The period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].
Step 1
The period T of a simple pendulum is given by the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where:
- T is the period of the pendulum,
- L is the length of the pendulum, and
- g is the acceleration due to gravity.
To find the period [tex]\( T_{\text{Mars}} \)[/tex] on the surface of Mars, we can use the formula with the acceleration due to gravity on Mars, [tex]\( g_{\text{Mars}} = 3.71 \, \text{m/s}^2 \)[/tex], and the same length L as on Earth.
We have the period [tex]\( T_{\text{Earth}} = 2.00 \, \text{s} \)[/tex] on Earth, and we need to find [tex]\( T_{\text{Mars}} \).[/tex]
Step 2
Using the formula, we get:
[tex]\[ T_{\text{Earth}} = 2\pi \sqrt{\frac{L}{g_{\text{Earth}}}} \][/tex]
Solving for L, we find:
[tex]\[ L = \left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}} \][/tex]
Now, using this value of L, we can find [tex]\( T_{\text{Mars}} \)[/tex] using the acceleration due to gravity on Mars:
[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{L}{g_{\text{Mars}}}} \][/tex]
Substituting the known values, we get:
[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{\left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}}}{g_{\text{Mars}}}} \][/tex]
Step 3
Simplifying:
[tex]\[ T_{\text{Mars}} = T_{\text{Earth}} \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}} \]Now, let's calculate \( T_{\text{Mars}} \):\[ T_{\text{Mars}} = 2.00 \times \sqrt{\frac{9.81}{3.71}} \]\[ T_{\text{Mars}} = 2.00 \times \sqrt{2.643} \]\[ T_{\text{Mars}} \approx 2.00 \times 1.626 \]\[ T_{\text{Mars}} \approx 3.25 \, \text{s} \][/tex]
So, the period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].
Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the amount of entropy produced, in kJ/K per kg of air, for the compression. What is the minimum theoretical work input, in kJ per kg of air, for an adiabatic compression from the given initial state to a final pressure of 10 bar?
Final answer:
The entropy production for a reversible adiabatic process of an ideal gas is zero. For the actual work input required for adiabatic compression, the formula involving pressures, volumes, and the heat capacity ratio is used, but cannot be calculated without additional information such as the specific volumes.
Explanation:
To determine the amount of entropy produced during an adiabatic compression of air modeled as an ideal gas, we need to recognize that, by definition, an adiabatic process is one in which no heat is transferred to or from the gas. Therefore, assuming the process is also reversible (which it must be, if we are to calculate a non-zero entropy production), the change in entropy (ΔS) for the process would actually be zero. However, as the conditions stated a rise in temperature during the compression, if any irreversibility were present in the real-world scenario, it would indeed generate entropy, but we need more information to calculate the precise amount for a real-world irreversible process.
The minimum theoretical work input for an adiabatic compression can be calculated using the first law of thermodynamics and the relation for adiabatic processes defined as PV^{γ} = constant, where γ (gamma) is the heat capacity ratio (Cp/Cv). For an ideal gas, the work done (W) on the air during adiabatic compression can be expressed using the formula W = (P2*V2 - P1*V1) / (γ - 1), where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively. However, since the volume is not given in the question, this calculation cannot be completed without further information.
The entropy produced per kg of air during the adiabatic compression is approximately [tex]\( 0.0348 \)[/tex] kJ/K, and the minimum theoretical work input for the compression is approximately [tex]\( 215.25 \)[/tex] kJ/kg of air.
The amount of entropy produced during the adiabatic compression process in a piston-cylinder assembly, where air is modeled as an ideal gas, can be determined using the following relationship for an adiabatic process:
[tex]\[ s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
where [tex]\( s \)[/tex] is the specific entropy, [tex]\( c_p \)[/tex] is the specific heat at constant pressure, [tex]\( T \)[/tex] is the absolute temperature, [tex]\( R \)[/tex] is the specific gas constant, and [tex]\( P \)[/tex] is the absolute pressure. The subscripts [tex]\( 1 \) and \( 2 \)[/tex] denote the initial and final states, respectively.
Given:
- Initial state: [tex]\( P_1 = 1 \) bar, \( T_1 = 300 \)[/tex] K
- Final state: [tex]\( P_2 = 10 \) bar, \( T_2 = 600 \)[/tex] K
- For air, [tex]\( c_p = 1.005 \) kJ/kg\K and \( R = 0.287 \) kJ/kg[/tex]
First, convert the pressures from bar to Pa (since the gas constant [tex]\( R \)[/tex] is in terms of Pa):
- [tex]\( P_1 = 1 \times 10^5 \) Pa[/tex]
- [tex]\( P_2 = 10 \times 10^5 \) Pa[/tex]
Now, calculate the entropy change:
[tex]\[ s_2 - s_1 = 1.005 \ln\left(\frac{600}{300}\right) - 0.287 \ln\left(\frac{10 \times 10^5}{1 \times 10^5}\right) \][/tex]
[tex]\[ s_2 - s_1 = 1.005 \ln(2) - 0.287 \ln(10) \][/tex]
[tex]\[ s_2 - s_1 = 1.005 \times 0.693 - 0.287 \times 2.303 \][/tex]
[tex]\[ s_2 - s_1 \approx 0.6965 - 0.6617 \][/tex]
[tex]\[ s_2 - s_1 \approx 0.0348 \text{ kJ/kg\K} \][/tex]
For the minimum theoretical work input during an adiabatic compression, we use the following equation for an ideal gas:
[tex]\[ w_{\text{in,min}} = \int_{V_1}^{V_2} P \, dV \][/tex]
For an adiabatic process, [tex]\( PV^\gamma = \text{constant} \), where \( \gamma \)[/tex] is the heat capacity ratio ([tex]\( c_p/c_v \)[/tex]). The work input can be calculated as:
[tex]\[ w_{\text{in,min}} = \frac{P_2V_2 - P_1V_1}{\gamma - 1} \][/tex]
Using the ideal gas law, [tex]\( PV = mRT \)[/tex], where [tex]\( m \)[/tex] is the mass of the gas, we can express [tex]\( V \)[/tex] in terms of [tex]\( P \) and \( T \)[/tex]:
[tex]\[ V = \frac{mRT}{P} \][/tex]
Since the mass [tex]\( m \)[/tex] cancels out, we can write:
[tex]\[ w_{\text{in,min}} = \frac{mRT_2 - mRT_1}{\gamma - 1} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{R(T_2 - T_1)}{\gamma - 1} \][/tex]
Given that [tex]\( \gamma = \frac{c_p}{c_v} = \frac{c_p}{c_p - R} \)[/tex], we can calculate \( \gamma \) for air:
[tex]\[ \gamma = \frac{1.005}{1.005 - 0.287} \approx 1.4 \][/tex]
Now, calculate the minimum work input:
[tex]\[ w_{\text{in,min}} = \frac{0.287(600 - 300)}{1.4 - 1} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{0.287 \times 300}{0.4} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{86.1}{0.4} \][/tex]
[tex]\[ w_{\text{in,min}} \approx 215.25 \text{ kJ/kg} \][/tex]
The concept of critical density is just like the idea of escape velocity for a projectile launched from the earth. An object launched with a velocity less than the escape velocity will fall back to the earth, and an object launched with one greater than the escape velocity can escape the earth's gravitational field. Find an expression for vesc, the escape velocity of a projectile launched from the earth's surface. Express your answer in terms of the universal gravitational constant G, the mass of the earth Me, and the radius of the earth Re.
