Answer:
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
[tex]my'' + \zeta y' + ky=0[/tex]
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
[tex]w - kL=0[/tex]
[tex]mg - kL=0[/tex]
[tex]4 - k\frac{1}{6}=0[/tex]
[tex]k = 4\times 6 =24[/tex]
The damping constant can be found by
[tex]F = -\zeta y'[/tex]
[tex]6 = 3\zeta[/tex]
[tex]\zeta = \frac{6}{3} = 2[/tex]
Finally, the mass m can be found by
[tex]w = 4[/tex]
[tex]mg=4[/tex]
[tex]m = \frac{4}{g}[/tex]
Where g is approximately 32 ft/s²
[tex]m = \frac{4}{32} = \frac{1}{8}[/tex]
Therefore, the required differential equation is
[tex]my'' + \zeta y' + ky=0[/tex]
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
The initial position is
[tex]y(0) = \frac{1}{2}[/tex]
The initial velocity is
[tex]y'(0) = 0[/tex]
We formulated the initial value problem for a damped spring-mass system:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0
with initial conditions x(0) = 0.5 ft and dx/dt(0) = 0 ft/s.
Let's break down the problem with the given data:
The mass weighs 4 lb.The spring stretches 2 inches under this weight.Additional displacement given is 6 inches in the positive direction.Viscous resistance is 6 lb when the velocity is 3 ft/s.First, find the spring constant k:
The weight of the mass stretches the spring by 2 inches (0.1667 feet).
The force exerted by the weight = 4 lb = mg
The spring force F = kx
So,
k = F/x k= 4 lb / 0.1667 ft k ≈ 24 lb/ftThe general form of the second-order differential equation governing the motion of the spring-mass-damper system is:
m*d²x/dt² + c*dx/dt + k*x = 0The viscous resistance given is 6 lb at 3 ft/s. Therefore, the damping coefficient c:
c = 6 lb / 3 ft/s c = 2 lb·s/ftThe initial conditions are displacement 6 inches (0.5 feet) and initial velocity 0:
x(0) = 0.5 ftCombining these elements, the initial value problem is:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center. The turntable is turned on and 0.8 s later it has an angular speed of 45 rpm. Assume the angular acceleration is constant and determine the following quantities for the ant 0.4 s after the turntable has been turned on. Express all quantities using appropriate mks units.
Complete Question
The complete is shown on the first uploaded image
Answer:
[tex]\alpha = 5.89 rad/s^2[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
[tex]v= 0.212m/s[/tex]
[tex]a_t= 0.5301 m/s[/tex]
[tex]a_r = 0.499 m/s[/tex]
[tex]a = 0.7279 m/s[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
Explanation:
From the question we are told that
mass of the ant is [tex]m_a = 0.8g = \frac{0.8}{1000} = 0.00018kg[/tex]
The distance from the center is [tex]d = 9cm = \frac{9}{100} = 0.09m[/tex]
The angular speed is [tex]w = 45rpm = 45 * \frac{2 \pi }{60} = 1.5 \pi[/tex]
The time taken to attain angular acceleration of 45rpm [tex]t_1 = 0.8s[/tex]
The time taken is [tex]t_2 = 0.4 s[/tex]
The angular acceleration is mathematically represented as
[tex]\alpha = \frac{w}{t}[/tex]
[tex]= \frac{1.5}{0.8}[/tex]
[tex]\alpha = 5.89 rad/s^2[/tex]
The angular velocity at time t= 0.4s is mathematically represented as
[tex]w__{0.4s}} = \alpha * t_2[/tex] Recall angular acceleration is constant
[tex]= 5.89 * 0.4[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
The linear velocity is mathematically represented as
[tex]v = w__{t_2}} * r[/tex]
[tex]= 2.36 * 0.09[/tex]
[tex]v= 0.212m/s[/tex]
The tangential acceleration is mathematically represented as
[tex]a_{t} = \alpha * r[/tex]
[tex]= 5.89 * 0.09[/tex]
[tex]a_t= 0.5301 m/s[/tex]
The radial acceleration is mathematically represented as
[tex]a_r = \frac{v^2}{r}[/tex]
[tex]= \frac{0.212^2}{0.09}[/tex]
[tex]a_r = 0.499 m/s[/tex]
The resultant velocity is mathematically represented as
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
[tex]= \sqrt{0.53^2 + 0.499^2}[/tex]
[tex]a = 0.7279 m/s[/tex]
The net force is mathematically represented as
[tex]F_{net} = 0.0008 * 0.7279[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
In a clear, coherent, paragraph-length answer, explain why the speed of the apple half way between the release and the maximum height is not half the initial speed. Include a qualitative justification for the speed being greater or less than half the initial speed. You may use no equations to justify your conclusion
Answer:
the speed of the apple half way between the release and the maximum height is not half the initial speed due to the acceleration due to gravity numerically equal to 9.81 m/s2. An acceleration in the direction of motion of a body will cause its velocity to increase, thus, this acceleration due to gravity increases the velocity of the apple so that it will be at its maximum speed before hitting the ground.
