When a doubled rope, i.e. two pieces of rope used together, is used, it stretches half as much as a single rope would under the same load. Correspondingly, a rope's stretch is directly proportional to its length. So, it will stretch twice as much when its length is doubled.
Explanation:The stretchiness of a rope under a load is a property called elasticity. When there are two ropes, they share the load between them. Therefore, each carries only half the load. Hence, a doubled rope would stretch half as much as a single rope when supporting the same weight. In this case, the single rope stretches by 20 cm under a load. Thus, the doubled rope would stretch by 10 cm.
Similarly, the amount a rope stretches is proportional to its length. If a 10 m rope stretches 60 cm under a certain load, a 20 m rope -- twice as long -- would stretch twice as much under the same load, or 120 cm.
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In the context of elasticity in physics, a doubled rope that halves each individual rope's load would stretch half as much. Therefore, if a 10m rope stretches by 20cm, a doubled rope would stretch by 10cm. If a 10m rope stretches by 60cm, a 20m rope under the same load would stretch twice as much, or 120cm.
Explanation:These questions refer to the concepts of strain and stress in physics, specifically within the context of elasticity. Strain refers to the deformation of a body due to the applied stress, in this case, the length of rope extending under the weight of the climber.
a. If a 10m climbing rope stretches by 20cm under a certain load, its strain is 0.02 (20cm/10m). When the load is shared by two ropes, each rope bears half of the load. Hence, each rope is likely to stretch half as much. So, the doubled rope would stretch by 10cm.
b. Stretching is directly proportional to the length of the rope. Therefore, if a 10m rope stretches by 60cm under a load, a 20m rope under the same load would stretch twice as much, which is 120cm or 1.2m.
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Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass=0.36 kg) and the other of metal (mass=0.82 kg) They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum
Answer:
a) T = 0.579 s , b) T = 0.579 s
Explanation:
The great advantage of wave mechanics is that the general equations for different systems are the same, what changes are the physical parameters involved, the equation of motion is
x = A cos (wt +φ)
Where w is the angular velocity that is this case for being a solid body is
w = √ (mg d / I)
Where I is the moment of inertia and d the distance to the pivot point
The moment of inertia for a ruler hold one end is
I = 1/12 M L²
The lost is related to the frequency and is with the angular velocity
T = 1 / f
w = 2π f
w = 2π / T
T = 2π / w
T = 2π √ I / mgd
For our case
d = L
T = 2π √(1/12 M L²) / M g L)
T = 2π √(L/(12 g))
a) wood suppose it is one meter long (L = 1m)
T = 2π √ (1 / (12 9.8))
T = 0.579 s
b) metal length (L = 1m)
T = 2pi RA (1 / (12 9.8))
T = 0.579 s
The period does not depend on mass but on length
Beaker A contains 100 mL of water at a temperature of 25 °C. Beaker B contains 100 mL of water at a temperature of 60 °C. Which of the following statement is true?
a. Beaker A has lower kinetic energy than beaker B
b. Beaker A has higher thermal energy than beaker B
c. Beaker A has higher potential energy than beaker B
d. Beaker A has lower potential energy than beaker B
e. Beaker A has higher kinetic energy than beaker B
Answer:
Only option A is correct. Beaker A has lower kinetic energy than beaker B.
Explanation:
Step 1: Data given
Beaker 1 has a volume of 100 mL at 25 °C
Beaker B has a volume of 100 mL at 60 °C
Thermal energy = m*c*T
Thermal energy beaker A = 100 grams*4.184 * 25°C
Thermal energy beaker B = 100 grams *4.184*60°C
⇒ Since both beakers contain the same amount of water, the thermal energy depends on the temperature.
Since beaker B has a higher temperature, it has a higher thermal energy than beaker A
When we heat a substance, its temperature rises and causes an increase in the kinetic energy of its constituent molecules. Temperature is, in fact, a measure of the kinetic energy of molecules.
This means beaker B has a higher kinetic energy than beaker A
Potential energy doesn't depend on temperature. this means the potential energy of beaker A and beaker B is the same.
a. Beaker A has lower kinetic energy than beaker B. This is correct.
b. Beaker A has higher thermal energy than beaker B. This is false.
c. Beaker A has higher potential energy than beaker B. This is false.
d. Beaker A has lower potential energy than beaker B. This is false
e. Beaker A has higher kinetic energy than beaker B. This is false.
Beaker A has lower kinetic energy than Beaker B because the temperature of a substance is directly proportional to its average kinetic energy. Thus, the water in Beaker B, which is at a higher temperature, has higher kinetic energy than the cooler water in Beaker A.
Explanation:The correct answer to your question is: a. Beaker A has lower kinetic energy than Beaker B. This is because the temperature of a substance is directly proportional to its average kinetic energy. Kinetic energy refers to the energy of movement, so in this case, it's the energy of the water molecules moving within each beaker. In Beaker B, the water is at a higher temperature, meaning the water molecules are moving more quickly, and therefore have higher kinetic energy. On the other hand, in Beaker A, the water is at a lower temperature, so the water molecules are moving more slowly, which means they have lower kinetic energy.
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For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?
Final answer:
The new pressure drop p2 will be equal to the original pressure drop p1 at the original mass flow rate.
Explanation:
The relation between the new pressure drop p2 and the original pressure drop p1 at the original mass flow rate can be determined using Poiseuille's equation. Poiseuille's equation states that the pressure drop p2 - p1 is equal to the product of the resistance R and the flow rate Q.
When the length of the pipe is doubled, the resistance R remains constant because the friction factor is constant at high Reynolds numbers. Therefore, the new pressure drop p2 will be equal to the original pressure drop p1.
In the vicinity of Earth's orbit around the Sun, the energy intensity of sunlight is about 1200 W/m2. What is the approximate magnitude of the electric field in the sunlight? (What you calculate is actually the "root-mean-square" or "rms" magnitude of the electric field, because in sunlight the magnitude of the electric field at a fixed location varies sinusoidally, and the intensity is proportional to E2.)
