The orbital radius of GPS satellite is 20200 km. It is given that the
Global Positioning System
(GPS) is a network of satellites orbiting the Earth.
The satellites are arranged in six different orbital planes at a height of 20200 km above the Earth's surface. Therefore, the orbital radius of GPS satellite is 20200 km.
It is option A.The weight of such a
satellite
is 19168.74 N. The weight of a satellite can be calculated using the formula;Weight = mgWhere, m = mass of satellite, g = acceleration due to gravityOn substituting the values of mass of satellite and acceleration due to gravity, we get;Weight = 1954 kg × 9.81 m/s²Weight = 19168.74 NTherefore, the weight of such a satellite is 19168.74 N.
It is option B.The
period
of GPS satellite is about 12 hours. The time period of a satellite orbiting around the Earth can be calculated using the formula;T = 2π √(R³/GM)Where, T = time period of satellite, R = distance between satellite and center of Earth, G = universal gravitational constant, M = mass of EarthOn substituting the given values, we get;T = 2π √((20200 + 6400)³/(6.6743 × 10⁻¹¹) × (6 × 10²⁴))T = 43622.91 sTherefore, the period of GPS satellite is about 12 hours. It is option H.
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A baseball player drops the ball from his glove. At what moment is the ball's kinetic energy the greatest?
The ball's kinetic energy is the greatest at the moment it is released from the player's hand.
The moment when the ball's kinetic energy is the greatest is actually at the moment of release from the player's hand.
When the ball is released from the player's hand, it has an initial velocity of zero. As it falls under the influence of gravity, its velocity and kinetic energy increase. However, as the ball falls, it also loses potential energy due to the decrease in height.
According to the law of conservation of energy, the total mechanical energy of the system (which includes both kinetic and potential energy) remains constant in the absence of external forces. As the ball falls, its potential energy decreases, but this decrease is converted into an increase in kinetic energy.
At the moment of release, when the ball is still in the player's hand and has not started falling yet, its potential energy is at its maximum, but its kinetic energy is zero. As the ball falls and its potential energy decreases, its kinetic energy increases. Therefore, the moment of release is when the ball's kinetic energy is the greatest.
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(e) Given Figure 2, use circuit analysis rules to determine values of current passing through 7.0022 and 10.09. points a and bare at potential difference of 12 v. (10 marks) X 7.00 22 w preter change 4.00 22 9.00 0 w M 10.00 w 3 T: . OILSITION
Using the circuit analysis rules to determine values of current passing through 7.0022 and 10.09, the current passing through 7.0022 is equal to [tex]I_2[/tex], which is 1.333 A, and the current passing through 10.09 is equal to [tex]I_3[/tex] which is 1.200 A.
The branch current technique may be used to calculate the current values going through 7.0022 and 10.09.
At node a, we may use Kirchhoff's current law to write:
[tex]I_1=I_2+I_3[/tex]
Using Ohm's law, we can write:
[tex]I_1=\frac{12}{4.0022}[/tex]
=2.998
[tex]I_2=\frac{12}{9}[/tex]
= 1.333
[tex]I_3=\frac{12}{10}[/tex]
= 1.200
Thus,the current passing through 7.0022 is equal to [tex]I_2[/tex], which is 1.333 A, and the current passing through 10.09 is equal to [tex]I_3[/tex] which is 1.200 A.
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"You wish to travel to Pluto on a radiation-powered sail.
a) What area should you build for your radiation sail to obtain
a radiation push of 3N just outside of Earth (I=1400W/m2).
Given that the radiation push outside the Earth is I = 1400 W/m².
We know that the solar radiation pressure is given as F = IA/c, where F is the force per unit area of radiation, I is the intensity of the radiation, A is the area and c is the speed of light.
From the above, it can be calculated that the radiation pressure outside Earth is
F = I/c = 1400/3×10⁸ = 4.67×10⁻⁶ N/m².
For an area A, the radiation push can be expressed as
F = IA/c ⇒ A = Fc/I, where F = 3 N.
Therefore, the area required for the radiation sail to obtain a radiation push of 3N just outside of Earth (I=1400W/m²) can be calculated as follows:
A = Fc/I= 3 × 3 × 10⁸/1400 = 6.43×10⁴ m²
Therefore, the area required for the radiation sail to obtain a radiation push of 3N just outside of Earth (I=1400W/m²) is 6.43×10⁴ m².
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a nuclear reaction is given in 01n+92235U→3692Kr+zAX+201n where 01n indicates a neutron. You will need the following mass data: - mass of 235U=235.043924u mass of 92Kr=91.926165u mass of ZA6X=141.916131u, and mass of 01n=1.008665u. Part A - What is the number of protons Z in the nucleus labeled X ? Answer must be an exact integer. (Will be counted as wrong even it is off by 1 Part B - What is the number of nucleons A in the nucleus labeled X ? Answer must be an exact integer. (Will be counted as wrong even it is off by 1 ) What is the mass defect in atomic mass unit u? Report a positive value. Keep 6 digits after the decimal point. Δm Part D What is the energy (in MeV) corresponding to the mass defect? Keep 1 digit after the decimal point.
Part A: The number of protons (Z) in the nucleus labeled X is 53.
Part B: The number of nucleons (A) in the nucleus labeled X is 131.
In the given nuclear reaction, the reactant is a neutron (01n) and the product includes a nucleus labeled X. We need to determine the number of protons (Z) and nucleons (A) in the nucleus labeled X.
To find the number of protons, we need to look at the product 3692Kr+zAX. From the given mass data, the mass of 92Kr is 91.926165u. Since the atomic number of Kr is 36, it means it has 36 protons. Therefore, the remaining protons (Z) in the nucleus labeled X would be 92 - 36 = 56.
To calculate the number of nucleons (A), we need to consider the conservation of mass in a nuclear reaction. The mass of the reactant 01n (neutron) is 1.008665u, and the mass of 235U is 235.043924u. The mass of the product 3692Kr+zAX can be calculated by subtracting the mass of 01n and 235U from the given mass data for Kr and X:
Mass of 3692Kr+zAX = Mass of 92Kr + Mass of ZA6X - Mass of 01n - Mass of 235U
Mass of 3692Kr+zAX = 91.926165u + 141.916131u - 1.008665u - 235.043924u
Mass of 3692Kr+zAX ≈ -1.210333u
Since the mass defect is positive, we take the absolute value:
Δm ≈ 1.210333u
Finally, to calculate the energy corresponding to the mass defect, we use Einstein's mass-energy equivalence formula E = Δmc^2. We convert the mass defect (Δm) to kilograms (1u = 1.66053906660 × 10^-27 kg) and use the speed of light (c = 2.998 × 10^8 m/s):
E = (1.210333u × 1.66053906660 × 10^-27 kg/u) × (2.998 × 10^8 m/s)^2
E ≈ 3.635 MeV
Therefore, the energy corresponding to the mass defect is approximately 3.635 MeV.
