u, m⁴₂He = 4.001506 u, and converting to kilograms (1 u = 1.66 × 10⁻²⁷ kg), the calculation gives:
Δm = (4.001506 u + 1.674929 u) - (2.016331 u + 3.016049 u)
≈ 0.643 u
K.E. = (0.643 u) × (1.66 × 10⁻²⁷ kg/u) × (3.00 × 10⁸ m/s)²
≈ 17.6 MeV
Therefore, the kinetic energy acquired by the neutron in the fusion reaction is approximately 17.6 MeV.
In the fusion reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, the neutron acquires a kinetic energy of approximately 17.6 MeV. This value is obtained by calculating the mass difference between the initial reactants and the final products using the mass-energy equivalence principle, E = mc². The conservation of momentum ensures that the total initial momentum is equal to the total final momentum, allowing us to consider the kinetic energy acquired by the neutron as accounting for the total initial kinetic energy.
Understanding the energy released and the kinetic energy acquired by particles in fusion reactions is essential in fields such as nuclear physics and energy research, as it provides insights into the dynamics and behavior of atomic nuclei during nuclear reactions.
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1.Find the force on a particle of mass m=1.70×10-27kg and charge q=1.60×10-19C if it enters a field B=5 mT with an initial speed of v=83.5 km/s. Assume the velocity is in the x direction and the magnetic field enters perpendicular to the screen. Also make a schematic drawing of these vectors. Don't forget to place your reference system.
2.Find the force on a straight conductor of length 0.3 m, which carries a current of 5 A in the negative z-direction. In that space there is a magnetic field given by the vector B=3.5×10-3Ti-3.5×10-3Tj . Make a schematic drawing of the situation. (We do not require precision in your drawing for the direction of the magnetic field, only approximate).
3.A conductor of length 2.5 m is located at z=0, x=4m with a current of 12 A in the -y direction. Find the magnetic field that exists in that region if the force on the conductor is F=-1.20×10-2N(-12i-12j).
4.A long thin wire carries a current I. A metal bar of length L is moving with a constant speed v as shown in the figure. Point a is a distance d from the wire a) Calculate the electromotive force induced in the bar. b) If the bar is replaced by a rectangular circuit of resistance R, what is the magnitude of the induced current in the circuit?
1. The force on the particle is 1.36 x 10^-14 N, schematic drawing shows velocity in x-direction, magnetic field entering perpendicular to the screen, and force perpendicular to both.
2. The force on the straight conductor is 5.25 x 10^-3 N, schematic drawing shows conductor in negative z-direction and magnetic field vectors approximately orthogonal to the conductor.
3. The magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.
4. a) The electromotive force induced in the bar is BLv. b) The magnitude of the induced current in the rectangular circuit is V/R.
1. The force on the particle can be calculated using the equation F = qvB, where q is the charge, v is the velocity, and B is the magnetic field. Plugging in the given values, the force is 1.36 x 10^-14 N. A schematic drawing would show the velocity vector in the x-direction, the magnetic field vector entering perpendicular to the screen, and the force vector perpendicular to both.
2. The force on the straight conductor can be calculated using the equation F = IL x B, where I is the current, L is the length of the conductor, and B is the magnetic field. Plugging in the given values, the force is 5.25 x 10^-3 N. A schematic drawing would show the conductor in the negative z-direction, with the magnetic field vectors shown approximately orthogonal to the conductor.
3. The magnetic field can be determined using the equation F = IL x B. Since the force is given as F = -1.20 x 10^-2 N (-12i - 12j), we can equate the force components to the corresponding components of the equation and solve for B. The resulting magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.
4. To calculate the electromotive force induced in the bar, we can use the equation emf = BLv, where B is the magnetic field, L is the length of the bar, and v is the speed of the bar. The magnitude of the induced current in the rectangular circuit can be calculated using Ohm's Law, I = V/R, where V is the electromotive force and R is the resistance of the circuit.
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Set 1: Gravitation and Planetary Motion NOTE. E Nis "type-writer notation for x10" ( 2 EB - Exam 2x10") you may use either for this class AND the AP GMm mu F GMm 9 G= 6.67 11 Nm /kg F = mg 9 GMm = mg GM 12 т GM V = 1 GM 9 GM V = - 21 T F 9 = mac T 1. A whale shark has a mass of 2.0 E4 kg and the blue whale has a mass of 1.5 E5 kg a. If the two whales are 1.5 m apart, what is the gravitational force between them? b. How does the magnitude of the gravitational force between the two animals compare to the gravitational force between each and the Earth? c. Explain why objects on Earth do not seem to be attracted 2. An asteroid with a mass of 1.5 E21 kg orbits at a distance 4E8 m from a planet with a mass of 6 E24 kg a. Determine the gravitational force on the asteroid. b. Determine the gravitational force on the planet. C Determine the orbital speed of the asteroid. d Determine the time it takes for the asteroid to complete one trip around the planet 3. A 2 2 14 kg comet moves with a velocity of 25 E4 m/s through Space. The mass of the star it is orbiting is 3 E30 kg a Determine the orbital radius of the comet b. Determine the angular momentum of the comet. (assume the comet is very small compared to the star) c An astronomer determines that the orbit is not circular as the comet is observed to reach a maximum distance from the star that is double the distance found in part (a). Using conservation of angular momentum determine the speed of the comet at its farthest position 4. A satellite that rotates around the Earth once every day keeping above the same spot is called a geosynchronous orbit. If the orbit is 3.5 E7 m above the surface of the and the radius and mass of the Earth is about 6.4 E6 m and 6.0 E24 kg respectively. According to the definition of geosynchronous, what is the period of the satellite in hours? seconds? a. Determine the speed of the satellite while in orbit b. Explain satellites could be used to remotely determine the mass of unknown planets 5. Two stars are orbiting each other in a binary star system. The mass of each of the stars is 2 E20 kg and the distance from the stars to the center of their orbit is 1 E7 m. a. Determine the gravitational force between the stars.. b. Determine the orbital speed of each star
In this set of questions, we are exploring the concepts of gravitation and planetary motion. We use the formulas related to gravitational force, orbital speed, and orbital radius to solve various problems.
Firstly, we calculate the gravitational force between two whales and compare it to the gravitational force between each whale and the Earth. Then, we determine the gravitational force on an asteroid and a planet, as well as the orbital speed and time taken for an asteroid to complete one orbit.
Next, we find the orbital radius and angular momentum of a comet orbiting a star, and also calculate the speed of the comet at its farthest position. Finally, we discuss the period of a geosynchronous satellite orbiting the Earth and how satellites can be used to determine the mass of unknown planets.
a. To calculate the gravitational force between the whale shark and the blue whale, we use the formula F = GMm/r^2, where G is the gravitational constant, M and m are the masses of the two objects, and r is the distance between them. Plugging in the values, we find the gravitational force between them.
b. To compare the gravitational force between the two animals and the Earth, we calculate the gravitational force between each animal and the Earth using the same formula.
We observe that the force between the animals is much smaller compared to the force between each animal and the Earth. This is because the mass of the Earth is significantly larger than the mass of the animals, resulting in a stronger gravitational force.
c. Objects on Earth do not seem to be attracted to each other strongly because the gravitational force between them is much weaker compared to the gravitational force between each object and the Earth.
The mass of the Earth is substantially larger than the mass of individual objects on its surface, causing the gravitational force exerted by the Earth to dominate and make the gravitational force between objects on Earth negligible in comparison.
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