The value of this acceleration implies that there is an increase of 9.8 m/s in its speed for every one seconds of fall. This means that at 1 sec when it starts to fall, the speed will be 9.8 m/s. By two seconds it becomes 19.62 m/s. It is clear to see that the velocity halfway will be more than its initial velocity.
When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5. When 900-nm light is incident on the same slit system the number is: a. 9b. 2c. 3d. 10e. 5
Answer:
Explanation:
width of central diffraction maxima = 2 λD / d₁
λ is wave length of light , D is screen distance and d₁ is slit width
width of each interference fringe = λD / d₂ , d₂ is slit separation.
No of interference fringe in central diffraction fringe
= width of central diffraction maxima / width of each interference fringe
= 2 λD / d₁ x λDd₂ / λD
No = 2 d₂ / d₁
No = 5
5 = 2 d₂ / d₁
Since this number does not depend upon wavelength so it will remain the same
No of required fringe will be 5 .
right option
e ) 5.
When 900-nm light is incident on the same slit system the number is 5.
The given parameters:
First wavelength of the incident light, λ₁ = 450 nmThe number of interference maxima, n = 5Second wavelength, λ₂ = 900 nmThe number of interference fringe in central diffraction fringe is calculated as follows;
[tex]n = \frac{width \ of \ central \ diffraction\ maxima}{ width \ of \ each \ interference\ fringe}\\\\n = \frac{2\pi \lambda D/d_1}{2\pi \lambda D/d_2} \\\\n = \frac{2d_2}{d_1}[/tex]
The number of number of interference fringe is independent of the wavelength.
Thus, when 900-nm light is incident on the same slit system the number is 5.
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If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself. ะเ50% Part (a) If you had a nic that you would have to use?
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.
a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?
b) What type of electromagnetic radiation would this be?
Given Information:
Wavelength = λ = 1.5×10⁻¹⁰ m
Required Information:
a) Frequency = f = ?
b) Type of electromagnetic radiation = ?
Answer:
a) Frequency = f = 2×10¹⁸ Hz
b) Type of electromagnetic radiation = X-rays
Explanation:
a) The frequency of the electromagnetic radiation is given by
f = c/ λ
Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s
f = 3×10⁸/1.5×10⁻¹⁰
f = 2×10¹⁸ Hz
Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.
b)
The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz
The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays
To observe atomic-sized features, we would need to use forms of radiation with similar wavelengths to the atom's size, such as X-rays. This is because observable detail with electromagnetic radiation is limited by wavelength; therefore, visible light cannot detect atoms due to their much smaller size.
To observe features that are around the size of atoms with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself. In the area of physics, this concept is a key part of the study of wave optics.
If we imagine atoms to be about 1.5 x 10^-10 m (or 0.1 nm) in size, because wavelength and size must be on similar scales, we'd need to use forms of radiation with similar wavelengths.
This falls within the range of X-rays on the electromagnetic spectrum. Therefore, to observe atomic-sized features, an appropriate method would be X-ray microscopy or other forms of X-ray imaging.
It's important to note that visible light can never detect individual atoms because atoms are much smaller than visible light's wavelength, reinforcing the concept that observable detail is limited by wavelength.
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An electron, moving south, enters a uniform magnetic field. Because of this uniform magnetic field, the electron curves upward. We can conclude that the magnetic field must have a component A. upward; B. downward; C. toward the north; D. toward the east; E. toward the south; F. toward the west.
Answer:
Towards the west
Explanation:
Magnetic force is the interaction between a moving charged particle and a magnetic field.
Magnetic force is given as
F = q (V × B)
Where F is the magnetic force
q is the charge
V is the velocity
B is the magnetic field
V×B means the cross product of the velocity and the magnetic field
NOTE:
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
So, if the electron is moving southward, then, it implies that the velocity of it motion is southward, so the electron is in the positive z-direction
Also, the electron is curved upward due to the magnetic field, this implies that the force field is directed up in the positive y direction.