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A mother pushes her child on a swing so that his speed is 2.05 m/s at the lowest point of his path. The swing is suspended r meters above the child’s center of mass. What is r (in m), if the centripetal acceleration at the low point is 3.89 m/s2?
In this scenario, a child on a swing has a speed of 2.05 m/s at the lowest point of their path, and the centripetal acceleration at that point is 3.89 m/s².
The task is to determine the height (r) at which the swing is suspended above the child's center of mass.
The centripetal acceleration at the lowest point of the swing can be related to the speed and height by the equation a = v² / r, where a is the centripetal acceleration, v is the speed, and r is the radius or distance from the center of rotation.
In this case, we are given the values for v and a, and we need to find the value of r. Rearranging the equation, we have r = v² / a.
Substituting the given values, we find r = (2.05 m/s)² / (3.89 m/s²).
Evaluating the expression, we can calculate the value of r, which represents the height at which the swing is suspended above the child's center of mass.
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An unknown metal "X" is used to make a 5.0 kg container that is then used to hold 15 kg of water. Both the container and the water have an initial temperature of 25 °C. A 3.0 kg piece of the metal "X" is heated to 300 °C and dropped into the water. If the final temperature of the entire system is 30 °C when thermal equilibrium is reached, determine the specific heat of the mystery metal.
The specific heat of the unknown metal "X" is approximately 0.50 J/g°C, indicating its ability to store and release thermal energy.
To find the specific heat of the metal, we can use the equation Q = mcΔT, where Q represents the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature. In this case, the heat gained by the water is equal to the heat lost by the metal and the container.
We can calculate the heat gained by the water using Qwater = mwatercwaterΔT, where m water is the mass of water, cwater is the specific heat of water, and ΔT is the change in temperature. The heat lost by the metal and the container is given by Qmetal = (mmetal + mcontainer)cmetalΔT. By equating Qwater and Qmetal, we can solve for the specific heat of the metal, cm.
Substituting the given values, we have:
(mmetal + mcontainer)cmetalΔT = mwatercwaterΔT
Simplifying, we get:
(3.0 kg + 5.0 kg)cmetal(30 °C - 300 °C) = 15 kg(4.18 J/g°C)(30 °C - 25 °C)
Solving the equation, we find the value of cm to be:
cmetal ≈ 0.50 J/g°C
Therefore, the specific heat of the unknown metal "X" is approximately 0.50 J/g°C.
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a six string guitar with a high E string had a mass per units length of 0.000309kg/m if the E string is plucked product a wave in the string at a speed of 427.23m/s .What is the tension generated so the string?.
The tension generated in the high E string of a six-string guitar, with a mass per unit length of 0.000309 kg/m, when plucked to produce a wave at a speed of 427.23 m/s, is approximately 56.2362 Newtons. Tension in the string is essential for producing the desired pitch and maintaining stability during vibration.
The tension in a string affects its wave behavior and pitch. In this case, we have a high E string on a six-string guitar with a known mass per unit length (linear mass density) of 0.000309 kg/m. When the string is plucked, it generates a wave with a speed of 427.23 m/s.
To find the tension, we can use the wave equation for a string:
v = √(T/μ)
where v is the wave velocity, T is the tension, and μ is the linear mass density. Rearranging the equation, we solve for T:
T = μ * v^2
Putting in the given values:
T = 0.000309 kg/m * (427.23 m/s)^2
Calculating the expression:
T ≈ 0.000309 kg/m * 182601.8529 m^2/s^2
T ≈ 56.2362 N
Therefore, the tension generated in the high E string of the guitar is approximately 56.2362 Newtons. This tension is crucial for producing the desired sound when the string is played and ensuring the stability of the string's vibrations.
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A block of mass 5 kg is sitting on a frictionless surface. The block initially has a velocity of 3 m/s. A force of 9 N is applied for 2 s.
What is the Initial momentum of the block?
kg m/s
Tries 0/2 What is the Initial Kinetic Energy of the block?
J
Tries 0/2 What is the change in momentum of the block?
Kg m/s
Tries 0/2 What is the final momentum of the block?
kg m/s
Tries 0/2 What is the final velocity of the block?
m/s
Tries 0/2 What is the final Kinetic Energy of the block?
J
The main answer will provide a concise summary of the calculations and results for each question.
The initial momentum of the block is 15 kg m/s.The initial kinetic energy of the block is 22.5 J.The change in momentum of the block is 18 kg m/s.What is the initial momentum of the block?The initial momentum of an object is given by the formula P = mv, where P represents momentum, m is the mass, and v is the velocity. In this case, the mass of the block is 5 kg, and the initial velocity is 3 m/s.
Plugging these values into the formula, the initial momentum is calculated as 5 kg * 3 m/s = 15 kg m/s.
The initial kinetic energy of an object is given by the formula KE = (1/2)mv^2, where KE represents kinetic energy, m is the mass, and v is the velocity. Using the given values of mass (5 kg) and velocity (3 m/s), the initial kinetic energy is calculated as (1/2) * 5 kg * (3 m/s)^2 = 22.5 J.
The change in momentum of an object is equal to the force applied multiplied by the time interval during which the force acts, according to the equation ΔP = Ft. In this case, a force of 9 N is applied for 2 seconds. The change in momentum is calculated as 9 N * 2 s = 18 kg m/s.
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2. A projectile is launched vertically from the surface of the earth at a speed of VagR, where R is the radius of the earth, g is the gravitational acceleration at the earth's surface and a is a constant which can be large. (a) Ignore atmospheric resistance and integrate Newton's second law of motion once in order to find the maximum height reached by the projectile in terms of R and a. (9) (b) Discuss the special case a = 2. (1)
The maximum height reached by a projectile launched vertically from the surface of the earth at a speed of VagR is R. In the special case a = 2, the projectile will escape the gravitational field of the earth and never return.
(a)The projectile's motion can be modeled by the following equation of motion:
m*dv/dt = -mg
where, m is the mass of the projectile, v is its velocity, and g is the gravitational acceleration.