Answer:
v = √ 2 G M/ [tex]R_{e}[/tex]
Explanation:
To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.
Initial
Em₀ = K + U₀
Final
[tex]Em_{f}[/tex] = [tex]U_{f}[/tex]
The kinetic energy is k = ½ m v²
The gravitational potential energy is U = - G m M / r
r is the distance measured from the center of the Earth
How energy is conserved
Em₀ = [tex]Em_{f}[/tex]
½ mv² - GmM / [tex]R_{e}[/tex] = -GmM / r
v² = 2 G M (1 / [tex]R_{e}[/tex] – 1 / r)
v = √ 2GM (1 / [tex]R_{e}[/tex] – 1 / r)
The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0
v = √ 2GM / [tex]R_{e}[/tex]
The escape velocity of an object from Earth is calculated by equating the object's kinetic energy to its gravitational potential energy, which yields the formula v_esc = sqrt(2GM_e/R_e), where G is the universal gravitational constant, M_e is Earth's mass, and R_e is Earth's radius.
The escape velocity (vesc) is the speed at which an object must travel to break free from the gravitational pull of a celestial body, in this case, Earth. To find the escape velocity, we use the principle that the kinetic energy at launch should be equal to the gravitational potential energy. The formula for escape velocity is derived using the equation KE = GPE, where KE stands for kinetic energy and GPE for gravitational potential energy. The kinetic energy of the object is given by (1/2)mvesc2 and the gravitational potential energy by -G(Mem)/Re, where m is the mass of the object, G is the universal gravitational constant, Me is the mass of the earth, and Re is the radius of the earth. Setting these two energies equal to each other and solving for vesc while cancelling out the mass of the object (since it appears on both sides of the equation), we obtain vesc = sqrt(2GMe/Re).
When radio waves try to pass through a city, they encounter thin vertical slits: the separations between the buildings. This causes the radio waves to diffract. In this problem, you will see how different wavelengths diffract as they pass through a city and relate this to reception for radios and cell phones. You will use the angle from the center of the central intensity maximum to the first intensity minimum as a measure of the width of the central maximum (where nearly all of the diffracted energy is found).a. Find the angle θ to the first minimum from thecenter of the central maximum (Express your answer in terms λ and a.):b. What is the angle θFM to the first minimum foran FM radio station with a frequency of 101mMHz? (Express your answer numerically indegrees to three significant figures. Note: Do not write youranswer in terms of trignometric functions. Evaluate any suchfunctions in your working.)c. What is the angle θcell for a cellular phonethat uses radiowaves with a frequency of 900MHz? (Express your answer indegrees to three significant figures.)d. What problem do you encounter in tryingto find the angle θAM for an AM radio stationwith frequency 1000kHz?i. The angle becomes zero.ii. The angle can be given only in radians.iii. To find the angle it would be necessary to takethe arcsine of a negative number.iv. To find the angle it would be necessary totake the arcsine of a number greater than one.
Final answer:
The diffraction of radio waves when passing through city buildings can be represented by single-slit diffraction. The angle to the first minimum can be calculated using the equation θ = sin-1(λ/a), but for AM radio with very large wavelengths, this calculation may not be valid as it could require taking the arcsine of a number greater than one.
Explanation:
When radio waves encounter thin vertical slits such as the spaces between buildings, they diffraction occurs. The property of diffraction can be analyzed using the concept of single-slit diffraction from wave optics. For a single-slit diffraction, the angle θ to the first minimum can be found using the equation θ = sin-1(λ/a), where λ is the wavelength of the wave and a is the width of the slit.
For an FM radio station with a frequency of 101 MHz, we would use the relationship between frequency (f), wavelength (λ), and the speed of light (c) to find the wavelength (λ = c/f) before calculating the angle using the aforementioned equation.
Similarly, for a cellular phone using radio waves with a frequency of 900 MHz, we again find the wavelength using the same relation and then calculate the angle θ to the first minimum.
However, for AM radio, the complication arises because the wavelengths for AM radio are considerably larger. This can lead to a scenario where the slit width is not narrow enough compared to the wavelength, and as a result, the angle θAM calculated using sin-1(λ/a) may result in taking the arcsine of a number greater than one, which is not possible and indicates that the first minimum may not occur.
A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the
horse going when it travels 15.3 m from where it started to accelerate.
Answer:
The horse is going at 12.72 m/s speed.
Explanation:
The initial speed of the horse (u) = 3 m/s
The acceleration of the horse (a)= 5 m/[tex]s^{2}[/tex]
The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m
Applying the second equation of motion to find out the time,
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]15.3=3t+2.5t^{2}[/tex]
[tex]2.5t^{2}+3t-15.3=0[/tex]
Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.
Now applying first equation of motion, to find out the final velocity,
[tex]v=u+at[/tex]
[tex]v=3+1.945*5[/tex]
[tex]v=3+9.72[/tex]
v=12.72 m/s
The horse travels at a speed of 12.72 m/s after covering the given distance.
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K. You set the bowl up so that as it cools to room temperature the heat first flows through a Carnot Engine. The soup has Cv= (33 J/K). Assume that the volume of the soup does not change.
1. What fraction of the total heat QH that is lost by the soup can be turned into useable work by the engine?
Answer:
Explanation:
Heat energy given out by the soup
= C_v x ( t₂ - t₁ )
= 33 x ( 340 - 300)
= 1320 J
This heat is given to Carnot engine . Efficiency of engine
= (340 - 300 ) / 340
= 40 / 340
2 / 17
This fraction of total heat given is converted into useable work by the engine.