Then,
V = V•k
F = F•j
Then, apply the theorem
F •j = q ( V•k × B•x)
Let x be the unknown
From vector k×i =j.
This shows that x = i
Then, the magnetic field point in the direction of positive x axis, which is towards the west
You can as well use the Fleming right hand rule
The thumb represent force
The index finger represent velocity
The middle finger represent field
A circular swimming pool has a diameter of 18 m. The circular side of the pool is 3 m high, and the depth of the water is 1.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m 3 kg/m^3.)How much work (in Joules) is required to: i. pump all of the water over the side? ii. pump all of the water out of an outlet 2m over the side?
To calculate the work required to pump water out of the swimming pool, physics principles such as the formula for work, volume of a cylinder, and weight of water are used to derive the necessary values for the complete calculation.
Explanation:The question requires the use of physics concepts to calculate work done in pumping water out of a swimming pool. Since we are dealing with the force of gravity, density of water, height to which water is raised, and volume of water, we must use the formula for work done (work = force x distance) alongside formulas for the volume of a cylinder (Volume = πr^2h) and the weight of the water (weight = mass x gravity).
To solve the problem, first calculate the volume of water in the pool using the volume formula for a cylinder, taking into account the water's depth. Then, calculate the mass by multiplying the volume by the density of water. The weight of the water is found by multiplying the mass by the acceleration due to gravity. Finally, calculate the work done by multiplying the weight of the water by the height to which it needs to be lifted, considering both scenarios: pumping over the side and pumping through an outlet 2 meters over the side.
Final answer:
Calculation of work required to pump water out of a circular swimming pool at different heights.
Explanation:
The work required to pump all the water over the side of the pool is:
Work = mgh = density x volume x gravity x height = 1000 kg/m³ x (pi x (9m)² x 1.5m) x 9.8 m/s² x 3m = 1222540 J
For pumping all water out of the outlet 2m over the side:
Work = mgh = density x volume x gravity x height = 1000 kg/m³ x (pi x (9m)²x 1.5m) x 9.8 m/s² x 5m = 2037567 J
A physical pendulum consists of a vertical board of of mass 6.12 kg, length 191 cm, and width 12 cm hanging from a horizontal, frictionless axle. A bullet of mass 143 g and a purely horizontal speed v impacts the pendulum at the bottom edge of the board. The board then makes a complete circle.
(a) If the bullet passed through the board, reducing its speed by a factor of 1/3, what is the minimum speed of the bullet?
(b) If the bullet embedded itself in the board, what is the minimum speed of the bullet?
Note (hint from instructor): It appears you may use Conservation of Energy to find angular speed. For potential energy: use center of mass. For kinetic energy, use impulse. Then, to find initial speed, use conservation of angular momentum.
This question involves applying the principles of Conservation of Energy and Angular Momentum to calculate the minimum speed of a bullet impacting a physical pendulum. The solution requires us to consider different scenarios: when the bullet passes through and when it embeds itself into the board.
Explanation:This problem involves the concepts of Conservation of Energy and Conservation of Angular Momentum.
(a) If the bullet passes through the board, we consider that the board moves by gaining kinetic energy (KE) from the bullet’s impact. The bullet's final speed is (1/3)*v, so the KE lost by the bullet will be (4/9)*mv². This energy is used for the board to perform a complete circle, which means it acquires a potential energy of (2*M*g*H), where H is the height when the board makes a complete circle (H=L/2). Equating these energies (4/9)*mv²-2*M*g*H = 0, we solve for v.
(b) If the bullet embeds itself, the system's final angular momentum equals its initial, which gives (1/2*M*L)v = (M+m)Lω. Then we equate the initial kinetic energy and potential energy during the complete circle, getting (1/2)(M+m)L²ω² = (M+m)gL. Solving for v will give us the minimum speed.
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Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.Which laser has its first maximum closer to the central maximum?
Answer:
The first laser
Explanation:
Using the equation
ym = m*λ*L/d
so for the first max for each one m =1
so for laser 1
y = (d/20)*(6.0m)/d
the d cancel out so we have
y = 6.0m /20 = 0.3
for laser 2 we do the same thing
y = 6.0m/15 = 0.4
y ( laser1 ) < y ( laser2 )
(1) hence first maxima of laser1 is closest to central maxima
Δy = 0.4m -0 .3 m = 0.1m
A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 48 feet across at its opening and 4 feet deep at its center, where should the receiver be placed
The receiver of the satellite dish, which is shaped like a paraboloid, should be positioned 9 feet from the base along the axis of symmetry for optimal signal reception.