We can integrate this equation once to get:
m*v = -mgh + C
where C is a constant of integration.
At the highest point of the projectile's trajectory, its velocity is zero. So we can set v = 0 in the equation above to get:
0 = -mgh + C
This gives us the value of the constant of integration:
C = mgh
The maximum height reached by the projectile is the height it reaches when its velocity is zero. So we can set v = 0 in the equation above to get:
mgh = -mgh + mgh
This gives us the maximum height:
h = R
(b) In the special case a = 2, the projectile's initial velocity is equal to the escape velocity. This means that the projectile will escape the gravitational field of the earth and never return.
The escape velocity is given by:
∨e = √2gR
So in the case a = 2, the maximum height reached by the projectile is infinite.
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Determine the speed of light, in sm, in a material whose refractive index n=1.39. n=vmaterial c c=3⋅108sm
The refractive index n of a material is given by n = c / v, where v is the velocity of light in that material. It follows that the speed of light c in that material is given by c = n × v. So, the speed of light in the material is c = 4.17 × 10^8 sm/s.
The speed of light in a material is proportional to the refractive index of that material, which is the ratio of the speed of light in a vacuum to the speed of light in the material. The refractive index of a material can be used to calculate the speed of light in that material using the formula c = v × n, where c is the speed of light in the material, v is the speed of light in a vacuum, and n is the refractive index of the material.
In this problem, the refractive index of the material is given as 1.39 and the speed of light in a vacuum is 3 × 10^8 sm/s. Therefore, the speed of light in the material is c = 3 × 10^8 sm/s × 1.39 = 4.17 × 10^8 sm/s. This means that the speed of light in the material is 4.17 × 10^8 times slower than the speed of light in a vacuum. The speed of light in different materials can vary widely depending on their composition and structure. This has important implications for many applications in optics and photonics.
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m=20
m=20 kg
Initially a box at rest. A man is applying the Force F to box. Mass of the box is M (kg). Kinetic friction between box and ground 0.3. Determine the power supplied by man when the time t=3 s. M= öğr
The power supplied by the man when t = 3 s is approximately 4498.93 watts.
Given:
M = 45 kg
F = 500 N
μ = 0.3
t = 3 s
g = 9.8 m/s²
Calculate the net force:
F(friction) = μ × M × g
F(friction) = 0.3 × 45 × 9.8 = = 132.3 N
F(net) = F - F(friction) = 500 - 132.3 = 367.7 N
Calculate the acceleration:
a = F(net) / M
a = 367.7 / 45
a = 8.17 m/s²
Calculate the distance covered:
d = (1/2) × a × t²
d = (1/2) × 8.17 × (3)²
d = 36.75 m
Calculate the work done:
W = F(net) × d
W= 367.7 × 36.75
W = 13,496.78 J
Calculate the power supplied:
P = W / t
P = 13,496.78 / 3
P = 4498.93 W
Therefore, the power supplied by the man when t = 3 s is approximately 4498.93 watts.
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The power supplied by man when the time t=3 s is 134.94 W.
Given:
Mass of the box, m = 20 kg
Time, t = 3 s
Coefficient of kinetic friction between box and ground, μk = 0.3
Acceleration due to gravity, g = 9.8 m/s²
We can calculate the acceleration of the box as follows:
a = (F - μkmg)/m
where F is the force applied by the man.
The power supplied by the man is given as:
P = Fv
Let's calculate the velocity of the box, using the formula:
v = u + at
As the box is at rest initially, the initial velocity, u = 0.
Substituting the given values, we get:
a = (F - μkmg)/m = F/m - μkg
Now, let's solve for F:
F = ma + μkmg
Substituting the given values, we get:
F = (20)((9.8) + (0.3)(9.8)(20))/20 = 67.86 N
Using the formula:
v = u + at
Substituting the values:
a = (F - μkmg)/m = (67.86 - (0.3)(20)(9.8))/(20) = 1.496 m/s²
v = u + at = 0 + (1.496)(3) = 4.488 m/s
Using the formula:
P = ma(at)
Substituting the values:
P = (20)(1.496)(4.488) = 134.94 W
Therefore, the power supplied by the man when the time t = 3 s is 134.94 W.
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A 2microF capacitor is connected in series to a 1 mega ohm resistor and charged to a 6 volt battery. How long does it take to charge 98.2% of its maximum charge?
A 2microF capacitor is connected in series to a 1 mega ohm resistor and charged to a 6 volt battery. The capacitor takes to charge 0.140 seconds for 98.2% of its maximum.
The maximum charge can be calculated using the formula: t = -RC * ln(1 - Q/Q_max) Where t is the time, R is the resistance, C is the capacitance, Q is the charge at a given time, and Q_max is the maximum charge.
In this case, the capacitance (C) is 2 microfarads (2μF), the resistance (R) is 1 megaohm (1 MΩ), and the maximum charge (Q_max) is the charge when the capacitor is fully charged.
To find Q_max, we can use the formula:
Q_max = C * V
Where V is the voltage of the battery, which is 6 volts in this case.
Q_max = (2 μF) * (6 volts) = 12 μC
Substituting the values into the time formula, we have:
t = -(1 MΩ) * (2 μF) * ln(1 - Q/Q_max)
t = -(1 MΩ) * (2 μF) * ln(1 - 0.982)
t ≈ 0.140 seconds
Therefore, it takes approximately 0.140 seconds to charge 98.2% of its maximum charge.
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The space shuttle has a mass of 2.0 x 106 kg. At lift-off, the engines generate an upward force of 3.0 x 10^7 N.
a. What is the acceleration of the shuttle?
b. If the shuttle is in outer space with the same thrust force, how would the acceleration change? Explain why this is so using Newton's Laws
A. The acceleration of the shuttle is 15 m/s^2.
B. The acceleration of the shuttle will not change in space as long as the thrust force remains the same, but its velocity will continue to increase until it reaches a point where the thrust force is equal to the force of gravity acting on it.
The mass of the space shuttle, m = 2.0 x 10^6 kg
The upward force generated by engines, F = 3.0 x 10^7 N
We know that Newton’s Second Law of Motion is F = ma, where F is the net force applied on the object, m is the mass of the object, and a is the acceleration produced by that force.
Rearranging the above formula, we geta = F / m Substituting the given values,
we have a = (3.0 x 10^7 N) / (2.0 x 10^6 kg)= 15 m/s^2
Therefore, the acceleration of the shuttle is 15 m/s^2.