Answer:
672.29 W/m²
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
I = Intensity of light = 1200 W/m²
[tex]E_m[/tex] = Maximum value electric field
Intensity of light is given by
[tex]I=\frac{1}{2}\epsilon_0cE_m^2\\\Rightarrow E_m=\sqrt{\frac{2I}{\epsilon_0c}}\\\Rightarrow E_m=\sqrt{\frac{2\times 1200}{8.85\times 10^{-12}\times 3\times 10^8}}\\\Rightarrow E_m=950.765\ N/C[/tex]
RMS value
[tex]E_r=\frac{E_m}{\sqrt2}\\\Rightarrow E_r=\frac{950.765}{\sqrt2}\\\Rightarrow E_r=672.29\ W/m^2[/tex]
The approximate magnitude of the electric field in the sunlight is 672.29 W/m²
A loop of wire enclosing an area of 0.19 m2 has a magnetic field passing through its plane at an angle to the plane. The component of the field perpendicular to the plane is 0.5 T and the component parallel to the plane is 0.8 T. What is the magnetic flux through this coil? Round the final answer to three decimal places.
To solve this exercise it is necessary to apply the concepts related to Magnetic Flow which is defined through the Gaus Law of Magnetism as the measure of the total magnetic field that passes through a given area.
Vectorially it can be defined as
[tex]\Phi = \vec{A}\vec{B}[/tex]
Where,
A = Area
B = Magnetic Field
A scalar mode can also be expressed as
[tex]\Phi = A*B Cos\theta[/tex]
Where,
[tex]\theta[/tex] is the angle between B and A, at this case the direction are perpendicular then
Our values are given as,
[tex]A = 0.19m^2[/tex]
[tex]B_1 = 0.5T[/tex]
[tex]B_2 = 0.8T[/tex]
The magnetic field component that interests us is the perpendicular therefore
[tex]\Phi = 0.19* 0.5cos(90)[/tex]
[tex]\Phi = 0.095Tm^2[/tex]
Therefore the magnetic flux through this coil is 0.095[tex]Tm^2[/tex]
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a mass of 1.9 x 10^−14 kg. How long does it take for all of the particles to settle to the bottom of the container?
To determine how long it takes for particles to settle at the bottom of a container, principles of sedimentation in physics are applied. The calculation considers gravitational forces, buoyancy, and drag forces. However, exact time estimation requires additional specific information on fluid characteristics.
Explanation:The question revolves around the concept of sedimentation, which is a physical process where particulate matter settles down at the bottom of a container due to gravity. To calculate how long it takes for spherical particles with a diameter of 2.5 μm (micrometers) and a mass of 1.9 x 10⁻¹⁴ kg to settle at the bottom of a 15-cm-tall container, we need to understand the principles of particle sedimentation in fluids, a topic studied in physics. This involves the forces acting on the particles, including gravitational forces, buoyancy, and drag forces, along with the properties of the fluid and the particles themselves. However, without a specific formula or further details on the characteristics of the fluid (e.g., viscosity) and assuming we ignore any air resistance or buoyancy effects, it's difficult to provide an exact time for sedimentation.
Therefore, it takes approximately [tex]\(4657.76\)[/tex] seconds for all the particles to settle to the bottom of the container.
To find the time it takes for all the particles to settle to the bottom of the container, we can use Stokes' law, which relates the settling velocity of particles to their size and density.
Stokes' law states: [tex]\( v = \frac{{2gr^2(\rho_p - \rho)}}{{9\eta}} \),[/tex] where:
- v is the settling velocity of the particle,
- g is the acceleration due to gravity [tex](\( 9.8 \, \text{m/s}^2 \)),[/tex]
- r is the radius of the particle [tex](\( 1.25 \times 10^{-6} \, \text{m} \)),[/tex]
- [tex]\( \rho_p \)[/tex]is the density of the particle[tex](\( \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3} = \frac{1.9 \times 10^{-14}}{\frac{4}{3}\pi(1.25 \times 10^{-6})^3} \)),[/tex]
- [tex]\( \rho \)[/tex] is the density of the fluid [tex](\( \rho = 1.2 \, \text{kg/m}^3 \)),[/tex] and
- [tex]\( \eta \)[/tex] is the viscosity of the fluid[tex](\( \eta = 1.81 \times 10^{-5} \, \text{Pa s} \)).[/tex]
After substituting the values into Stokes' law and solving for the settling velocity, we found it to be approximately[tex]\(3.22 \times 10^{-5} \, \text{m/s}\).[/tex]
Then, using the formula [tex]\(t = \frac{h}{v}\), where \(h\)[/tex]is the height of the container (0.15 m), we calculated the time it takes for the particles to settle to the bottom:
[tex]\[ t \approx 4657.76 \, \text{s} \][/tex]
An air-filled capacitor stores a potential energy of 6 mJ due to its charge. It is accidentally filled with water in such a way as not to discharge its plates. How much energy does it continue to store after it is filled? (The dielectric constant for water is 78 and for air it is 1.0006.
Answer:
The potential energy in the water will be 0.0769 mJ
Explanation:
We know that energy stored in the capacitor due to charge is given by [tex]U=\frac{Q^2}{2C}[/tex]
From the relation we can see that potential energy is inversely proportional to the capacitance
And we also know that capacitance is directly proportional to the dielectric constant
So the new potential energy will be [tex]=\frac{6}{78}=0.0769mJ[/tex]
So the potential energy in the water will be 0.0769 mJ
Final answer:
When a capacitor filled with air and charged is filled with water without discharging its plates, the energy stored increases due to the higher dielectric constant of water. The formula used is U' = K_water * U, leading to the capacitor storing 468 mJ of energy after being filled with water.
Explanation:
The energy stored in a capacitor is given by the formula U = ½ CV^2, where U is the stored energy, C is the capacitance and V is the voltage across the capacitor. When a dielectric is introduced between the plates of a capacitor that is charged but disconnected from any voltage supply, the capacitance of the capacitor increases by a factor equal to the dielectric constant of the material inserted, K. Since the charge Q remains constant, and since C increases while U is directly proportional to C, the energy stored in the capacitor will increase as well.