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield, as we saw in Figure 4.2. These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm.
a. A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s.
Estimate the net force on the person if he or she is wearing a seat belt and if the air
bag deploys.
b. Estimate the net force that ultimately stops the person if he or she is not restrained
by a seat belt or air bag.
c. How does these two forces compare to the person's weight
Answer:
a) [tex]F_{belt}[/tex] = 1.35 10⁴ N , b) F = 1.35 10⁶ N, c) F / [tex]F_{belt}[/tex] = 100
Explanation:
a) Let's start by calculating the aceleration it takes for the person with a seat belt and air bag to stop, the approximate distance between the floor and the steering wheel is about 50 cm, let's use kinematics to calculate the acceleration
v² = v₀² - 2 a x
0 = v₀² - 2 a x
a = v₀² / 2 x
a = 15² / (2 0.50)
a = 225 m / s²
We calculate the force with Newton's second law
F = m a
F = 60 225
[tex]F_{belt}[/tex] = 1.35 10⁴ N
b) we perform the same calculation for a person without a belt
a = v₀² / 2x
a = 15² / (2 0.005)
a = 22500 m / s²
F = m a
F = 60 22500
F = 1.35 10⁶ N
c) let's calculate the relationship between these two forces
F / [tex]F_{belt}[/tex] = 1.35 10⁶ / 1.35 10⁴
F / [tex]F_{belt}[/tex] = 10²
a)A certain medical machine emits x-rays with a minimum wavelength of 0.024 nm. One day, the machine has an electrical problem and the voltage applied to the x-ray tube decreases to 76% of its normal value. Now what is the minimum x-ray wavelength produced by the machine?
lmin = ___________nm
(b) What is the maximum x-ray energy this machine (with electrical problems) can produce?
Emax =___________ eV
(c) The atomic number of an element is 82. According to the Bohr model, what is the energy of a Ka x-ray photon?
E =_______________ eV
Answer:
(a) 0.032 nm
(b) 39,235 eV
(c) 70,267.8 eV
Explanation:
(a) The energy of a photon can be calculated using:
E = hc/λ equation (1)
where:
h = 4.13*10^-15 eV.s
c = 3*10^8 m/s
λ = 0.024*10^-9 m
Thus:
E = (4.13*10^-15)*(3*10^8)/0.024*10^-9 = 51,625 eV
Then we calculate 76% of this estimated energy and determine the new wavelength:
[tex]E_{new} = 0.76*51625 = 39,235 eV[/tex]
Using equation (1) to determine the new wavelength:
λ[tex]_{new} = \frac{h*c}{E_{new} }[/tex]
λ[tex]_{new}[/tex] = (4.13*10^-15)*(3*10^8)/39235 = 3.15*10^-11 m = 0.032 nm
(b) As calculated in part (a), the maximum x-ray energy this machine can produce is [tex]E_{new} = 0.76*51625 = 39,235 eV[/tex]
(c) The energy of a Ka x-ray photon can be estimated using:
[tex]E_{ka} = (10.2 eV)*(Z-1)^{2}[/tex]
where Z is the atomic number = 84.
[tex]E_{ka} = (10.2 eV)*(84-1)^{2}[/tex] = 70,267.8 eV
A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m per meter when the shaft rotates. The shaft is to be used for applying a torque of 0.1 N m to actuate a switch. What is the maximum length of shaft that may be used if the shear stress in the shaft is not to exceed 50 MPa?
Answer:
2.5m
Explanation:
Torque is defined as the rotational effect of a force on a body.
The torque T for the maximum shear stress is given as 0.1 Nm
Frictional torque is the torque caused by a frictional force
The frictional torque F is given as 0.04 Nm/m
The maximum length of the shaft is thus given as
L = T / F
= 0.1/0.04
L= 2.5 m
Two blocks of ice , one four times as heavy as the other, aren at rest on a frozen lake. a person pushes each block the same distance d.ignore friction and assume that an equal force F is exerted on each block. which of the following statements is true about the kinetic energy of the heavier block after the push?A) it is smaller than kinetic energy of the lighter blockB) it is equal to the kinetic energy of the lighter blockC) it is larger than the kinetic energy of the lighter blockD) it cannot be determined without knowing the force and the mass of each block
Answer:
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Explanation:
Hi there!
The work done on each block is calculated as follows:
W = F · d
Since the two blocks were pushed the same distance with the same force, the work done on each object is the same.
Using the work-energy theorem, we know that the work done on an object is equal to its change in kinetic energy (KE):
W = ΔKE
W = final KE - initial KE
Since the objects are at rest, initial KE = 0, then:
W = final KE
Since the work done on each block is the same, so will be its final kinetic energy.
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Final answer:
When an equal force is applied over the same distance to two blocks of ice, with one being four times heavier than the other, the kinetic energy gained by both blocks is equal, because the work done on them is the same.
Explanation:
The question involves two blocks of ice on a frozen lake, with one block being four times as heavy as the other. When an equal force F is exerted on each block, pushing them the same distance d, we are asked which statement is true about the kinetic energy of the heavier block after the push compared to the kinetic energy of the lighter block. Since the work done (work = force × distance) on both blocks is the same and work done on an object is equal to the change in its kinetic energy (Work-Energy Principle), both blocks will have the same increase in kinetic energy. Therefore, without considering initial kinetic energies (since both start from rest), the correct answer is that it is equal to the kinetic energy of the lighter block (Option B). This is because the amount of work done on both blocks is the same, leading to an equal increase in kinetic energy.
You are in a submarine and are at the surface of the ocean but out in the deep sea. There is a big storm and you want to dive down deep enough so that you avoid the turbulence of the sea. You notice the distance between successive waves (wave length) is 20 meters. How far down do you need to dive down to not feel the effect of the waves?
One of the maritime principles that relate the turbulence and wavelength of the waves is called the "depth of 1/2 wavelength" which is also usually referred to as the floor of the wave: A point of depth in which There is no movement. There if a submarine is found, it can be unbalanced and steadily navigate.
If the wavelength is 20 meters, then it must be submerged 10 meters (20/2) to avoid turbulence.
Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge q1) at the origin and the other sphere (charge q2) at x = +0.400 m. Available to you are a third sphere with net charge q3 = 3.00×10−6 C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the x-axis at x = 0.200 m; you measure the net force on it to be 6.50 N in the +x-direction. Then you move the third sphere to x = +0.600 m and measure the net force on it now to be 3.50 N in the +x-direction.
Part A) Calculate q1.
Part B) Calculate q2.