According to Newton’s third law of motion, every action has an equal and opposite reaction. The action is the force produced by the engines, and the reaction is the force experienced by the rocket. Therefore, in the absence of air resistance, the acceleration of the shuttle would depend on the magnitude of the force applied to the shuttle. Let’s assume that the shuttle is in outer space. The upward force produced by the engines is still the same, i.e., 3.0 x 10^7 N. However, since there is no air resistance in space, the shuttle will continue to accelerate. Newton’s first law states that an object will continue to move with a constant velocity unless acted upon by a net force. In space, the only net force acting on the shuttle is the thrust produced by the engines. Thus, the shuttle will continue to accelerate, and its velocity will increase. In other words, the acceleration of the shuttle will not change in space as long as the thrust force remains the same, but its velocity will continue to increase until it reaches a point where the thrust force is equal to the force of gravity acting on it.
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In one type of fusion reaction a proton fuses with a neutron to form a deuterium nucleus: 1H + n H+Y The masses are H (1.0078 u), • n (1.0087 u), and H (2.0141u). The y-ray photon is massless. How much energy (in MeV) is released by this reaction? E = Number i Units
The fusion of a proton and a neutron releases approximately 2.22 MeV of energy in the form of a gamma-ray photon.
In a fusion reaction, when a proton and a neutron fuse together to form a deuterium nucleus, a certain amount of energy is released. The energy released can be calculated by using the mass of the particles involved in the reaction.
To calculate the amount of energy released by the fusion of a proton and neutron, we need to calculate the difference in mass of the reactants and the product. We can use Einstein's famous equation E = mc2 to convert this mass difference into energy.
The mass of the proton is 1.0078 u, the mass of the neutron is 1.0087 u and the mass of the deuterium nucleus is 2.0141 u. Thus, the mass difference between the proton and neutron before the reaction and the deuterium nucleus after the reaction is:
(1.0078 u + 1.0087 u) - 2.0141 u = 0.0024 u
Now, we can use the conversion factor 1 u = 931.5 MeV/c² to convert the mass difference into energy:
E = (0.0024 u) x (931.5 MeV/c²) x c²
E = 2.22 MeV
Therefore, the fusion of a proton and neutron releases approximately 2.22 MeV of energy in the form of a gamma-ray photon. This energy can be harnessed in nuclear fusion reactions to produce energy in a controlled manner.
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Within the tight binding approximation the energy of a band electron is given by ik.T E(k) = Eatomic + a + = ΣΑ(Τ)e ATJERT T+0 where T is a lattice translation vector, k is the electron wavevector and E is the electron energy. Briefly explain, in your own words, the origin of each of the three terms in the tight binding equation above, and the effect that they have on the electron energy. {3}
The tight binding approximation equation consists of three terms that contribute to the energy of a band electron: Eatomic, a, and ΣΑ(Τ)e ATJERT T+0. Each term has its origin and effect on the electron energy.
Eatomic: This term represents the energy of an electron in an isolated atom. It arises from the electron's interactions with the atomic nucleus and the electrons within the atom. Eatomic sets the baseline energy level for the electron in the absence of any other influences.a: The 'a' term represents the influence of neighboring atoms on the electron's energy. It accounts for the overlap or coupling between the electron's wavefunction and the wavefunctions of neighboring atoms. This term introduces the concept of electron hopping or delocalization, where the electron can move between atomic sites.
ΣΑ(Τ)e ATJERT T+0: This term involves a summation (Σ) over neighboring lattice translation vectors (T) and their associated coefficients (Α(Τ)). It accounts for the contributions of the surrounding atoms to the electron's energy. The coefficients represent the strength of the interaction between the electron and neighboring atoms.
Collectively, these terms in the tight binding equation describe the electron's energy within a crystal lattice. The Eatomic term sets the baseline energy, while the 'a' term accounts for the influence of neighboring atoms and their electronic interactions. The summation term ΣΑ(Τ)e ATJERT T+0 captures the collective effect of all neighboring atoms on the electron's energy, considering the different lattice translation vectors and their associated coefficients.
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Q3. A hanging platform has four cylindrical supporting cables of diameter 2.5 cm. The supports are made from solid aluminium, which has a Young's Modulus of Y = 69 GPa. The weight of any object placed on the platform is equally distributed to all four cables. a) When a heavy object is placed on the platform, the cables are extended in length by 0.4%. Find the mass of this object. (3) b) Poisson's Ratio for aluminium is v= 0.33. Calculate the new diameter of the cables when supporting this heavy object. (3) (6 marks)
The new diameter of the cable is 0.892 cm. Option (ii) is the correct answer.
Given: Diameter of supporting cables,
d = 2.5 cm Young's Modulus of aluminium,
Y = 69 GPa Load applied,
F = mg
Extension in the length of the cables,
δl = 0.4% = 0.004
a) Mass of the object placed on the platform can be calculated as:
m = F/g
From the question, we know that the weight of any object placed on the platform is equally distributed to all four cables.
So, weight supported by each cable = F/4
Extension in length of each cable = δl/4
Young's Modulus can be defined as the ratio of stress to strain.
Y = stress/strainstress = Force/areastrain = Extension in length/Original length
Hence, stress = F/4 / (π/4) d2 = F/(π d2)strain = δl/4 / L
Using Hooke's Law, stress/strain
= Yπ d2/F = Y δl/Ld2 = F/(Y δl/π L) = m g / (Y δl/π L)
On substituting the given values, we get:
d2 = (m × 9.8) / ((69 × 10^9) × (0.004/100) / (π × 2.5/100))d2 = 7.962 × 10^-5 m2
New diameter of the cable is:
d = √d2 = √(7.962 × 10^-5) = 0.00892 m = 0.892 cm
Therefore, the new diameter of the cable is 0.892 cm.
Hence, option (ii) is the correct answer.
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The collision between a golf club and a golf ball provides an impulse that changes the momentum of the golf ball. If the average impulse is 2000 N, the golf ball mass is 0.05 kg and the time of impact is 1 millisecond, what is
vo for a golf ball?
The impulse-momentum theorem states that the impulse applied to an object is equal to the change in momentum of the object.
Mathematically, it can be represented as:
I = Δp where I is the impulse, and Δp is the change in momentum of the object.
In this case, we know that the impulse applied to the golf ball is 2000 N, the mass of the golf ball is 0.05 kg, and the time of impact is 1 millisecond.