For this particular question, the initial energy is 6 mJ and the dielectric constant for water is 78. After filling the capacitor with water, the new energy stored U' can be found using the dielectric constant. However, since dielectric constant for air is approximately 1, and capacitance is directly proportional to dielectric constant, we can simply multiply the original energy by the dielectric constant of water to find the energy stored after it is filled.
U' = K_water * U
= 78 * 6 mJ
= 468 mJ
Therefore, the energy the capacitor continues to store after it is filled with water is 468 mJ.
A brass statue with a mass of 0.40 kg and a density of 8.00×103kg/m3 is suspended from a string. When the statue is completely submerged in an unknown liquid, the string tension is 3.2 N. What is the density of the liquid?
Answer:
ρ = 1469 kg/m³
Explanation:
given,
mass of statue = 0.4 Kg
density of statue = 8 x 10³ kg/m³
tension in the string = 3.2 N
density of the fluid = ?
Volume of the statue
[tex]V = \dfrac{0.4}{8\times 10^3}[/tex]
V = 5 x 10⁻⁵ m³
W = ρ g V
W = ρ x 9.8 x 5 x 10⁻⁵
now, tension on the string will be equal to
T = mg - W
3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵
ρ x 9.8 x 5 x 10⁻⁵ = 0.72
ρ = 1469 kg/m³
In a CD player, a CD starts from rest and accelerates at a rate of . Suppose the CD has radius 1212 cm. At time 0.15 sec after it started spinning, what is the magnitude of the linear acceleration∣ for a point on its outer rim?
Answer:
[tex]a =29.54\ m/s^2[/tex]
Explanation:
given,
radius of CD player = 12 cm
assume rate of acceleration = 100 rad/s²
times = 0.15 s
now,
tangential acceleration
[tex]a_t = \alpha r[/tex]
[tex]a_t = 100 \times 0.12[/tex]
[tex]a_t = 12 m/s^2[/tex]
now using equation
v = v₀ + a_t x t
v =0+ 12 x 0.15
v = 1.8 m/s
now, radial acceleration
[tex]a_r = \dfrac{v^2}{r}[/tex]
[tex]a_r = \dfrac{1.8^2}{0.12}[/tex]
[tex]a_r =27\ m.s^2[/tex]
now
acceleration
[tex]a = \sqrt{a_r^2+a_t^2}[/tex]
[tex]a = \sqrt{27^2+12^2}[/tex]
[tex]a =29.54\ m/s^2[/tex]
Two ponies of equal mass are initially at diametrically opposite points on the rim of a large horizontal turntable that is turning freely on a vertical, frictionless axle through its center. The ponies simultaneously start walking toward each other across the turntable. As they walk, what happens to the angular speed of the turntable? (A) It increases(B) It decreases(C) It stays constantConsider the ponies-turntable system in this process is the angular momentum of the system conserved?(A) Yes(B) No
Answer:A
Explanation:
Given
Two ponies are at Extreme end of turntable with mass m
suppose turntable is moving with angular velocity [tex]\omega [/tex]
Moment of inertia of Turntable and two ponies
[tex]I=I_o+2mr_0^2[/tex]
let say at any time t they are at a distance of r from center such [tex]r<r_0[/tex]
Moment of inertia at that instant
[tex]I'=I_0+2mr'^2[/tex]
[tex]I'<I_0[/tex]
conserving angular momentum as net Torque is zero
[tex]I\omega =I'\omega '[/tex]
[tex]\omega '=\frac{I}{I'}\times \omega [/tex]
[tex]\omega '>\omega [/tex]
Thus we can say that angular velocity increases as they move towards each other.
The angular speed of the turntable increases as the ponies walk towards each other due to conservation of angular momentum. The angular momentum remains constant as there are no external forces acting upon the system.
Explanation:As the two ponies of equal mass walk toward each other across the turntable, the angular speed of the system increases. This is because the moment of inertia of the system decreases as the ponies move closer to the center, and according to the principle of conservation of angular momentum, a decrease in the moment of inertia must be accompanied by an increase in angular speed to keep the system's angular momentum constant.
In this scenario, yes, the angular momentum of the system is conserved. There are no external forces acting on the system, thus the sum of the angular momenta of the ponies and the turntable stays the same.
This principle can be exemplified by the behavior of figure skaters as they increase their rotation speed by bringing their arms in closer to the body, decreasing their moment of inertia and consequently increasing their angular velocity, to conserve their angular momentum.
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A 65-kg ice hockey goalie, originally at rest, catches a 0.145-kg hockey puck slapped at him at a velocity of 35 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came.What would the final velocities of the goalie and the puck be in this case? Assume that the collision is completely elastic.v goalie = _____ m / sv puck = ______ m / s
Answer:
The velocity of the goalie = 0.156 m/s
The velocity of the puck is 34.85 m/s, but in the reverse direction (-34.85 m/s)
Explanation:
The total momentum before equals the totam momentum after the collision for objects with mass m1 and m2
m1*vi1 + m2 vi2 = m1 vf1 + m2 vf2
⇒ with vi = the initial velocity
⇒ with vf = the final velocity
initial momentum = final momentum
An elastic collision is one in which the total kinetic energy of the two colliding objects is the same before and after the collision. For an elastic collision, kinetic energy is conserved. That is:
0.5 m1 vi1² + 0.5 m2 vi2² = 0.5 m1 vf1² + 0.5 m2 vf2²
Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:
vf1 = [(m1 - m2) vi1 + 2m2 vi2]/[m1 + m2]
vf2 = [2m1vi1 − (m1 - m2) vi2]/[m1 + m2]
⇒ with m1 = the mass of the goalie = 65 kg
⇒ with m2 = the mass of the puck = 0.145 kg
⇒ with vi1 = the initial velocity of the goalie 0 m/s
⇒ with vi2 = the initial velocity of the puck = 35 m/s
⇒ with vf1 = The final velocity of the goalie
=> vf1 = [(65 - 0.145)0 + 2(0.145)35]/[65.145]
= 0.1558 ≈ 0.156 m/s. Since it's positve this is in the direction of the puck
=> vf2 = [0 − (65 - 0.145)35]/[65.145] = -34.85 m/s
The velocity of the puck is 34.85 m/s, in the reverse direction. (Has a negative sign)
An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that apply. a. The stored electric field energy can be greater than the stored magnetic field energy. b. As the capacitor is discharging, the current is increasing. c. As the capacitor is charging, the current is increasing. d. The stored electric field energy can be less than the stored magnetic field energy. e. The stored electric field energy can be equal to the stored magnetic field energy.