Answer:
a) q₁ = 15. 28 10⁻⁶C, b) q₂ = 5.64 10⁻⁶ C
Explanation:
For this exercise we use Newton's second law where force is Coulomb's electric force
Case 1. Distance (x₁ = 0.200 m) from the third sphere
F₁ = F₁₃ - F₂₃
F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²
F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²
Case2 Distance (x₂ = 0.6 m) from the third sphere
F₂ = F₁₃ + F₂₃
F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²
F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²
The distance is between the spheres, in the annex you can see the configuration of the charge and forces
Let's replace the values
F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²
F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²
6.50 = 674. 25 10³ (q₁ –q₂)
3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)
We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second
q₁ = q₂ + 6.50 / 674 10³
3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04
1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶
q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶
q₂ 27,777 = 156,557 10⁻⁶
q₂ = 156.557 10-6 /27.777
q₂ = 5.636 10⁻⁶ C
We look for q1 in the other equation
q₁ = q₂ + 6.50 / 674 10³
q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶
q₁ = 15. 28 10⁻⁶C
Final answer:
To find the charges on the two spheres, we can use Coulomb's law. Calculations show that q1 is approximately 4.333 × 10^-8 C and q2 is approximately 1.111 × 10^-7 C.
Explanation:
To determine the charges on the two spheres, we can use Coulomb's law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Let's calculate the charges:
Part A) Calculating q1:
Using the information given, we can set up the following equation:
F1 = k * (q1 * q3) / (0.200)², where F1 is the net force at x = 0.200 m.
Substituting the given values, we get:
6.50 = (9 × 10^9) * (q1 * (3 × 10^-6)) / (0.200)².
Simplifying the equation, we find that q1 = 4.333 × 10^-8 C.
Part B) Calculating q2:
Using the same equation, but with F2, the net force at x = 0.600 m, we have:
3.50 = (9 × 10^9) * (q2 * (3 × 10^-6)) / (0.600)².
Simplifying, we find that q2 = 1.111 × 10^-7 C.#SPJ11
Select the statement that correctly completes the description of phase difference.
Phase difference describes:
O the difference in the phase angle between any two waves at any given position along the waves.
O the shift between the positions of corresponding crests of two waves of the same frequency.
O the difference in the frequencies of two waves at a given time.
O the displacement of a wave particle from its undisturbed position at the origin.
Phase difference denotes the difference in phase angle between two waves at a given point, occurring when waves are separated by a whole number of multiples of wavelengths.
Explanation:Phase difference describes the difference in the phase angle between any two waves at any given position along the waves. When the waves have the same frequency and the difference in their path lengths is an integer multiple of the wavelength, the waves are said to be in phase. This means these points are separated by a whole number multiple of whole wave cycles or wavelengths. For example, sound waves can illustrate a phase shift when they have different path lengths. It is also important to understand that the wavelength is defined as the distance between any two adjacent points that are in phase.
The mean diameters of Mars and Earth are 6.9 x 10^3 km and 1.3 x 10^4 km, respectively. The mass of Mars is 0.11 times Earth’s mass.
a) What is the ratio of the mean density (mass per unit volume) of Mars to that of Earth?
b) What is the value of the gravitational acceleration on Mars?
c) What is the escape speed on Mars?
Final answer:
a) The ratio of the mean density of Mars to that of Earth is 0.11 times the ratio of the volume of Earth to the volume of Mars. b) The value of the gravitational acceleration on Mars is 3.7 m/s². c) The escape speed on Mars is 5.03 km/s.
Explanation:
a) To find the ratio of the mean density of Mars to that of Earth, we need to divide the mass of Mars by the volume of Mars and divide the mass of Earth by the volume of Earth. The mean density is given by:
Mean density (Mars) = mass (Mars) / volume (Mars)
Mean density (Earth) = mass (Earth) / volume (Earth)
Substituting the given values, we have:
Mean density (Mars) = (0.11 x mass (Earth)) / volume (Mars)
Mean density (Earth) = mass (Earth) / volume (Earth)
Dividing these two equations, we get the ratio of the mean densities as:
Ratio of mean density (Mars to Earth) = (0.11 x mass (Earth)) / volume (Mars) / (mass (Earth) / volume (Earth))
Simplifying, the ratio of mean densities is 0.11 times the ratio of the volume of Earth to the volume of Mars.
b) The value of the gravitational acceleration on Mars can be found using Newton's law of gravitation. The formula for gravitational acceleration is:
Gravitational acceleration = (Gravitational constant * mass of Mars) / radius of Mars^2
Substituting the given values, we have:
Gravitational acceleration on Mars = (6.67 x 10^-11 N m^2/kg^2 * 6.418 x 10^23 kg) / (3.38 x 10^6 m)^2 = 3.7 m/s^2
c) The escape speed on Mars can be found using the formula:
Escape speed = sqrt(2 x Gravitational constant x mass of Mars / radius of Mars)
Substituting the given values, we have:
Escape speed on Mars = sqrt(2 x 6.67 x 10^-11 N m^2/kg^2 x 6.418 x 10^23 kg / 3.38 x 106 m) = 5.03 km/s
A neutral copper block is polarized as shown in the figure below, due to an electric field made by external charges (notshown). Which arrow (a–j) in the figure below best indicates the direction of the net electric field at location B, which isinside the copper block?
Answer:
As point B is located inside the copper block so net electric field at point B is j.
Explanation:
Consider the figure attached below. The net electric field at location B,that is inside the copper block is zero because when a conductor is charged or placed in an electric field of external charges, net charge lies on the surface of conductor and there is no electric field inside the conductor. As point B is located inside the copper block so net electric field at point B is zero as well direction of net electric field at point B is zero.
Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same length (1.00 m) at 0°C. The coefficients of volume expansion for steel and invar are 3.6 × 10-5 /°C and 2.7 × 10-6 /°C respectively.What is their difference in length, in meters, at 20.5°C ?Repeat the calculation for two 30.0-m-long surveyor’s tapes
Answer:
Difference in Length of steel is 0.000246m
Difference in Length of invar is 0.00001845m
Difference in Length of steel surveyor's tape is 0.00738m
Difference in Length of invar surveyor's tape is 0.0005535m
Explanation:
Linear expansivity of steel is volume expansivity ÷ 3
Linear expansivity of steel = 0.000036/°C ÷ 3 = 0.000012/°C
Difference in Length of steel= Linear expansivity × initial length × temperature change
= 0.000012 × 1 × (20.5-0)
= 0.000012×1×20.5 = 0.000246m
Linear expansivity of invar = volume expansivity of invar ÷ 3
Linear expansivity of invar= 0.0000027/°C ÷ 3= 0.0000009/°C
Difference in Length of invar = 0.0000009×1×(20.5-0) = 0.00001845m
Difference in Length of steel surveyor's tape= 0.000012×30×(20.5-0) = 0.00738m
Difference in Length of invar surveyor's tape = 0.0000009×30×(20.5-0) = 0.0005535m
The difference in length between a steel and invar meter stick at 20.5°C is 6.83 mm, and for two 30-m-long surveyor's tapes, the difference at the same temperature is 20.5 cm.