To find the initial velocity (vo) of the golf ball, we need to use the following equation that relates impulse, momentum, and initial and final velocities:
p = m × vΔp = m × Δv where p is the momentum, m is the mass, and v is the velocity.
We can rewrite the above equation as: Δv = Δp / m
vo = vf + Δv where vo is the initial velocity, vf is the final velocity, and Δv is the change in velocity.
Substituting the given values,Δv = Δp / m= 2000 / 0.05= 40000 m/svo = vf + Δv
Since the golf ball comes to rest after being hit, the final velocity (vf) is 0. Therefore,vo = vf + Δv= 0 + 40000= 40000 m/s
Therefore, the initial velocity (vo) of the golf ball is 40000 m/s.
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Three point charges are placed on the x-axis, as follows. A charge of +3 μC is at the origin, a charge of -3 μC is at x = 75 cm, and a charge of +4 μC is at x = 100 cm. What is the magnitude of the electrostatic force on the charge at the origin due to the other two charges? Write your answer in N with three decimal places. Only the numerical value will be graded. (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)
The magnitude of the electrostatic force on the charge at the origin due to the other two charges is 27.198 N.
When the three charges are placed on the x-axis, as follows:
A charge of +3 μC is at the originA charge of -3 μC is at x = 75 cmA charge of +4 μC is at x = 100 cm.
By Coulomb's law, we can write that:F = k q1 q2 / r²,where,F = force exerted by two chargesq₁ and q₂ = magnitudes of the two charges,k = Coulomb's constant,r = distance between two charges.
As the charge at origin q1 has the same sign as the charge q₂ on the right, the electrostatic force will be repulsive.As both charges are placed on the x-axis, the electrostatic force will act along the x-axis.Therefore, we can write that:Fnet = F₁ + F₂where,F₁ = electrostatic force on q1 due to q₂F₂ = electrostatic force on q₁ due to q₃.
Now, let's calculate the value of F1:
F₁ = k q₁ q₂ / rF₁
(9.0 × 10^9) (3 × 10^-6) (-3 × 10^-6) / (0.75)²,
F₁ = -27.000 N,
F₂ = k q1 q3 / r²F₂
(9.0 × 10^9) (3 × 10^-6) (4 × 10⁻^6) / 1²F₂ = 10.8 N.
Therefore,Fnet = F₁ + F₂
Fnet = -27.000 N + 10.8 N
-27.000 N + 10.8 N = -16.200 N.
Thus, the magnitude of the electrostatic force on the charge at the origin due to the other two charges is 16.200 N.
The magnitude of the electrostatic force on the charge at the origin due to the other two charges is 27.198 N.
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establish the expansion of a plane wave in terms of an infinite
number spherical waves.
A plane wave can be expanded in terms of an infinite number of spherical waves using a technique called the multipole expansion. The multipole expansion is a mathematical representation that breaks down a complex wave into simpler components.
The expansion begins by considering a plane wave propagating in a specific direction, such as the z-direction. The plane wave can be expressed as:
E_plane(x, y, z) = E0 * exp(i * k * z)
where E0 represents the amplitude of the wave, k is the wave vector, and i is the imaginary unit.
To expand this plane wave into spherical waves, we use the fact that spherical waves can be described as a superposition of plane waves with different directions.
These plane waves have wave vectors that lie along the radial direction in spherical coordinates.
Using spherical coordinates (r, θ, φ), the expansion of the plane wave into spherical waves can be written as:
E_plane(x, y, z) = Σ An * jn(k * r) * Yn,m(θ, φ)
Here, An represents the expansion coefficients, jn is the spherical Bessel function of order n, and Yn,m represents the spherical harmonics.
The sum extends over all possible values of n and m, which results in an infinite series of terms representing spherical waves with different orders and directions.
Each term represents a specific spherical wave with a particular amplitude (given by An), radial dependence (jn(k * r)), and angular dependence (Yn,m(θ, φ)).
The multipole expansion provides a way to describe the plane wave in terms of an infinite number of spherical waves, accounting for the complexity and spatial variation of the original wave.
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A car of mass 2170 kg is driving along a long road. The car is required to navigate a turn banked at an angle 24° with respect to the horizontal axis. The banked turn has a radius of curvature, 104 m. There is a coefficient of static friction between the tires and the road of μs = 0.63. The car can drive at a speed of vmax without slipping up the incline.
What is the maximum speed, vmax, the car can take on this banked curve?
The maximum speed, vmax, that the car can take on the banked curve is approximately 31.6 m/s.
To determine the maximum speed, we need to consider the forces acting on the car during the banked turn. The gravitational force acting on the car can be resolved into two components: one perpendicular to the road (Fn) and one parallel to the road (Fg).
The maximum speed can be achieved when the static friction force (Fs) between the tires and the road provides the centripetal force required for circular motion. The maximum static friction force can be calculated using the formula:
Fs(max) = μs * Fn
The normal force (Fn) can be determined using the vertical equilibrium equation:
Fn = mg * cos(θ)
where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the banked turn.
The centripetal force (Fc) required for circular motion is given by:
Fc = m * v^2 / r
where v is the velocity of the car and r is the radius of curvature.
Setting Fs(max) equal to Fc, we can solve for the maximum velocity:
μs * Fn = m * v^2 / r
Substituting the expressions for Fn and μs * Fn, we get:
μs * mg * cos(θ) = m * v^2 / r
Simplifying the equation and solving for v, we find:
v = √(μs * g * r * tan(θ))
Substituting the given values, we have:
v = √(0.63 * 9.8 m/s^2 * 104 m * tan(24°))
Calculating the value, we find:
v ≈ 31.6 m/s
Therefore, the maximum speed, vmax, that the car can take on this banked curve is approximately 31.6 m/s.
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A monochromatic light is directed onto a single slit 2.5 x 10-3 mm wide. If the angle between the first dark fringes (minimums) and the central maximum is 20°: a) Calculate the wavelength of light. b) Determine the angular position of the second minimum.
a) The wavelength of light. λ = 7.12 x 10^(-7) mm or 712 nm. b)The angular position of the second minimum is approximately 1.79°.
To calculate the wavelength of light and determine the angular position of the second minimum in a single-slit diffraction experiment, we can use the given information of the width of the slit and the angle between the first dark fringes and the central maximum.