Explanation:This question is related to the properties of a parabolic dish (or paraboloid). The satellite dish is a 3D shape formed by rotating a parabola along its axis of symmetry. For a parabola, the focus is located a certain distance from the vertex, and this distance is defined by the depth of the dish.
The formula for the focus (F) in a parabolic mirror/dish is given by: F = D^2/16d where D is the diameter of the dish (48ft in this case) and d is the depth of the dish (4ft in this case). By substitifying these values, we obtain F = 48^2/(16*4) = 144/16 = 9 feet.
Therefore, to ensure the strongest reception of signals, the receiver of the satellite dish should be positioned 9 feet from the base along the axis of symmetry.
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The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified the conservation of momentum in inertial reference frames. d) supported the relationship between mass and energy. e) indicated that the speed of light is the same in all inertial reference frames.
Answer:
e) indicated that the speed of light is the same in all inertial reference frames.
Explanation:
In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium aether. They believed that even the space is not empty and filled with aether.
Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.
It thus proved that the speed of light remains same in all inertial frames.
Also, it became a base for the special theory of relativity by Einstein.
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
Answer:
check the diagram in the attachment below.
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Explanation:
question
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
ANSWER;
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Answer:
the potential difference across the membrane of this single segment is zero.
ΔVmem=0
Explanation:
When the switch is closed, the capacitor ([tex]C_{mem}\\[/tex]) will be short. It means the current will not flow through the resistor([tex]R_{mem[/tex]). Rather current will only flow through the capacitor as there is no resistance. Due to zero resistance, the voltage across the capacitor will also be zero. So, the potential difference across the membrane of this single segment is zero.
ΔVmem=0
Albert is piloting his spaceship, heading east with a speed of 0.92 cc relative to Earth. Albert's ship sends a light beam in the forward (eastward) direction, which travels away from his ship at a speed cc. Meanwhile, Isaac is piloting his ship in the westward direction, also at 0.92 cc, toward Albert's ship
With what speed does Isaac see Albert's light beam pass his ship?
Answer:
He sees the light as 1c
Explanation:
According to relativity, the speed of light is the same in all inertial frame of reference.
If we were to add the velocities as applicable to a normal moving bodies, the relative speed of the light beam will exceed c which will break relativistic law since nothing can go past the speed of light.
What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffraction grating has 700 lines per mm?
Answer:
The wavelength is 3500 nm.
Explanation:
d= [tex]\frac{1}{700 lines per mm} = 0.007mm = 7000 nm[/tex]
n= 1
θ= 30°
λ= unknown
Solution:
d sinθ = nλ
λ = [tex]\frac{7000 nm sin 30}{1}[/tex]
λ = 3500 nm
the triceps muscle in the back of the upper arm extends the forearm. this muscle in a professional boxer exerts a force of 2.00×103 n with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 121 rad/s2. what is the moment of inertia of the boxer's forearm?
Answer: 0.512 kgm²
Explanation:
Given
Force, F = 2*10^3 N
Angular acceleration, α = 121 rad/s²
Lever arm, r(⊥) = 3.1 cm = 3.1*10^-2 m
τ = r(⊥) * F
Also,
τ = Iα
Using the first equation, we have
τ = r(⊥) * F
τ = 0.031 * 2*10^3
τ = 62 Nm
Now we calculate for the inertia using the second equation
τ = Iα, making I subject of formula, we have
I = τ / α, on substituting, we have
I = 62 / 121
I = 0.512 kgm²
Thus, the moment of inertia of the boxers forearm is 0.512 kgm²
To practice Problem-Solving Strategy 25.1 for charge interaction problems. A proton and an alpha particle are momentarily at rest at a distance rrr from each other. They then begin to move apart. Find the speed of the proton by the time the distance between the proton and the alpha particle doubles. Both particles are positively charged. The charge and the mass of the proton are, respectively, eee and mmm. The charge and the mass of the alpha particle are, respectively, 2e2e and 4m4m.
By using the principle of energy conservation and accounting for the charges and masses of the proton and alpha particle, we can establish that the speed of the proton when the distance between the particles doubles is given by the square root of four times Coulomb's constant times the charge of the proton squared, divided by the mass of the proton and the original distance.