Answer:
a. The stored electric field energy can be greater than the stored magnetic field energy.
b. As the capacitor is discharging, the current is increasing.
d. The stored electric field energy can be less than the stored magnetic field energy.
e. The stored electric field energy can be equal to the stored magnetic field energy.
Explanation:
We know that total energy of capacitor and inductor system given as
[tex]TE=\dfrac{LI^2}{2}+\dfrac{CV^2}{2}[/tex]
L=Inductance ,I =current
C=Capacitance ,V=Voltage
The total energy of the system will be remain conserve.If energy in the inductor increases then the energy in the capacitor will decrease and vice -versa.
The values of stored energy in the capacitor and inductor can be equal ,greater or less than to each other.But the summation will be constant always.
Therefore the following option are correct
a ,b , d , e
In a series LC circuit, energy oscillates between the capacitor and inductor. Statements a, b, d, and e are correct as there are moments when the electric field energy can be greater than, less than, or equal to the magnetic field energy. Statement c is incorrect because the current decreases as the capacitor charges.
When an inductor is connected in series with a fully charged capacitor, the behavior of the system can be analyzed by considering the energy stored in both components.
Statement a: This is true. The stored electric field energy can be greater than the stored magnetic field energy at any given instant depending on the phase of oscillation.Statement b: This is also true. As the capacitor discharges, the current in the circuit initially increases because the inductor resists changes in current.Statement c: This statement is false. As the capacitor is charging, the current actually decreases because energy is being stored back in the capacitor.Statement d: This is true as well. The stored electric field energy can be less than the stored magnetic field energy at any point in time due to the oscillatory nature of the energy transfer.Statement e: This statement is correct. There are instants where the stored electric field energy can be equal to the stored magnetic field energy during the oscillation cycle.Overall, the series LC circuit exhibits an oscillatory energy transfer between the electric field of the capacitor and the magnetic field of the inductor.
PROBLEM A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? STRATEGY The angular acceleration is constant, so this problem just requires substituting given values into the proper equations. SOLUTION
Answer:
11 radians or 1.7507 revolutions
9 rad/s
7.27565 revolutions
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
t = Time taken
Equation of rotational motion
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\times 2+\frac{1}{2}\times 3.5\times 2^2\\\Rightarrow \theta=11\ rad=\frac{11}{2\pi}=1.7507\ rev[/tex]
Angle the wheel rotates in the given time is 11 radians or 1.7507 revolutions
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=2+3.5\times 2\\\Rightarrow \omega_f=9\ rad/s[/tex]
The angular speed of the wheel at the given time is 9 rad/s
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{(9\times 2)^2-2^2}{2\times 3.5}\\\Rightarrow \theta=45.71428\ rad=\frac{45.71428}{2\pi}=7.27565\ rev[/tex]
The number of revolutions if the final angular speed doubles is 7.27565 revolutions
This question involves the concepts of the equations of motion for angular motion.
(a) The wheel rotates through the angle "11 rad" (OR) "1.75 rev".
(b) The angular speed at t = 2 s is "9 rad/s".
(c) The angular displacement for double speed is "45.71 rad" (OR) "7.28 rev".
(a)
The angular displacement can be found using the second equation of motion for angular motion.
[tex]\theta = \omega_it+\frac{1}{2}\alpha t^2[/tex]
where,
θ = angular displacement = ?
ωi = initial angular speed = 2 rad/s
t = time interval = 2 s
α = angular acceleration = 3.5 rad/s²
Therefore,
[tex]\theta = (2\ rad/s)(2\ s)+\frac{1}{2}(3.5\ rad/s^2)(2\ s)^2\\\theta = 4\ rad\ +\ 7\ rad[/tex]
θ = 11 rad
[tex]\theta = (11\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ = 1.75 rev
(b)
The angular speed can be found using the first equation of motion for angular motion.
[tex]\omega_f = \omega_i+\alpha t\\\omega_f = 2\ rad/s + (3.5\ rad/s^2)(2\ s)\\[/tex]
ωf = 9 rad/s
(c)
The angular displacement can be found using the third equation of motion for angular motion after we double the value of ωf.
[tex]2\alpha \theta = \omega_f^2-\omega_i^2\\2(3.5\ rad/s^2)\theta = (18\ rad/s)^2-(2\ rad/s)^2\\\\\theta = \frac{320\ rad^2/s^2}{7\ rad/s^2}\\\\[/tex]
θ = 45.71 rad
[tex]\theta = (45.71\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ = 7.28 rev
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The attached picture shows the angular equations of motion.
A toroid with a square cross section 3.0 cm ✕ 3.0 cm has an inner radius of 25.1 cm. It is wound with 600 turns of wire, and it carries a current of 3.0 A.
What is the strength of the magnetic field (in T) at the center of the square cross section?
Answer:
B = 1.353 x 10⁻³ T
Explanation:
The Magnetic field within a toroid is given by
B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.
To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,
r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.
b = 25.1 + 3 = 28.1 cm
a = 25.1 cm
r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m
B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266
B = 1.353 x 10⁻³ T
The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T
The strength of the magnetic field at the center of the square cross section of a given toroid, calculated using Ampere's Law, is approximately 0.00269 T.
Explanation:The magnetic field B produced by a current in a toroidal solenoid is described by Ampere's Law. The formula to calculate magnetic field (B) in a toroid is B = μ₀IN / (2πr), where:
μ₀ = permeability of free space = 4π x 10⁻⁷ T m/A,I = current passing through the wire = 3.0 A,N = total turns of the wire = 600 turns, andr = radius from the center of the toroid to the wire = 25.1 cm its inner radius + 1.5 cm half of the cross-section = 26.6 cm = 0.266 m.Substituting these values in the formula, we get:
B = (4π x 10⁻⁷ T m/A) * (600) * (3 A) / (2π * 0.266 m) = 0.00269 T.