Explanation:When a steel meter stick and an invar meter stick both expand due to an increase in temperature, we can determine their difference in length using the coefficients of linear expansion for each material.
The difference in length at 20.5°C can be determined using the formula:
ΔL = αLΔT
For steel, the coefficient of linear expansion (α) is 3.6 × 10^-5 /°C, and for invar it is 2.7 × 10^-6 /°C. Plugging in the values:
ΔL_steel = 3.6 × 10^-5 /°C × 1.00 m × 20.5 °C = 7.38 × 10^-3 mΔL_invar = 2.7 × 10^-6 /°C × 1.00 m × 20.5 °C = 5.54 × 10^-4 mThe difference in length is therefore 7.38 × 10^-3 m - 5.54 × 10^-4 m = 6.83 × 10^-3 m or 6.83 mm.
For two 30.00-m-long surveyor’s tapes:
ΔL_steel = 3.6 × 10^-5 /°C × 30.00 m × 20.5 °C = 2.21 × 10^-1 mΔL_invar = 2.7 × 10^-6 /°C × 30.00 m × 20.5 °C = 1.66 × 10^-2 mThe difference in length for the surveyor's tapes is 2.21 × 10^-1 m - 1.66 × 10^-2 m = 2.05 × 10^-1 m or 20.5 cm.
A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 375 N/m; the other end of the spring is fixed in place. The cookie has a kinetic energy of 19.0 J as it passes through the position where the spring is unstretched. As the cookie slides, a frictional force of magnitude 10.0 N acts on it. (a) How far will the cookie slide from the position where the spring is unstretched before coming momentarily to rest? m (b) What will be the kinetic energy of the cookie as it slides back through the position where the spring is unstretched?
Answer:
The conservation of energy should be used to answer this question.
a)
At the position where the spring is unstretched, the elastic potential energy of the spring is zero.
[tex]K_1 + U_1 - W_f = K_2 +U_2\\K_1 - W_f = U_2[/tex]
since [tex]U_1[/tex] and [tex]K_2[/tex] is equal to zero.
[tex]W_f = F_fx\\\\U_2 = \frac{1}{2}kx^2\\\\19 - (10)x = \frac{1}{2}(375)x^2\\\\375x^2 + 20x - 38 = 0[/tex]
The roots of this quadratic equation can be solved by using discriminant.
[tex]\Delta = b^2 - 4ac\\x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}[/tex]
[tex]x_1 = -0.346\\x_2 = 0.292[/tex]
We should use the positive root, so
x = 0.292 m.
b)
We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.
[tex]K_2 + U_2 - W_f = K_3 + U_3\\U_2 - W_f = K_3[/tex]
since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.
[tex]\frac{1}{2}kx^2 - F_fx = K_3\\K_3 = 15.987 - 2.92 = 13.067 J[/tex]
Explanation:
In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.
In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).
What is the difference between transverse and longitudinal waves?
a. Only longitudinal waves transmit matter.
b. Plane waves are transverse waves while spherical waves are longitudinal.
c. Only transverse waves transmit energy.
d. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.
e. Mechanical waves are transverse waves while electromagnetic waves are longitudinal.
To explain how transverse and longitudinal waves work, let us give two examples for each particular case.
In the case of transverse waves, the displacement of the medium is PERPENDICULAR to the direction of the wave. One way to visualize this effect is when you have a rope and between two people the rope is shaken horizontally. The shift is done from top to bottom. This phenomenon is common to see it in solids but rarely in liquids and gases. A common application usually occurs in electromagnetic radiation.
On the other hand in the longitudinal waves the displacement of the medium is PARALLEL to the direction of propagation of the wave. A clear example of this phenomenon is when a Slinky is pushed along a table where each of the rings will also move. From practice, sound waves enclose the definition of longitudinal wave displacement.
Therefore the correct answer is:
C. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.
Transverse waves have a perpendicular disturbance while longitudinal waves have a parallel disturbance.
Explanation:A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. For example, when you ripple a string up and down, you create a transverse wave. But when you push and pull a slinky back and forth, you create a longitudinal wave.
The intensity level of a power mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are mowing their lawn 20 m from your open bedroom window. What is the intensity level in your bedroom? The intensity level of a power mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are mowing their lawn 20 m from your open bedroom window. What is the intensity level in your bedroom?
A. 50 dB
B. 400 dB
C. 104 dB
D. 80 dB
E. 40 dB
Answer:
β₂ = 74 dB, The answer is D which is the closest
Explanation:
The definition and intensity is the power per unit area
I = P / A
P = I A
The emitted power is constant whereby the energy is distributed over the surface of a sphere
A = 4π R²
We can also write it in two points
P = I₁ A₁ = I₂ A₂
I₁ / I₂ = A₂ / A₁
I₁ / I₂ = 4π R₂² / 4π R₁²
I₁ / I₂ = R₂² / R₁²
The definition of decibels is
β = 10 log (I / I₀)
Let's write this equation for the two given points
m = 1m
β₁ = 10 log (I₁ / I₀)
m = 20m
β₂ = 10 log (I₂ / I₀)
Let's eliminate I₀
β₁ - β₂ = 10 log (I₁ / I₀) - 10 log (I₂ / I₀) = 10 (log (I₁ / I₀) –log (I₂ / I₀))
β₁ - β₂ = 10 log (I₁ / I₂)
β₁ - β₂ = 10 log (R₂² / R₁²)
Let's calculate
100 –β₂ = 10 log (20²/1²)
β₂ = 100 - 10 log 400
β₂ = 100 - 26.0
β₂ = 74 dB
The answer is D which is the closest
The intensity level in the bedroom is approximately 74 dB.