Explanation:To solve the problem, we can apply the principle of energy conservation. In the initial state, the system has potential energy, which is due to the electrostatic interactions between the proton and the alpha particle. When the particles begin to move apart, some of this potential energy is converted into kinetic energy. Given that both particles are positively charged, the potential energy U of the interaction is given by Coulomb's law:
U = k * Q1 * Q2 / r
where Q1 and Q2 are the charges of the particles, r is the distance between them and k = 8.99 * 10^9 N * m^2/C^2 is Coulomb's constant. The alpha particle has 2e, and the proton has e, leading to the potential energy of the system being: U_initial = 2 * k * e^2 / r.
When the distance between the proton and the alpha particle doubles, the potential energy becomes: U_final = k * e^2 / (2r).
The change in potential energy (delta U) thus equals the kinetic energy of the proton (as the alpha particle is more massive, its kinetic energy can be neglected): delta U = U_initial - U_final = K_proton = (1/2) * m * v^2, where m and v are the mass and speed of the proton, respectively. Solving this equation for v, we get that the speed of the proton is: v = sqrt((4 * k * e^2)/(m * r)).
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What is the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth? Assume λ = 506 nm and a pupil diameter is 4.90 mm. (In this problem, you may use the Rayleigh criterion for the limiting angle of resolution of an eye.)
Answer:
the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Explanation:
Using Rayleigh criterion for the limiting angle of resolution of an eye
[tex]\theta = \frac{1.22\lambda }{D } \\ \\ \theta = \frac{1.22*506 *10^{-9} }{4.90*10^{-3}m}[/tex]
[tex]\theta = 1.2598*10^{-4}[/tex] rad
[tex]\theta = 125.98*10^{-6} \ rad[/tex]
Thus; the separation between the two sources is expressed as:
[tex]\theta = \frac{y}{L} \\ \\ y = L \theta \\ \\ y = (250*10^3 )(125.98*10^{-6} \ rad) \\ \\ y = 31.495 \ m[/tex]
Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Final answer:
The approximate size of the smallest object that astronauts can resolve by eye when orbiting 250 km above the Earth is approximately 628 meters.
Explanation:
The approximate size of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth can be determined using the Rayleigh criterion. According to the Rayleigh criterion, the resolution of the eye is determined by the angle of resolution, which is given by:
angle of resolution = 1.22 * (wavelength / pupil diameter)
Using the given values, we can calculate the angle of resolution:
angle of resolution = 1.22 * (506 nm / 4.90 mm) = 1.2611957657 x 10^-3 radians
To determine the approximate size of the smallest object, we need to find the linear size corresponding to this angular resolution at a distance of 250 km:
linear size = 2 * distance * tan(angle of resolution)
linear size = 2 * 250000 m * tan(1.2611957657 x 10^-3 radians) = 628.1079000855 m
Therefore, the approximate size of the smallest object that astronauts can resolve by eye when they are orbiting 250 km above the Earth is approximately 628 meters.
g If the primary coil of wire on a transformer is kept the same and the number of turns of wire on the secondary is increased, how will this affect the voltage observed at the secondary? There is no voltage measured on the secondary coil since it is only connected to a resistor and not connected to a battery. The voltage will stay the same. The voltage will increase. The voltage will decrease.
Answer:
The voltage will increase.
Explanation:
Increason the number of coil in the secondary will increase the voltage in a transformer.
After a party the host pours the remnants of several bottles of wine into a jug. He then inserts a cork with a 2.25-cm diameter into the jug, placing it in direct contact with the wine. He is amazed when he pounds the cork into place that the bottom of the jug, with a 16.5-cm diameter, breaks away. show answer No Attempt Find the force that the liquid exerts on the bottom of the bottle when the host is pounding the cork with a 150-N force. Ignore the force caused by the weight of the wine.
Answer:
F = 8066.67 N
Explanation:
The extra force exerted on the bottom of the jug can be expressed as the pressure generated by the cork multiplied by the area of the bottom. We can also obtain the Pressure P by dividing the force F1 applied to the cork by it's area A1.
Thus;
F = PA2 = (F1/A1) x (A2)
F = (F1/(πd1²/4)) x (πd2²/4)
π/4 will cancel out to give;
F = F1(d2/d1)²
F = 150(16.5/2.25)
F = 8066.67 N
Using Pascal's Principle, the force the host applies to the cork is transferred through the wine to the bottom of the jug. To calculate the force exerted on the jug's bottom, we find the cross-sectional areas of the cork and bottom of the jug and apply the principle that pressure remains constant throughout the fluid. The amplified force is then calculated from these values.
To calculate the force the liquid exerts on the bottom of the bottle when the host pounds the cork with a 150-N force, we first consider Pascal's Principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls of the containing vessel. The key to solving this problem is to recognize that the force exerted by pounding the cork is amplified by the liquid because the area at the bottom of the jug is larger than the area of the cork.