So the strength of the magnetic field at the center of the square cross section is approximately 0.00269 T.
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You can, in an emergency, start a manual transmission car by putting it in neutral, letting the car roll down a hill to pick up speed, then putting it in gear and quickly letting out the clutch. If the car needs to be moving at 3.5 m/s for this to work, how high a hill do you need? (You can ignore friction and drag)
Answer:
The hill should be not less than 0.625 m high
Explanation:
This problem can be solved by using the principle of conservation of mechanical energy. In the absence of friction, the total mechanical energy is conserved. That means that
[tex]E_m=U+K[/tex] is constant, being U the potential energy and K the kinetic energy
[tex]U=mgh[/tex]
[tex]K=\frac{mv^2}{2}[/tex]
When the car is in the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.
The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.
[tex]mgh=\frac{mv^2}{2}[/tex]
We can solve for h:
[tex]h=\frac{v^2}{2g}=\frac{3.5^2}{2(9.8)}=0.625m[/tex]
The hill should be not less than 0.625 m high
The hill should have a height of 0.625 m and above.
What is Height?This is defined as the measurement of the vertical position of a body.
Total mechanical energy = Potential energy + kinetic energy.
Potential energy = mgh
Kinetic energy = 1/2mv²
We can infer that:
mgh = 1/2mv²
h = v² / 2g
= (3.5)² / 2(9.8)
= 0.625m.
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A transformer has 500 primary turns and 9.7 secondary turns.
(a) If Vp is 120 V (rms), what is Vs with an open circuit?
(b) If the secondary now has a resistive load of 13 Ω, what is the current in the primary and secondary?
Answer: a) 2.33V b) Ip = 0.0035A, Is = 0.179A
Explanation:
The relationship formula between the Voltage, Number of turns and current in a transformer is given as
Vp/Vs = Np/Ns = Is/Ip
Vp is voltage in the primary coil
Vs is voltage in the secondary coil
Ns is number of turns in the secondary coil
Np is number of turns in the primary coil
Is is current in the secondary coil
Ip is current in the primary coil
a) substituting the values in the formula
Vp/Vs = Np/Ns
120/Vs = 500/9.7
Vs = 120×9.7/500
Vs = 2.33V
b) since secondary now has a resistive load of 13 Ω, Using ohm's law to get current in the secondary
V=IR
Is = Vs/R
Is =2.33/13
Is = 0.179A
To get Ip, we will use the relationship Np/Ns = Is/Ip
500/9.7= 0.179/Ip
Ip = 9.7×0.179/500
Ip = 0.0035A
how does the gravitational force between two objects change if the distance between the objects is cut in half?
A. the gravitational force doubles
B. the gravitational force decreases by half
C. the gravitational force decreases by a factor of 4
D. the gravitational force increases by a factor of 4
D. the gravitational force increases by a factor of 4
Explanation:
The magnitude of the gravitational force between two objects is given by:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
Here let's call F the initial gravitational force between the two objects. Later, the distance between the objects is halved, so the new distance is:
[tex]r'=\frac{r}{2}[/tex]
Substituting into the equation, we find the new force:
[tex]F'=G\frac{m_1 m_2}{(r/2)^2}=4(G\frac{m_1 m_2}{r^2})=4F[/tex]
Therefore, the gravitational force increases by a factor of 4.
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Final answer:
The gravitational force between two objects doubles if the distance between them is cut in half.
Explanation:
The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that if the distance between the objects is cut in half, the force will increase by a factor of 4.
For example, let's say the original distance is 10 units. If we cut it in half, the new distance would be 5 units. The force would increase by a factor of (10/5)² = 4, meaning the gravitational force would quadruple.
So the correct answer is, A. the gravitational force doubles.
A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° north of east. If he walked straight home,
(a)-how far would he have to walk?
(b)-In what direction would he have to walk?
Answer:
a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º
This is 8º north west
Explanation:
This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately
To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.
First vector A = 2 to 20º north west
Measured from the positive x axis is θ = 180 -20 = 160º
We use trigonometry to find the components
Cos 20 = Aₓ / A
sin 20 = [tex]A_{y}[/tex] / A
Aₓ = A cos 160 = 2 cos 160
[tex]A_{y}[/tex] = A sin160 = 2 sin160
Aₓ = -1,879 mi
[tex]A_{y}[/tex] = 0.684 mi
Second vector B = 4 mi 10º west of the south
Angle θ = 270 - 10 = 260º
cos 2600 = Bₓ / B
sin 260 = [tex]B_{y}[/tex] / B
Bₓ = B cos 260
[tex]B_{y}[/tex] = B sin 260
Bₓ = 4 cos 260
[tex]B_{y}[/tex] = 4 sin 260
Bₓ = -0.6946mi
[tex]B_{y}[/tex] = - 3,939 mi
Third vector C = 3 mi to 15 north east
cos 15 = Cₓ / C
sin15 = [tex]C_{y}[/tex] / C
Cₓ = C cos 15
[tex]C_{y}[/tex] = C sin15
Cₓ = 3 cos 15
[tex]C_{y}[/tex] = 3 sin 15
Cₓ = 2,898 mi
[tex]C_{y}[/tex] = 0.7765 mi
Now we can find the final position of the person
X = Aₓ + Bₓ + Cₓ
X = -1.879 -0.6949 + 2.898
X = 0.3241 mi
Y = [tex]A_{y}[/tex] + [tex]B_{y}[/tex] + [tex]C_{y}[/tex]
Y = 0.684 - 3.939 +0.7765
Y = -2.4785 mi
a) We use Pythagoras' theorem
R = √ (x2 + y2)
R = √ (0.3241 2 + (-2.4785) 2)
R = 2.4996 mi
R = 2.5 mi
b) let's use trigonometry
Tan θ = y / x
Tanθ = -2.4785 / 0.3241
θ = tan⁻¹ (-7,647)
θ = -82
Measured from the positive side of the x axis is Te = 360 - 82 = 278º
(90-82) south east
To return to your case you must walk in the opposite direction or Te = 98º
This is 8º north west
Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 65 m below. How much time elapses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately 1530 m/s.)