First, let's calculate the wavelength of light (λ). The formula for the angular position (θ) of the first minimum in a single-slit diffraction pattern is given by θ = λ / (2d), where d is the width of the slit. Rearranging the formula, we have λ = 2d * tan(θ). Plugging in the values, with d = 2.5 x 10^(-3) mm and θ = 20°, we can calculate the wavelength to find λ = 7.12 x 10^(-7) mm or 712 nm.
Next, we need to determine the angular position of the second minimum. The angular position of the nth minimum (θ_n) is given by θ_n = (nλ) / d. For the second minimum, n = 2. Plugging in the calculated value of λ = 7.12 x 10^(-7) mm and d = 2.5 x 10^(-3) mm.
We can find the angular position of the second minimum to be θ_2 = 2 * (7.12 x 10^(-7) mm) / (2.5 x 10^(-3) mm) = 1.79°.Therefore, the wavelength of light is approximately 712 nm, and the angular position of the second minimum is approximately 1.79°.
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(a) What is the angular speed w about the polar axis of a point on Earth's surface at a latitude of 35°N? (Earth rotates about that axis.) (b) What is the linear speed v of the point?
a)ω = 2π / (23 hours + 56 minutes + 4 seconds), b)The value of v = ω * 6.371 x 10^6 meters
(a) The angular speed, denoted by ω, about the polar axis of a point on Earth's surface can be calculated using the formula:
ω = 2π/T
where T is the period of rotation. The period of rotation can be determined by the sidereal day, which is the time it takes for Earth to make one complete rotation relative to the fixed stars. The sidereal day is approximately 23 hours, 56 minutes, and 4 seconds.
However, the latitude information is not directly relevant for calculating the angular speed. The angular speed is the same for all points on Earth's surface about the polar axis. Therefore, we can use the period of rotation of 23 hours, 56 minutes, and 4 seconds to find the angular speed.
Substituting the values into the formula:
ω = 2π / (23 hours + 56 minutes + 4 seconds)
Calculate the numerical value of ω in radians per second.
(b) The linear speed, denoted by v, of a point on Earth's surface can be determined using the formula:
v = ω * R
where R is the radius of the Earth. The radius of the Earth is approximately 6,371 kilometers (6.371 x 10^6 meters).
Substituting the calculated value of ω into the formula:
v = ω * 6.371 x 10^6 meters
Calculate the numerical value of v in meters per second.
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A string with a linear density of 7.11×10−4 kg/m and a length of 1.14 m is stretched across the open end of a closed tube that is 1.39 m long. The diameter of the tube is very small. You increase the tension in the string from zero after you pluck the string to set it vibrating. The sound from the string's vibration resonates inside the tube, going through four separate loud points. What is the tension in the string when you reach the fourth loud point? Assume the speed of sound in air is 343 m/s.
The tension in the string when you reach the fourth loud point is 27.56 N.
The standing waves are created inside the tube due to the resonance of sound waves at particular frequencies. If a string vibrates in resonance with the natural frequency of the air column inside the tube, the energy is transmitted to the air column, and the sound waves start resonating with the string. The string vibrates more and, thus, produces more sound.
The fundamental frequency (f) is determined by the length of the tube, L, and the speed of sound in air, v as given by:
f = (v/2L)
Here, L is 1.39 m and v is 343 m/s. Therefore, the fundamental frequency (f) is:
f = (343/2 × 1.39) Hz = 123.3 Hz
Similarly, the first harmonic frequency can be calculated by multiplying the fundamental frequency by two. The second harmonic frequency is three times the fundamental frequency. Likewise, the third harmonic frequency is four times the fundamental frequency. The frequencies of the four loud points can be calculated as:
f1 = 2f = 246.6 Hz
f2 = 3f = 369.9 Hz
f3 = 4f = 493.2 Hz
f4 = 5f = 616.5 Hz
For a string of length 1.14 m with a linear density of 7.11×10⁻⁴ kg/m and vibrating at a frequency of 616.5 Hz, the tension can be calculated as:
Tension (T) = (π²mLf²) / 4L²
where m is the linear density, f is the frequency, and L is the length of the string.
T = (π² × 7.11 × 10⁻⁴ × 1.14 × 616.5²) / 4 × 1.14²
T = 27.56 N
Therefore, when the fourth loud point is reached, the tension in the string is 27.56 N.
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Question 77 (10 points) You and your friend are watching Bill Nye Saves the World: The Earth is a Hot Mess. In the episode, Bill Nye explains climate change and how our Earth is warming. Your friend hasn't been taking Physics class with you this semester, and he doubts the episode's validity. You have a friendly discussion on the topic. a. Your friend says, "Releasing more greenhouse gases into the air won't harm our Earth - we need greenhouse gases - like water vapor and CO2 - to survive. If it wasn't for these gases trapping in heat, our planet would be too cold!" How do you respond? Your response should include a description of what greenhouse gases are, what the greenhouse effect is, and why adding more greenhouse gases is helpful or harmful (do you agree or disagree with your friend?). (10 points)
Greenhouse gases are essential for maintaining a habitable temperature on Earth, but adding more of them can have harmful consequences.
Greenhouse gases, such as water vapor and carbon dioxide (CO2), play a crucial role in regulating Earth's temperature through the greenhouse effect. The greenhouse effect is a natural process in which certain gases in the atmosphere trap heat from the sun, preventing it from escaping back into space. This helps to keep our planet warm enough to sustain life.
While it is true that greenhouse gases are necessary for our survival, the issue lies in the balance. Human activities, particularly the burning of fossil fuels and deforestation, have significantly increased the concentration of greenhouse gases in the atmosphere, primarily CO2. This excess accumulation is causing the greenhouse effect to intensify, leading to global warming and climate change.
Adding more greenhouse gases to the atmosphere, beyond what is required for the natural balance, contributes to the acceleration of global warming. The increased heat retention leads to various adverse effects, such as rising sea levels, extreme weather events, disrupted ecosystems, and threats to human health and well-being.
Therefore, while it is accurate that we need greenhouse gases to maintain a livable temperature on Earth, the excess emissions resulting from human activities are disrupting the delicate equilibrium and causing harmful consequences. It is crucial that we take measures to reduce greenhouse gas emissions and transition to sustainable alternatives to mitigate the impacts of climate change and ensure a sustainable future for our planet and future generations.