Explanation:The intensity level of sound decreases as the distance from the source increases. With each doubling of distance, the sound intensity decreases by 6 dB. In this case, the sound mowers are at a distance of 20 m from your bedroom window, which is 20 times the distance of 1.0 m where the intensity level is 100 dB. Therefore, the intensity level in your bedroom would be 100 dB - (6 dB x log2(20)) = 100 dB - 6 dB x 4.32 = 100 dB - 25.92 dB ≈ 74 dB.
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2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane- strain fracture toughness of 98.9 MPa √ m and a yield strength of 860 MPa . The flaw size resolution limit of the flaw de tection appa ratus is 3.0 mm . If the design stress is one- half the yield strength and the value of Y is 1.0, determine whether a critical flaw for this plate is subject to detection.
Answer:
the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Explanation:
This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.
[tex]ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\ ac=0.0168m\\ac=16.8mm[/tex]
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 95 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t) = A cos(ωt – φ), with the positive y-axis pointing upward. At time t = 0 the mass is observed to be at a distance d = 0.35 m below its equilibrium height with an upward speed of v0 = 6 m/s.
Answer:
The angular frequency of spring-mass system is 7.95 rad/s.
The phase angle is 65.11 degree.
Explanation:
Given data:
The mass is, [tex]m=1.5 \;\rm kg[/tex].
The value of spring constant is, [tex]k=95 \;\rm N/m[/tex].
Distance covered at t=0 is, [tex]y (t=0) = 0.35 \;\rm m[/tex]
Upward speed is, [tex]v_{0}=6 \;\rm m/s[/tex].
The position of mass is,
[tex]y(t)=Acos(\omega t-\phi)[/tex]
Here, A is amplitude of vibration, [tex]\omega[/tex] is the angular frequency, t is time and [tex]\phi[/tex] is the phase angle.
(a)
The expression for the angular frequency of spring-mass system is:
[tex]\omega =\sqrt\frac{k}{m}[/tex]
Substitute the values as,
[tex]\omega =\sqrt\dfrac{95}{1.5}\\[/tex]
[tex]\omega = 7.95 \;\rm rad/s[/tex]
Thus, the value of angular frequency is 7.95 rad/s.
(b)
The position of mass is, [tex]y(t)=Acos(\omega t-\phi)[/tex].
Differentiating the above equation to obtain the speed as,
[tex]y(t)=Acos(\omega t-\phi)\\y'(t)= v_{0}=-A\omega sin(-\phi)[/tex]
At time t=0,
[tex]y(t=0)=A cos(\omega (0)-\phi)\\y(t=0)=A cos(-\phi)\\45= A cos(-\phi) .....................................................(1)[/tex]
Speed of system is,
[tex]6=-A\omega sin(-\phi) ............................................(2)[/tex]
Taking ratio of equation (2) and (1) as,
[tex]\dfrac{-A \omega sin(- \phi)}{Acos(\phi)} = \dfrac{6}{0.35} \\\\7.95 \times tan(\phi) = \dfrac{6}{0.35} \\\phi = tan^{-1}(\dfrac{6}{0.35 \times 7.95}) \\\phi =65.11 \;\rm degree[/tex]
Thus, the phase angle is 65.11 degree.
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The question involves solving for properties of simple harmonic motion, such as force exerted by a spring, equilibrium position, oscillation amplitude, and maximum velocity for a mass-spring system. These are common problems in high school physics involving mechanics and wave motion.
Explanation:The student's question pertains to simple harmonic motion (SHM) exhibited by a mass attached to a vertical spring. Given the mass m, the spring constant k, the displacement d below the equilibrium position, and the upward speed v0, we are tasked with discussing characteristics such as force exerted by the spring, the new equilibrium position, the amplitude of oscillations, and the maximum velocity of the mass.
Step-By-Step Explanation:To find the force exerted by the spring at the lowest point, apply Hooke's law F = -kx, where x is the displacement from the equilibrium position.To determine the new equilibrium position of the spring when a mass is attached, use the force balance where the spring force equals the gravitational force mg.Calculate the amplitude of oscillations by considering the total energy in the system or the maximum displacement of the mass from its equilibrium position.The maximum velocity can be found by using conservation of energy principles or by referring to the maximum speed in SHM as v_max = Aω, where ω is the angular frequency.The solutions to these problems involve a blend of mechanics and oscillatory motion principles and are emblematic of problems found in high school or introductory college-level physics classes.
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Two coils, held in fixed positions, have a mutual inductance of M = 0.0034 H. The current in the first coil is I(t) = I0sin(ωt), where I0 = 5.4 A, ω = 143 rad/s.
M = 0.0034 H I0 = 5.4 A ω = 143 rad/s
(a) Express the magnitude of the induced emf in the second coil, ε2, in terms of M and I.
(b) Express the magnitude of ε2 in terms of M, Io, and ω.
(c) Express the maximum value of |ε2|, εmax, in terms of M, I0, and ω.
(d) Calculate the numerical value of εmax in V.
The magnitude of the induced emf in the second coil can be expressed using the mutual inductance, maximum current in the first coil, and angular frequency.
Explanation:(a) The magnitude of the induced emf in the second coil, ε2, can be expressed as ε2 = M * (dI2/dt), where M is the mutual inductance and dI2/dt is the rate of change of current in the second coil.
(b) The magnitude of ε2 can also be expressed as ε2 = M * I0 * ω * cos(ωt), where I0 is the maximum current in the first coil and ω is the angular frequency.
(c) The maximum value of |ε2|, εmax, can be calculated by taking the maximum value of the function ε2 = M * I0 * ω * cos(ωt) over one period.
(d) To calculate the numerical value of εmax, substitute the values of M, I0, and ω into the equation ε2 = M * I0 * ω * cos(ωt) and evaluate it at the maximum value of cos(ωt), which is 1.
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An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which is used for hunting and self-defense. To perform this feat, approximately 80% of its body is filled with organs made up by electrocytes. These electrocytes act as self-charging capacitors and are lined up so that a current of ions can easily flow through them.
a) How much charge flows through the electrocytes in that amount of time?
b) If each electrocyte can maintain a potential of 100 mV, how many electrocytes must be in series to produce the maximum shock?
c) How much energy is released when the electric eel delivers a shock?
d) With the given information, estimate the equivalent capacitance of all the electrocyte cells in the electric eel.