To solve this problem it is necessary to apply the concepts related to the kinematic equations of motion.
By definition we know that speed can be expressed as
[tex]v= \frac{d}{t}[/tex]
Where,
t = time
d = distance
For this specific case we know that the speed traveled corresponds to the round trip, therefore
[tex]v = \frac{2d}{t}[/tex]
Re-arrange to find t (The time taken to here the echo)
[tex]t = \frac{2d}{v}[/tex]
Replacing with our values we have that the distance is equal to 65m
[tex]t = \frac{2*65}{1530}[/tex]
[tex]t = 0.085s[/tex]
Therefore the time before the dolphin hears the echoes of the clicks is 0.085s
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground.How fast was the arrow shot?
To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.
By definition we know that velocity is defined as the change of position due to time, therefore
[tex]V = \frac{d}{t}[/tex]
Where,
d = Distance
t = Time
Speed can also be expressed in vector form through its components [tex]V_x[/tex] and [tex]V_y[/tex]
In the case of the horizontal component X, we have to
[tex]V_x = \frac{d}{t}[/tex]
Here d means the horizontal displacement, then
[tex]t = \frac{d}{V_x}[/tex]
[tex]t = \frac{67}{V_x}[/tex]
At the same time we have that the vertical component of the velocity is
[tex]V_y = gt[/tex]
Here,
g = Gravity
Therefore using the relation previously found we have that
[tex]V_y = g \frac{67}{V_x}[/tex]
The relationship between the two velocities and the angle can be expressed through the Tangent, therefore
[tex]tan\theta = \frac{V_y}{V_x}[/tex]
[tex]tan \theta = \frac{g \frac{67}{V_x} }{V_x}[/tex]
[tex]tan 3 = \frac{9.8\frac{67}{V_x} }{V_x}[/tex]
[tex]tan 3 = \frac{9.8*67}{V_x^2}[/tex]
[tex]V_x^2 = \frac{9.8*67}{tan 3}[/tex]
[tex]V_x= \sqrt{ \frac{9.8*67}{tan 3}}[/tex]
[tex]V_x = 111.93m/s \hat{i}[/tex]
This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,
Then [tex]V_y,[/tex]
[tex]V_y = g \frac{67}{V_x}[/tex]
[tex]V_y = 9.8*\frac{67}{111.93}[/tex]
[tex]V_y = 5.866m/s\hat{j}[/tex]
[tex]|\vec{V}| = \sqrt{111.93^2+5.866^2}[/tex]
[tex]|\vec{V}| = 112.084m/s[/tex]
Final answer:
To calculate the initial speed of the arrow in a projectile motion problem, one needs to find the time of flight using the horizontal range and the final impact angle and then solve for the initial horizontal and vertical velocity components.
Explanation:
To determine the initial speed of the arrow shot from the bow, we'll use the principle of projectile motion. First, we need to find the time of flight. The arrow makes a 3.00° angle with the ground upon impact, and since it was shot parallel to the ground, it maintains a constant horizontal velocity throughout its flight. So, we can calculate the initial horizontal speed using this final impact angle.
The horizontal range (R) is given by R = v0x * t, where v0x is the initial horizontal velocity and t is the time of flight. The arrow landed 67.0 m away, so R is 67.0 m.
To find t, we will use the vertical motion equation under constant acceleration due to gravity (g = 9.8 m/s2). We know the final vertical velocity component (vfy) can be related to the initial vertical velocity component (v0y = 0 since it was shot parallel to the ground) by vfy = v0y + g*t. We can also calculate vfy using the impact angle: vfy = tan(impact angle) * v0x. Combining these, we can solve for t and subsequently for the initial speed v0.
A bungee jumper, whose mass is 85 kg, jumps from a tall building. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 6.8 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.
Answer:
K= 290.28 N/m
Explanation:
Given: mass= 85 kg
the time taken to reach point two more times in 6.8 s.
2×t= 6.8 sec
t= 6.8/2= 3.4 sec
then, the time period for oscillation is
[tex]t= 2\pi\sqrt{\frac{m}{k} }[/tex]
Here K= spring constant
m= mass of jumper
⇒[tex]K= \frac{4\pi^2m}{t^2}[/tex]
now plugging the values we get
[tex]K= \frac{4\pi^2\times85}{3.4^2}[/tex]
K= 290.28 N/m
A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s about the y axis in a region where a 1.00-T magnetic field is directed along the x axis. The time t = 0 is chosen to be at an instant when the plane of the coil is perpendicular to the direction of B with arrow.
(a) Calculate the maximum induced emf in the coil.
V
(b) Calculate the maximum rate of change of magnetic flux through the coil.
Wb/s
(c) Calculate the induced emf at t = 0.050 0 s.
V
(d) Calculate the torque exerted by the magnetic field on the coil at the instant when the emf is a maximum.
N
Answer:
Explanation:
N = 65
Area, A = 0.1 x 0.2 = 0.02 m^2
R = 10 ohm
ω = 29.5 rad/s
B = 1 T
(a) at t = 0
e = N x B x A x ω
e = 65 x 1 x 0.02 x 29.5
e = 38.35 V
(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.
Ф = 38.35 Wb/s
(c) e = NBAω Sinωt
e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)
e = 38.174 V
(d) Maximum torque
τ = M B Sin 90
τ = N i A B
τ = N e A B / R
τ = 65 x 38.35 x 0.02 x 1 / 10
τ = 5 Nm
Find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1)to (3,-3). The amount of work done is ___?