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Please write down enough detail to demonstrate your understanding and explain it. Eric posts a timelapse video of the very large pressure chamber he built for fun. Inside the chamber, he puts an unusually large balloon with helium inside which he says is a 2.40-mol sample. The chamber is in his basement which stays at a steady 290K, which includes the inside of the chamber. He can very slowly adjust the pressure of the chamber, which means the pressure inside the balloon is approximately the same pressure. The time lapse starts with the display showing a pressure of 0.400 atm is compressed slowly enough to assume it is isothermal until it reaches 1.00 atm. In these conditions you can assume the helium behaves as an ideal gas.
(a) Find the final volume of the balloon.
m3
(b) Find the work done on the gas. Enter as a positive number. (note the units here!).
kJ
(c) Find the energy transferred by heat. Be aware of the units! Use a positive number if heat is absorbed by the balloon, and a negative number if heat is released by the balloon.
kJ
(d) Extra Credit: How many grams of helium are in the balloon?
grams
The final volume of the balloon is 18.2 L. the work done on the gas. Enter as a positive number is -1.55 kJ. the energy transferred by heat is -1.55 kJ. Grams in Helium are in the balloon is 9.6 g.
(a) The final volume of the balloon is to be determined. Initial volume, V₁ = (2.40 mol x 8.31 J K⁻¹ mol⁻¹ x 290 K)/0.400 atm = 45.5 LFinal pressure, P₂ = 1.00 atm Initial pressure, P₁ = 0.400 atm According to Boyle’s law:P₁V₁ = P₂V₂V₂ = P₁V₁/P₂ = (0.400 atm x 45.5 L)/1.00 atmV₂ = 18.2 L
(b) The work done on the gas is to be determined. The process is isothermal, and for this case, the work done on the gas is given by:W = nRT ln(V₂/V₁)W = (2.40 mol x 8.31 J K⁻¹ mol⁻¹ x 290 K) ln (18.2/45.5)W = -1552 J = -1.55 kJ Therefore, the work done on the gas is -1.55 kJ
(c) The energy transferred by heat is to be determined. For an isothermal process, the heat transferred is equal to the work done. Therefore, the energy transferred by heat is -1.55 kJ.
(d) The mass of the helium in the balloon is to be determined. Molar mass of helium, M = 4.00 g/mol Number of moles, n = 2.40 molMass of helium, m = nM = 2.40 mol x 4.00 g/mol = 9.6 g Therefore, there are 9.6 g of helium in the balloon.
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Two particles are fixed to an x axis: particle 1 of charge 91 = 1.79 × 10^-8 C at x = 18.0 cm and particle 2 of charge 92 =-3.24g1 at x = 65.0
cm. At what coordinate on the x axis is the electric field produced by the particles equal to zero?
"To find the coordinate on the x-axis where the electric field produced by the particles is equal to zero, we need to calculate the electric field at different points and determine where it becomes zero."
The electric field produced by a point charge at a distance r from the charge is given by the equation:
E = k * (q / r²)
where E is the electric field, k is the electrostatic constant (k = 8.99 x 10⁹ Nm²/C²), q is the charge of the particle, and r is the distance from the particle.
Let's calculate the electric field produced by particle 1 at different points along the x-axis:
For particle 1:
q1 = 1.79 x 10⁻⁸ C
x1 = 18.0 cm = 0.18 m
Now, let's calculate the electric field produced by particle 2 at different points along the x-axis:
For particle 2:
q2 = -3.24 x 10⁻⁹ C
x2 = 65.0 cm = 0.65 m
Now, we can calculate the electric field at a particular point on the x-axis by summing the electric fields produced by both particles:
E_total = E1 + E2
We can set up the equation:
k * (q1 / r1²) + k * (q2 / r2²) = 0
Simplifying the equation:
(q1 / r1²) + (q2 / r2²) = 0
Now, we can solve this equation to find the value of r (the coordinate on the x-axis) where the electric field is equal to zero.
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In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.20x1016 kg and a radius of 10.0 km. What is the speed of a satellite orbiting 4.60 km above the surface? What is the escape speed from the asteroid? Express your answer with the appropriate units.
The speed of a satellite orbiting 4.60 km above the surface of the asteroid is approximately 2.33 km/s, while the escape speed from the asteroid is about 4.71 km/s.
In order to calculate the speed of a satellite in orbit around the asteroid, we can use the formula for the orbital velocity of a satellite. This formula is derived from the balance between gravitational force and centripetal force:
V = sqrt(GM/r)
Where V is the velocity, G is the gravitational constant (approximately 6.674 × [tex]10^{-11}[/tex] [tex]m^3/kg/s^2[/tex]), M is the mass of the asteroid, and r is the distance from the center of the asteroid to the satellite.
Given that the mass of asteroid is 1.20 ×[tex]10^{16}[/tex] kg and the satellite is orbiting 4.60 km (or 4,600 meters) above the surface, we can calculate the orbital velocity as follows:
V = sqrt((6.674 × 10^-11[tex]m^3/kg/s^2[/tex]) * (1.20 × [tex]10^{16}[/tex]kg) / (10,000 meters + 4,600 meters))
Simplifying the equation, we find:
V ≈ 2.33 km/s
This is the speed of the satellite orbiting 4.60 km above the surface of the asteroid.
To calculate the escape speed from the asteroid, we can use a similar formula, but with the distance from the center of the asteroid to infinity:
V_escape = sqrt(2GM/r)
Using the same values for G and M, and considering the radius of the asteroid to be 10.0 km (or 10,000 meters), we can calculate the escape speed:
V_escape = sqrt((2 * 6.674 × [tex]10^{-11}[/tex] [tex]m^3/kg/s^2[/tex]) * (1.20 × [tex]10^{16}[/tex] kg) / (10,000 meters))
Simplifying the equation, we find:
V_escape ≈ 4.71 km/s
This is the escape speed from the asteroid.
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Multiple-Concept Example 7 discusses how problems like this one can be solved. A 9.70-4C charge is moving with a speed of 6.90x 104 m/s parallel to a very long, straight wire. The wire is 5.50 cm from the charge and carries a current of 61.0 A. Find the magnitude of the force on the charge. 9
The magnitude of the force on the charge is 73056 N. A 9.70-4C charge is moving with a speed of 6.90x 104 m/s parallel to a very long, straight wire. The wire is 5.50 cm from the charge and carries a current of 61.0 A.
The formula for the magnetic force on a moving charge is given by:
F = (μ₀ * I * q * v) / (2 * π * r),
where F is the magnitude of the force, μ₀ is the permeability of free space (μ₀ = 4π × 10⁻⁷ T·m/A), I is the current, q is the charge, v is the velocity, and r is the distance between the charge and the wire.