Answer:
[tex]2\times 10^{-3}\ C[/tex]
6000
1.2 J
[tex]3.33\times 10^{-6}\ F[/tex]
Explanation:
I = Current = 1 A
t = Time = 2 ms
n = Number of electrocyte
V = Voltage = 100 mV
Charge is given by
[tex]Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C[/tex]
The charge flowing through the electrocytes in that amount of time is [tex]2\times 10^{-3}\ C[/tex]
The maximum potential is given by
[tex]V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000[/tex]
The number of electrolytes is 6000
Energy is given by
[tex]E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J[/tex]
The energy released when the electric eel delivers a shock is 1.2 J
Equivalent capacitance is given by
[tex]C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F[/tex]
The equivalent capacitance of all the electrocyte cells in the electric eel is [tex]3.33\times 10^{-6}\ F[/tex]
a) The charge that flows through the electrocytes is [tex]\( 2 \text{ mC} \).[/tex]
b) 6000 electrolytes must be in series to produce the maximum shock. c) the energy released when the electric eel delivers a shock is [tex]\( 0.6 \text{ J} \).[/tex]
d) the equivalent capacitance of all the electrocyte cells in the electric eel is approximately [tex]\( 3.33 \mu\text{F} \)[/tex]
To answer the questions about the electric eel's shock, we'll use the provided information and relevant physics formulas.
Part (a): We know:
- Current [tex]\( I = 1 \text{ A} \)[/tex]
- Time [tex]\( t = 2 \text{ ms} = 2 \times 10^{-3} \text{ s} \)[/tex]
The charge Q that flows can be found using the relation:
[tex]\[ Q = I \times t \][/tex]
Plugging in the values:
[tex]\[ Q = 1 \text{ A} \times 2 \times 10^{-3} \text{ s} \][/tex]
[tex]\[ Q = 2 \times 10^{-3} \text{ C} \][/tex]
[tex]\[ Q = 2 \text{ mC} \][/tex]
Part (b): We know:
- Maximum potential [tex]\( V = 600 \text{ V} \)[/tex]
- Potential of each electrocyte [tex]\( V_{\text{single}} = 100 \text{ mV} = 0.1 \text{ V} \)[/tex]
The number of electrocytes n in series required to produce 600 V can be found by:
[tex]\[ n = \frac{V}{V_{\text{single}}} \][/tex]
Plugging in the values:
[tex]\[ n = \frac{600 \text{ V}}{0.1 \text{ V}} \][/tex]
[tex]\[ n = 6000 \][/tex]
Part (c): We know:
- Voltage [tex]\( V = 600 \text{ V} \)[/tex]
- Charge [tex]\( Q = 2 \text{ mC} = 2 \times 10^{-3} \text{ C} \)[/tex]
The energy E released can be found using the relation:
[tex]\[ E = \frac{1}{2} Q V \][/tex]
Plugging in the values:
[tex]\[ E = \frac{1}{2} \times 2 \times 10^{-3} \text{ C} \times 600 \text{ V} \][/tex]
[tex]\[ E = \frac{1}{2} \times 1.2 \text{ J} \][/tex]
[tex]\[ E = 0.6 \text{ J} \][/tex]
Part (d): Given:
- Total voltage [tex]\( V = 600 \text{ V} \)[/tex]
- Charge [tex]\( Q = 2 \times 10^{-3} \text{ C} \)[/tex]
The capacitance C can be found using the relation:
[tex]\[ C = \frac{Q}{V} \][/tex]
Plugging in the values:
[tex]\[ C = \frac{2 \times 10^{-3} \text{ C}}{600 \text{ V}} \][/tex]
[tex]\[ C = \frac{2 \times 10^{-3}}{600} \text{ F} \][/tex]
[tex]\[ C = \frac{1}{300} \times 10^{-3} \text{ F} \][/tex]
[tex]\[ C \approx 3.33 \times 10^{-6} \text{ F} \][/tex]
[tex]\[ C = 3.33 \mu\text{F} \][/tex]
Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5 A to the right and is located along the line y = 0.420 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m
[tex]\dfrac{F}{L}= \dfrac{KI_1I_2}{d}[/tex]
[tex]I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}[/tex]
[tex]I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}[/tex]
I₂ = 32.67 A
distance where the magnetic field is zero
[tex]\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}[/tex]
[tex]y_1 = 0.248\ m[/tex]
there the distance at which the magnetic field is zero in the two wire is at 0.248 m.
2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yards) and continued to do so all day. Ship B was sailing east at 9 knots and continued to do so all day. The visibility was 5 nautical miles. Did the ships ever sight each other?
Answer:
No
Explanation:
Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):
A = 12 - 12t
B = 9t
Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t
[tex]d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}[/tex]
For the ships to sight each other, distance must be 5 or smaller
[tex] d \leq 5[/tex]
[tex]\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5[/tex]
[tex](12 - 12t)^2 + (9t)^2 \leq 25[/tex]
[tex]144t^2 - 288t + 144 + 81t^2 - 25 \leq 0[/tex]
[tex]225t^2 - 288t + 119 \leq 0[/tex]
[tex](15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0[/tex]
[tex](15t^2 - 9.6)^2 + 26.84 \leq 0[/tex]
Since [tex](15t^2 - 9.6)^2 \geq 0[/tex] then
[tex](15t^2 - 9.6)^2 + 26.84 > 0[/tex]
So our equation has no solution, the answer is no, the 2 ships never sight each other.
While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 17m , and you use your watch to find that each loop around takes 26sWhat is your speed?Express your answer to two significant figures and include the appropriate units.What is the magnitude of your acceleration?Express your answer to two significant figures and include the appropriate units.What is the ratio of your weight at the top of the ride to your weight while standing on the ground?Express your answer using two significant figures.
Answer:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Explanation:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed theCase Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10-s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. (a) Determine the emf induced in the loop. (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
The magnitude of the induced emf in the loop can be calculated using Faraday's law of electromagnetic induction. The direction of the induced current that would oppose the change in flux can be determined using Lenz's law and the right-hand rule, which in this case would be clockwise.
Explanation:To answer your questions: (a) emf induced in the loop and (b) the direction of the current induced in the loop, we must understand Faraday's law of electromagnetic induction and Lenz's law.
Using Faraday's law, the magnitude of the induced emf is given as ε = |døm/dt|, where øm is the magnetic flux.
The magnetic flux through a loop is given by the product of the magnetic field B, the area A (Which can be calculated using the formula for the area of a circle, A = πr², where r is the radius of the loop), and the cosine of the angle Ө between the magnetic field lines and the perpendicular to the plane of the loop.
However, since the loop is perpendicular to the magnetic field, cosӨ = 1. Also, because the magnetic field B is spatially uniform and changes uniformly with time, we can find the average magnetic field over the 0.1 s interval as B = (200mT + 300mT) / 2 = 250mT = 0.25 T. Therefore, the change in magnetic flux over the time interval is Δøm = BA = (0.25 T)(π(0.05 m)²). Substituting these into Faraday's law gives us the induced emf.