Answer:
amount of work done is[tex] W = \frac{-4}{3}[/tex]
Explanation:
Formula for work done by force field
[tex]W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt[/tex]
where
r(t) is parametrization of line
as it is straight line so
[tex]r(t) = 1- t) r_o + tr_1 0 \leq t \leq 1[/tex]
thus,
r(t) = (1-t)(-1,1) + t(3,-3)
= (-1+t,1-t) + (3t - 3t)
= (-1+t +3t, 1-t-3t)
r(t) = (4t -1, 1- 4t)
r'(t) = (4,-4)
putting value in above integral
[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \int_{0}^{1} (-(4t -1)^2 , 2-8t).(4,-4) dt[/tex]
[tex]= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt[/tex]
[tex] =\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt[/tex]
[tex]= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}[/tex]
[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \frac{-4}{3}[/tex]
To find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1) to (3,-3), we compute the line integral of the dot product of the force and displacement vectors along the path. Performing the integrations using the limits set by the line's endpoints gives us the work done.
Explanation:The task involves physics concepts in vector calculus. Specifically, we're looking at the work done by a force vector over a path in two-dimensional space. Here, the force is described by the vector function F = xyi +(y-x)j. We need to calculate the work done by this force as it moves along a straight line path from (-1,1) to (3,-3).
Work done by a force in moving an object is given by the line integral of force · displacement. In mathematical terms, this definition reads as W = ∫F.dr. The computation requires us to take the dot product of the force and differential displacement vectors along the path.
The direction of displacement as we go from (-1,1) to (3,-3) is simply the path's tangent. Therefore, dr = dx i + dy j, where (dx,dy) is the differential displacement vector along the line. Since the line is straight, (dx, dy) = (3 - (-1), -3 - 1) = (4, -4) up to a constant. Then dot product becomes F · dr = (xy - (y-x))*(4).
Now, we take the line integral of this product over the path. The limits of integration, derived from the line's endpoints, are x = -1 to x = 3. We carry out the integration and simplify to get the work done by the force over the straight line. Note that the straight-line path simplifies the calculation. For paths with more complex shapes or forces that vary nonlinearly with displacement, such integrations may require specialist mathematical techniques.
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The nearest known exoplanets (planets beyond the solar system) are around 20 light-years away. What would have to be the minimum diameter of an optical telescope to resolve a Jupiter-sized planet at that distance using light of wavelength 600 nm? (Express your answer to two significant figures.)
To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as
[tex]\Delta \theta = 1.22\frac{\lambda}{d}[/tex]
Where,
\lambda = Wavelength
d = Optical Diameter
[tex]\theta =[/tex] Angular resolution
In turn you can calculate the angle through the diameter and the arc length, that is,
[tex]\Delta \theta = \frac{x}{D}[/tex]
Where,
x = The length of the arc
D = Distance
From known data we know that Jupiter's diameter is,
[tex]x_J = 1.43*10^8m[/tex]
[tex]D = 20*9.4608*10^{15}[/tex]
[tex]\lambda = 600*10^{-9}m[/tex]
Replacing we have that,
[tex]\frac{x}{D} = 1.22\frac{\lambda}{d}[/tex]
[tex]\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}[/tex]
Re-arrange to find d,
[tex]d = 968.5m = 0.968Km[/tex]
Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.
wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,
Answer:
[tex]T_1 =677224.40\ N[/tex]
Explanation:
given,
mass of the both ball = 5 Kg
length of rod = 2 L
where L = 0.55 m
angular speed = 45.6 rev/s
ω = 45.6 x 2 π
ω = 286.51 rad/s
v₁ = r₁ ω₁
v₁ =0.55 x 286.51 = 157.58 m/s
v₂ = r₂ ω₂
v₂ = 1.10 x 286.51 = 315.161 m/s
finding tension on the first half of the rod
r₁ = 0.55 r₂ = 2 x r₁ = 1.10
[tex]T_1 = m (\dfrac{v_1^2}{r_1}+\dfrac{v_2^2}{r_2})[/tex]
[tex]T_1 = 5 (\dfrac{157.58^2}{0.55_1}+\dfrac{315.161^2}{1.1})[/tex]
[tex]T_1 =677224.40\ N[/tex]
Suppose a yo-yo has a center shaft that has a 0.200 cm radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.70 m/s2, what is the angular acceleration of the yo-yo in rad/s2? rad/s2 (b) What is the angular velocity in rad/s after 0.750 s if it starts from rest? rad/s (c) The outside radius of the yo-yo is 3.30 cm. What is the tangential acceleration in m/s2 of a point on its edge?
Final answer:
The angular acceleration of the yo-yo is 850 rad/s², the angular velocity after 0.750 s is 637.5 rad/s, and the tangential acceleration of a point on its edge is 28.05 m/s².
Explanation:
To solve the yo-yo problem:
(a) The angular acceleration (α) can be found by using the linear acceleration (a) of the yo-yo and the radius (r) of its center shaft. The formula is α = a/r. Given that the yo-yo accelerates at 1.70 m/s2 and the radius of the center shaft is 0.200 cm (0.002 m), the angular acceleration is α = 1.70 m/s2 / 0.002 m = 850 rad/s2.
(b) The angular velocity (ω) after a certain time can be calculated using the formula ω = αt, assuming it starts from rest. So after 0.750 s, the angular velocity is ω = 850 rad/s2 * 0.750 s = 637.5 rad/s.
(c) To find the tangential acceleration (at) of a point on the edge of the yo-yo, with an outside radius of 3.30 cm (0.033 m), we use the formula at = α * R. Therefore, the tangential acceleration at the edge is at = 850 rad/s2 * 0.033 m = 28.05 m/s2.
Which of the following is characteristic of a nuclear reaction?
I. Electrons in atomic orbitals are involved in the breaking and forming of bonds
II. Elements are converted from one to another
III. Reaction rates are typically not affected by catalysts
IV. Atoms are rearranged by the breaking and forming of chemical bonds
a. I and II
b. I, II, and III
c. Ill and IV
d. II and III
e. I, II, III and IV
Answer:
d
Explanation:
Nuclear reactions unlike chemical reactions do not involve electrons or bonds.The main use for the reaction is neutrons. In nuclear there is only turning the element into another element and releasing a large amount of energy. So catalyst are also redundant in nuclear reactions.
Therefore, points III and 3 are true, and rest are all wrong. therefore option d is correct
In a nuclear reaction, elements are converted from one to another and the reaction rates are typically not affected by catalysts. Characteristics such as electrons in atomic orbitals being involved in breaking and forming bonds, and atoms being rearranged by breaking and forming of chemical bonds, pertain to chemical reactions, not nuclear ones.