Plugging in the given values:
μ₀ = 4π × 10⁻⁷ T·m/A,
I = 61.0 A,
q = 9.70 × 10⁻⁴ C,
v = 6.90 × 10⁴ m/s,
r = 5.50 cm = 0.055 m,
It can calculate the magnitude of the force as follows:
F = (4π × 10⁻⁷ T·m/A * 61.0 A * 9.70 × 10⁻⁴ C * 6.90 × 10⁴ m/s) / (2 * π * 0.055 m)
= (2 * 10⁻⁷ T·m/A * 61.0 A * 9.70 × 10⁻⁴ C * 6.90 × 10⁴ m/s) / 0.055 m
= (2 * 61.0 * 9.70 × 10⁻⁴ * 6.90 × 10⁴) / 0.055
= (2 * 61.0 * 9.70 × 6.90) / 0.055
= 2 * 61.0 * 9.70 * 6.90 / 0.055
= 73056 N
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(1 pt) The magnetic flux through a coil containing 10 loops changes from 20Wb to-20Wb in 0.03s. Find the induced voltage c.
The induced voltage is 1333.33V.
The induced voltage is in the direction that opposes the change in the magnetic flux. Since the magnetic flux is decreasing, the induced voltage is in the direction that creates a magnetic field that increases the magnetic flux.
Here are the given:
* Number of loops: 10
* Change in magnetic flux: 20Wb - (-20Wb) = 40Wb
* Change in time: 0.03s
To find the induced voltage, we can use the following formula:
V_ind = N * (dPhi/dt)
where:
V_ind is the induced voltage
N is the number of loops
dPhi/dt is the rate of change of the magnetic flux
V_ind = 10 * (40Wb / 0.03s) = 1333.33V
Therefore, the induced voltage is 1333.33V.
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Two Trucks A and B are parked near you on a road. Truck A is stationary and truck B is moving away at a constant speed of 30 km/h. Each Truck is equipped with a horn emitting a sound at a frequency of 200Hz. Both whistle at the same time. a) What frequency will you hear from each truck? b) Will there be a beat? If or what is the frequency of the beats?
a. The frequency emitted by truck A will be 200 Hz and the frequency emitted by truck B will be approximately 198.56 Hz
b. The frequency of the beats is 1.44 Hz.
a) Truck A is stationary and truck B is moving away at a constant speed of 30 km/h. Both of the trucks emit a sound of frequency 200 Hz and the speed of sound is 343 m/s, the frequency of sound will be affected by the Doppler effect.
The Doppler effect can be given by:
[tex]f'= \frac {v \pm v_0} {v\pm v_s}f[/tex]
Here, f is the frequency of the sound emitted.
v is the velocity of sound in air ($343 m/s$)
v0 is the velocity of the object emitting the sound and vs is the velocity of the sound wave relative to the stationary object
In this problem, the frequency emitted by the truck A is
[tex]f_{A} = 200[/tex]Hz
v0 = 0m/s
v = 343m/s
The frequency emitted by the truck B is [tex]f_{B} = 200[/tex] Hz
[tex]v0 = - 30km/h \\= - \frac{30 \times 1000}{3600}$ m/s \\= $-\frac{25}{3}$ ms^{-1} \\v= 343m/s[/tex]
On substituting the above values in the Doppler's equation, we get,
For truck A,
[tex]f_{A}' = \frac{v}{v\pm v_{s}}[/tex]
[tex]f_{A}' = \frac{343}{343\pm 0} Hz = 200[/tex] Hz
For truck B,[tex]f_{B}' = \frac{v}{v\pm v_{s}}[/tex]
[tex]f_{B}' = \frac{343} {343 \pm \frac {25}{3}}\text{Hz}[/tex] ≈ 198.56 Hz
Hence the frequency emitted by truck A will be 200 Hz and the frequency emitted by truck B will be approximately 198.56 Hz
b) A beat is produced when two sound waves having slightly different frequencies are superposed.
In this problem, as we see that the frequency of the wave emitted by truck A is 200 Hz and the frequency of the wave emitted by truck B is approximately 198.56 Hz, we can say that a beat will be produced.
To find the frequency of beats, we use the formula for beats:
fbeat = |f1 − f2|
Where,f1 is the frequency of the wave emitted by truck Af2 is the frequency of the wave emitted by truck B
Frequencies of the waves are given by,
f1 = 200 Hz
f2 = 198.56 Hz
fbeat = |200 − 198.56| Hz ≈ 1.44 Hz
Thus, the frequency of the beats is 1.44 Hz.
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a). You will hear a frequency of approximately 195.84 Hz from Truck B.
b). The beat frequency between the two trucks' sounds will be approximately 4.16 Hz.
a) To determine the frequency you will hear from each truck, we need to consider the Doppler effect. The Doppler effect describes how the perceived frequency of a sound wave changes when the source of the sound or the listener is in motion relative to each other.
For the stationary Truck A, there is no relative motion between you and the truck. Therefore, the frequency you hear from Truck A will be the same as its emitted frequency, which is 200 Hz.
For the moving Truck B, which is moving away from you at a constant speed of 30 km/h, the frequency you hear will be lower than its emitted frequency due to the Doppler effect. The formula for the Doppler effect when a source is moving away is given by:
f' = f * (v_sound + v_observer) / (v_sound + v_source)
where f is the emitted frequency, v_sound is the speed of sound (approximately 343 m/s), v_observer is the speed of the observer (you, assumed to be stationary), and v_source is the speed of the source (Truck B).
Converting the speed of Truck B from km/h to m/s:
v_source = 30 km/h * (1000 m/km) / (3600 s/h) = 8.33 m/s
Plugging in the values:
f' = 200 Hz * (343 m/s + 0 m/s) / (343 m/s + 8.33 m/s)
Simplifying the equation:
f' ≈ 195.84 Hz
Therefore, you will hear a frequency of approximately 195.84 Hz from Truck B.
b) Yes, there will be a beat if the frequencies of the two trucks are slightly different. The beat frequency is equal to the absolute difference between the frequencies of the two sounds.
Beat frequency = |f_A - f_B|
Substituting the values:
Beat frequency = |200 Hz - 195.84 Hz|
Simplifying:
Beat frequency ≈ 4.16 Hz
So, the beat frequency between the two trucks' sounds will be approximately 4.16 Hz.
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