According to Lenz's law, the direction of the induced emf (and hence the current) is such that it produces a magnetic field that opposes the change in the magnetic flux. Since the original field is directed out of the page and is increasing, the induced current must flow in a way that it produces a field into the page. Hence, according to the right-hand rule, the current would flow clockwise when viewed from above.
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By applying Faraday's Law of electromagnetic induction, the induced electromotive force (emf) in the loop is calculated to be approximately 7.85 mV. Following Lenz's Law, the induced current flows in a clockwise direction to oppose the increase in the magnetic field.
To address the student's question, we can use Faraday's Law of electromagnetic induction which states that the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.
(a) To find the induced emf, use the formula:
E = -dΦ/dt
where Φ is the magnetic flux given by:
Φ = BA cos(θ)
B is the magnetic field's magnitude, A is the area of the loop, and θ is the angle between the field and normal to the loop. In this case, θ is 0 degrees because the field is perpendicular to the plane of the loop.
Φ = B(πr²)
Given that r = 0.05 m, Binitial = 200 mT, Bfinal = 300 mT, and Δt = 0.10 s, we have:
ΔΦ = π(0.05)²(300 mT - 200 mT)
ΔΦ = π(0.05)²(100 mT) = π(0.0025 m²)(100 x 10⁻³ T) = π(0.00025 T·m²)
E = -ΔΦ/Δt = -π(0.00025)/0.10 = -0.00785 V
The magnitude of the induced emf is approximately |0.00785| V or 7.85 mV.
(b) According to Lenz's Law, the direction of the induced current will be such that it opposes the change in flux. Since the magnetic field is increasing and directed out of the page, the induced current will create a magnetic field into the page to oppose the increase. Using the right-hand rule, we can determine that the current flows in a clockwise direction when viewed from above.
A car travels 40 miles in 30 minutes.
a) What is the average velocity in kilometers/hour?
b) If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)?
c) When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping?
d) What is the average acceleration of the car (in m/s2) during braking?
Answer:
a) v = 160.9 Km / h, b) K = 9,988 10⁵ J , c) d = 335.2 m , d) a = - 2.9796 m / s²
Explanation:
a) The average speed is defined as the distance traveled in the time interval
v = d / t
Let's reduce the magnitudes to the SI system
t = 30 min (60 s / 1min) = 1800 s
d = 50 mile (1609 m / 1mile) = 80450 m
v = 80450/1800
v = 44.6944 m / s
v = 44.6944 m / s (1 km / 1000m) (3600 s / 1h)
v = 160.9 Km / h
b) W = 2 ton (1000 kg / 1 ton) = 1000 kg
K = ½ m v²
K = ½ 1000 44.694²
K = 9,988 10⁵ J
c) take t = 15 s to stop
v = v₀ - at
v = 0
a = v₀ / t
a = 44,694 / 15
a = 2.9796 m / s²
v² = v₀² - 2 a d
v = 0
d = v₀² / 2 a
d = 44.694² / (2 2.9796)
d = 335.2 m
d) the average acceleration is the change of speed in the time interval
a = (v - v₀) / (t -t₀)
a = (0 - 44.694) / 15
a = - 2.9796 m / s²
The St. Louis Arch has a height of 192 m. Suppose that a stunt woman of mass 84 kg jumps off the top of the arch with an elastic band attached to her feet. She reaches the ground at zero speed. The acceleration of gravity is 9.81 m/s 2 . Find her kinetic energy after 2.6 s of the flight. Assume the elastic band has no length and obeys Hooke’s Law. Answer in units of kJ
Final answer:
The kinetic energy of the stunt woman after 2.6 s of flight is 27578.835 kJ.
Explanation:
To find the kinetic energy of the stunt woman after 2.6 s of flight, we can use the formula for kinetic energy:
KE = 0.5 mv^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
First, let's find the velocity of the stunt woman after 2.6 s of flight. We can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time (2.6 s).
Substituting the values, we get:
v = 0 + (-9.81) * 2.6
v = -25.446 m/s
Since the stunt woman reaches the ground at zero speed, her final velocity is 0 m/s. Therefore, her velocity is -25.446 m/s after 2.6 s of flight.
Now, let's plug the values of mass (84 kg) and velocity (-25.446 m/s) into the formula for kinetic energy:
KE = 0.5 * 84 * (-25.446)^2
KE = 0.5 * 84 * 650.701716
KE = 27578.835 kJ
Therefore, the stunt woman's kinetic energy after 2.6 s of flight is 27578.835 kJ.
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 210 N applied to its edge causes the wheel to have an angular acceleration of 0.932 rad/s2. What is the moment of inertia of the wheel? (Pick the answer closest to the true value.)A. 27.3 kg m2B. 42.4 kg m2C. 54.9 kg m2D. 74.4 kg m2E. 98.5 kg m2
Answer:
Moment of inertia will be [tex]I=74.356kgm^2[/tex]
So option (d) will be the correct answer
Explanation:
We have given radius of solid cylinder r = 0.330 m
Constant tangential force F = 210 N
Angular acceleration [tex]\alpha =0.932rad/sec^2[/tex]
We know that torque [tex]\tau =Fr=210\times 0.330=69.3Nm[/tex]
We also know that torque is given by [tex]\tau =I\alpha[/tex]
So [tex]69.3=I\times 0.932[/tex]
[tex]I=74.356kgm^2[/tex]
So option (d) will be the correct answer
What is the difference between the states of phase equilibrium and metastability?
Answer:
The main difference is that the metastable state is not a state of equilibrium, but a state of non-equilibrium that is maintained for a long time. A metastable state is when the system approaches equilibrium in a very slow manner.
And on the other hand the phase equilibrium as the name says is a state of equilibrium in which there are more than two phases coexisting.
The state of phase is description about the substance existing in equilibrium and metastability is non-equilibrium state of substance.
The given problem is based on the major difference between the states of phase of equilibrium and the metastability. These two are the concepts of chemical equilibrium, when there is subsequent change in the phase of one substance, with respect to the other substance.
The metastable state is not a state of equilibrium, but a state of non-equilibrium that is maintained for a long time. A metastable state is when the system approaches equilibrium in a very slow manner.And on the other hand the phase equilibrium as the name says is a state of equilibrium in which there are more than two phases coexisting.
Thus, we can conclude that the state of phase is description about the substance existing in equilibrium and metastability is non-equilibrium state of substance.
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