The force on the cork is given by:
F1 = 150 N, and the cross-sectional area of the cork, A1, can be calculated using the cork's diameter:
A1 = π*(d1/2)^2 = π*(2.25 cm/2)^2
The cross-sectional area of the bottom of the jug, A2, is:
A2 = π*(d2/2)^2 = π*(16.5 cm/2)^2
According to Pascal's Principle:
P1 = P2
where P1 is the pressure applied to the cork and P2 is the pressure on the bottom of the jug. Thus,
P1 = F1/A1
Therefore, the force on the bottom of the jug, F2, can be calculated:
F2 = P2*A2 = (F1/A1)*A2
By inserting the values and calculating F2, we determine the force that the liquid exerts on the bottom of the bottle.
A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the body is pulled 1 m to the left, stretching the spring, and set in motion with an initial velocity of 20 m/s to the left. (a) Find x(t) in the form Upper C cosine (omega 0 t minus alpha ). (b) Find the amplitude and the period of motion of the body.
The problem refers to an object undergoing simple harmonic motion, initially displaced and set in motion. The displacement and the period of motion can be calculated using information of spring constant, mass, initial velocity, and principles of oscillations.
Explanation:The question refers to a classic setup for simple harmonic motion (SHM), where a body of known mass is attached to a spring with a known force constant. The force constant is the ratio of the force applied to the spring and the resulting extension; we can calculate it as k = F/x = 9N/0.2m = 45 N/m. This accounts for the spring's stiffness.
As SHM is in effect here, the displacement of the body from the equilibrium position can be represented by the equation x(t) = A cos(ωt + α). In this case, the body is initially displaced 1m to the left (A = 1m) and given an initial velocity to the left (which determines α).
Angular frequency ω can be determined by the formula √(k/m), where k is the spring constant and m is the mass of the body (noting to convert grams to kilograms for SI units). Hence, ω = √(45 N/m / 0.2 kg) = 15 rad/s.
On the other hand, initial velocity is related to α by the equation v = -Aωsinα, thus α can be calculated as such. Here, it is worth mentioning that the amplitude is the maximum displacement (1m) and the period of motion, given by T = 2π/ω, refers to the time taken for one complete oscillation.
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A circular loop of wire with area A Bz of B By of B Determine the component Bz B . Determine the component By Bx of B? k . Determine the component Bx j , and k^ i , j^ A , i^ I , A U = - Mu - B is negative. The magnitude of the magnetic field is B0=15D/IA Determine the vector magnetic moment of the current loop. Express your answer in terms of the variables I D is a positive constant, and for this orientation of the loop the magnetic potential energy, where B vector is given by -z -direction toward the origin, a current I is circulating clockwise around the loop. The torque produced by an external magnetic field B? z -axis looking in the xy -plane. As viewed along the z A lies in the xy
Find the given attachment
Three nested coffee filters fall, quickly reaching a terminal speed of ~ 1.7 m/s. If the air resistance force on a single falling coffee filter was F1, how large is the air resistance force on the three falling filters? 1) F1 2) 3*F1 3) (1/3)*F1 4) not enough information
Final answer:
The air resistance force on three nested coffee filters falling is 3*F1.
Explanation:
The air resistance force on three nested coffee filters falling is 3*F1.
When the coffee filters fall, they quickly reach a terminal speed of approximately 1.7 m/s. Terminal velocity occurs when the air resistance force equals the weight of the object. Since the filters are nested, the air resistance force on a single falling filter is F1. Therefore, assuming the air resistance force is the same for each filter, the air resistance force on the three falling filters is 3*F1.