Explanation:The correct characteristics of a nuclear reaction are options II and III. In a nuclear reaction, elements are indeed converted from one to another (II). This happens as the structure of an atom's nucleus changes. Furthermore, nuclear reaction rates are typically not affected by catalysts (III). This contrasts with chemical reactions, where catalysts can significantly speed up reaction rates. As for options I and IV, they are characteristic of chemical reactions, not nuclear ones. In chemical reactions, electrons in atomic orbitals are involved in the breaking and forming of bonds (I), and atoms are rearranged by the breaking and forming of chemical bonds (IV). Therefore, the answer is (d) II and III.
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A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.
a. The force of the golf club acting on the ball.
b. The force of the ball moving forward through the air.
c. The force of gravity acting on the ball.
d. All of the above.
Final answer:
The correct answer to the question is (c) The force of gravity acting on the ball. Once the golf ball is in flight and air resistance is neglected, gravity is the only force acting upon it, influencing its parabolic trajectory.
Explanation:
The question asks which forces act on a golf ball while it is in flight, given that air resistance is neglected. The correct option in this case is (c) The force of gravity acting on the ball. When a golf ball is hit and is flying through the air, the force of the golf club is no longer acting on it once it has left contact with the club. There is no force of the ball moving forward through the air; this is a description of motion, not a force. Therefore, the primary force acting on the golf ball once it is in the air is the force of gravity, which is the gravitational pull exerted by the Earth on the ball, causing it to follow a parabolic trajectory. The gravity is also responsible for the ball's downward acceleration at all points in its flight, as it is the only force acting on the ball if we ignore air resistance.
Final answer:
When a golf ball is hit and flying through the air, with air resistance neglected, the only force acting on it is gravity. The force from the golf club is present only during the impact, and the motion of the ball through the air is not a force but the result of the initial momentum given to the ball.
Explanation:
The question addresses the forces acting on a golf ball after it is hit by a golf club. Considering air resistance is neglected, the correct answer is (c) the force of gravity acting on the ball. Once the ball is in flight, the only force acting on it is the gravitational pull of the Earth, which affects the motion of the ball, causing it to follow a parabolic trajectory until it lands. The force of the golf club is only present during the brief contact with the ball, and after that, it doesn't affect the ball's flight. The idea of the force of the ball moving forward through the air is not a force but rather a consequence of the change in the momentum of the golf ball when the club strikes it, resulting in the velocity of the golf ball as it leaves the club.
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic field of magnitude 0.10 T . When the field is oriented normal to the slab's rectangular face, a Hall emf of 20 mV is measured across the slab's width. The probe is then placed in a magnetic field of unknown magnitude B, and a Hall emf of 69 mV is measured. Determine B assuming that the angle θ between the unknown field and the plane of the slab's rectangular face is
(a) θ = 90
(b) θ = 60
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, [tex]V_{Hall} = 20\ mV = 0.02\ V[/tex]
Magnetic Field, B = 0.10 T
Hall emf, [tex]V'_{Hall} = 69\ mV = 0.069\ V[/tex]
Now,
Drift velocity, [tex]v_{d} = \frac{V_{Hall}}{B}[/tex]
[tex]v_{d} = \frac{0.02}{0.10} = 0.2\ m/s[/tex]
Now, the expression for the electric field is given by:
[tex]E_{Hall} = Bv_{d}sin\theta[/tex] (1)
And
[tex]E_{Hall} = V_{Hall}d[/tex]
Thus eqn (1) becomes
[tex]V_{Hall}d = dBv_{d}sin\theta[/tex]
where
d = distance
[tex]B = \frac{V_{Hall}}{v_{d}sin\theta}[/tex] (2)
(a) When [tex]\theta = 90^{\circ}[/tex]
[tex]B = \frac{0.069}{0.2\times sin90} = 0.345\ T[/tex]
(b) When [tex]\theta = 60^{\circ}[/tex]
[tex]B = \frac{0.069}{0.2\times sin60} = 0.398\ T[/tex]
The Eiffel Tower is built of wrought iron approximately 300 m tall.Estimate how much its height changes between January (averagetemperature of 2o C) and July (average temperature of25o C). Ignore the angles of the iron beams and treatthe tower as a vertical beam.
The concepts required to solve this problem are the thermodynamic expressions of expansion and linear expansion.
In mathematical terms the dilation of a body can be expressed as
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
Where,
[tex]L_0 =[/tex] Initial Length
[tex]\alpha =[/tex] Thermal coefficient of linear expansion
[tex]\Delta T =[/tex] Change in Length
Our values are given as
[tex]L_0 = 300m[/tex]
[tex]T_f = 25\°C[/tex]
[tex]T_i = 2\°C[/tex]
[tex]\alpha = 12*10^{-6}C^{-1} (Iron)[/tex]
Replacing at the equation we have,
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
[tex]\Delta L = (300)(12*10^{{-6})(25-2)[/tex]
[tex]\Delta L = 0.0828m[/tex]
Therefore the change in the height is 8.28cm
The Eiffel Tower's height is subject to changes due to thermal expansion. In fact, it's around 8.3 cm taller in July than in January, with the temperature rise contributing to this change.
Explanation:The height change of the Eiffel Tower due to thermal expansion can be estimated using the linear expansion formula: ΔL = α*L*ΔT. Here, ΔL is the change in length, α is the thermal expansion coefficient, L is the original length, and ΔT is the change in temperature. Wrought iron has an expansion coefficient of approximately 12*10^-6 °C^-1. The height of the Eiffel Tower is approximately 300 m. The difference in average temperatures is approximately 23°C (25°C-2°C).
Let's plug these values into the formula: ΔL = 12*10^-6 °C^-1 * 300m * 23°C, which gives us a change in height of approximately 0.083m, or 8.3 cm. Therefore, the Eiffel Tower is an estimated 8.3 cm taller in July than it is in January, due to the effects of thermal expansion.
Learn more about Thermal expansion here:https://brainly.com/question/30242448
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