A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and land at ground level on the other side. He has done this stunt many times and approaches it with confidence. Unfortunately, the motor-cycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the rolling friction of his rubber tires (coefficient 0.02) is not negligible. Does he survive, or does he become croco-dile food? Justify your answer by calculating the distance he travels through the air after leaving the end of the ramp
Answer:
He becomes a croco-dile food
Explanation:
From the question we are told that
The height is h = 2.0 m
The angle is [tex]\theta = 20^o[/tex]
The distance is [tex]w = 10m[/tex]
The speed is [tex]u = 11 m/s[/tex]
The coefficient of static friction is [tex]\mu = 0.02[/tex]
At equilibrium the forces acting on the motorcycle are mathematically represented as
[tex]ma = mgsin \theta + F_f[/tex]
where [tex]F_f[/tex] is the frictional force mathematically represented as
[tex]F_f =\mu F_x =\mu mgcos \theta[/tex]
where [tex]F_x[/tex] is the horizontal component of the force
substituting into the equation
[tex]ma = mgsin \theta + \mu mg cos \theta[/tex]
[tex]ma =mg (sin \theta + \mu cos \theta )[/tex]
making a the subject of the formula
[tex]a = g(sin \theta = \mu cos \theta )[/tex]
substituting values
[tex]a = 9.8 (sin(20) + (0.02 ) cos (20 ))[/tex]
[tex]= 3.54 m/s^2[/tex]
Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as
[tex]sin \theta = \frac{h}{l}[/tex]
making [tex]l[/tex] the subject
[tex]l = \frac{h}{sin \theta }[/tex]
substituting values
[tex]l = \frac{2}{sin (20)}[/tex]
[tex]l = 5.85m[/tex]
Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as
[tex]v^2 =u^2 + 2 (-a)l[/tex]
The negative a means it is moving against gravity
substituting values
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
[tex]v= \sqrt{79.582}[/tex]
[tex]= 8.92m/s[/tex]
The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is
Initial velocity along the x-axis which is mathematically evaluated as
[tex]v_x = vcos 20^o[/tex]
substituting values
[tex]v_x = 8.92 * cos (20)[/tex]
[tex]= 8.38 m/s[/tex]
Initial velocity along the y-axis which is mathematically evaluated as
[tex]v_y = vsin\theta[/tex]
substituting values
[tex]v_y = 8.90 sin (20)[/tex]
[tex]= 3.05 m/s[/tex]
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2} a_y t^2[/tex]
where [tex]y_o[/tex] is the initial height,
[tex]u_y[/tex] is the initial velocity in the y-axis
[tex]a_y[/tex] is the initial acceleration in the y axis with a constant value of ([tex]g = 9.8 m/s^2[/tex])
at the y= 0 which is when the height above ground is zero
Substituting values
[tex]0 = 2 + (3.05)t - 0.5 (9.8)t^2[/tex]
The negative sign is because the acceleration is moving against the motion
[tex]-(4.9)t^2 + (2.79)t + 2m = 0[/tex]
Solving using quadratic formula
[tex]\frac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex]
substituting values
[tex]\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}[/tex]
[tex]t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}[/tex]
[tex]t = -0.39s \ or \ t = 1.02s[/tex]
since in this case time cannot be negative
[tex]t = 1.02s[/tex]
At this time the position the motorcycle along the x-axis is mathematically evaluated as
[tex]x = u_x t[/tex]
x [tex]=8.38 *1.02[/tex]
[tex]x =8.54m[/tex]
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool
If the horizontal distance x covered by the motorcyclist is greater than or equal to 10 meters, he survives; otherwise, he does not, From the given calculations, x = 8.21 m, so he becomes crocodile food.
According to question, the height of motorcycle daredevil is h = 2.0 m, the angle of the ramp is θ = 20°, the distance of the wide pool is w = 10 m, the speed of the motorcycle daredevil is u = 11 m/s, and the coefficient of static friction between the ramp and the tyres of the motorcycle daredevil is μ = 0.02.
At equilibrium, the forces acting on the motorcycle are equated as:
ma = mgsinθ + f
here, m = mass of the motorcycle, a is the acceleration of the motorcycle, g is the acceelration due to gravity (g = 9.8 m/s²), and f is the frictional force between the ramp and the tyres of the motorcycle daredevil mathematically represented using normal reaction N as
f =μ N = μ mgcosθ
substituting into the equation
ma = mg sinθ + μ mg cosθ
ma =mg (sinθ + μ cosθ)
⇒ a = g(sinθ + μ cosθ)
⇒ a = 9.8 (sin(20°) + (0.02) cos (20°))
⇒ a = 3.54 m/s²
Now, for the ramp, we can say that:
[tex]sin \theta = \frac{h}{l}[/tex]
[tex]\therefore l = \frac{h}{sin \theta}[/tex]
substituting the above values, we get:
[tex]l = \frac{2}{sin20^{\circ}}[/tex]
or, l = 5.85m
Using Newton equation of motion, the final velocity of the motorcycle at the end of the ramp, can be given as:
[tex]v^2 =u^2 + 2 (-a)l[/tex]
As the motorcycle is moving against gravity, the acceleration (a) is taken negative. From the above values, we get:
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
⇒ [tex]v= \sqrt{79.582} \hspace{0.8 mm} m/s[/tex]
or, v = 8.92m/s
The motion of the motorcycle is a projectile motion. The initial velocity at the beginning of the pool (end of ramp) is composed of two component which are the x, and the y components respectively.
Initial velocity along the x-axis which is [tex]v_x[/tex] = v cos 20°
substituting values, we get:
[tex]v_x[/tex] = 8.92 cos (20°)
or, v = 8.38 m/s
Initial velocity along the y-axis which [tex]v_y[/tex] = v sin 20°
substituting values, we get:
[tex]v_y[/tex] = 8.90 sin (20)
[tex]v_y[/tex] = 3.05 m/s
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2}a_y t^2[/tex]
where y₀ is the initial height, [tex]u_y[/tex] is the initial velocity in the y-axis, and [tex]a_y[/tex] is the initial acceleration in the y-axis with a constant value of (g = 9.8 m/s^2)
At y = 0
0 = 2 + (3.05) t - 0.5 (9.8) t²
The negative sign is because the acceleration is moving against the motion:
- (4.9) t² + (2.79) t + 2m = 0
To solve for t, we can use the quadratic formula:
[tex]t = \frac{- b \hspace{0.5} \pm \sqrt{(b^2 - 4ac)}}{2a}[/tex]
In this case, a = 4.9, b = -2.79, and c = -2.
Plugging in these values, we get:
[tex]t = \frac{(2.79 \pm \sqrt{(-2.79)^2 - 4(4.9)(-2)}}{2(4.9)}[/tex]
Solving for both possible values of t, we get:
t ≈ [tex]\frac{(2.79 + 6.858)}{9.8}[/tex] ≈ 0.98 seconds
t ≈ [tex]\frac{(2.79 - 6.858)}{9.8}[/tex] ≈ -0.41 seconds
Since time cannot be negative,
∴ t ≈ 0.98 seconds
Hence displacement along the x-coordinate will be:
x = [tex]u_x[/tex] t
⇒ x =8.38 × 0.98 m
x = 8.21 m
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool.
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor.What frequency is required?
Answer:
Frequency required will be 2421.127 kHz
Explanation:
We have given inductance [tex]L=0.450H=0.45\times 10^{-3}H[/tex]
Current in the inductor [tex]i=1.90mA=1.90\times 10^{-3}A[/tex]
Voltage v = 13 volt
Inductive reactance of the circuit [tex]X_l=\frac{v}{i}[/tex]
[tex]X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm[/tex]
We know that
[tex]X_l=\omega L=2\pi fL[/tex]
[tex]2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10[/tex]
f = 2421.127 kHz
3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed and acceleration of her swing. She has a device that measures the linear velocity of the bat during the swing. Her bat is 0.85 m long. On her first swing, the linear velocity of the bat just prior to the swing is 0 m/s, and 0.15 s later, just prior to ball contact is 16 m/s. What is the average angular acceleration of the end of her bat during the swing?
Answer:
The average angular acceleration is [tex]\alpha =125.487 rad /s^2[/tex]
Explanation:
From the question we are told that
From the question we are told that
The length of the bat is [tex]l = 0.85m[/tex] \
The initial linear velocity is [tex]u = 0 m/s[/tex]
The time is [tex]t = 0.15s[/tex]
The velocity at t is [tex]v = 16 m/s[/tex]
Generally average angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_o}{t}[/tex]
Where [tex]w_f[/tex] is the finial angular velocity which is mathematically evaluated as
[tex]w_f = \frac{v}{l}[/tex]
[tex]w_f = \frac{16}{0.85}[/tex]
[tex]= 18.823 rad/s[/tex]
and [tex]w_o[/tex] is the initial angular velocity which is zero since initial linear velocity is zero
So
[tex]\alpha = \frac{18.823 - 0}{0.15}[/tex]
[tex]\alpha =125.487 rad /s^2[/tex]
A 325-turn circular-loop coil 9.4 cm in diameter is initially aligned so that its axis is parallel to Earth’s magnetic field. In 2.48 ms the coil is flipped so that its axis is perpendicular to Earth’s magnetic field. If an average voltage of 0.17 V is thereby induced in the coil, what is the value of Earth’s magnetic field at that location? Answer in units of µT.
Answer:
[tex]B = 187\ \mu T[/tex]
Explanation:
Given,
Number of turns, N = 325
Diameter of coil, d = 9.4 cm
time,t = 2.48 m s
average voltage,ε = 0.17 V
Earth magnetic field, B = ?
We know that
[tex]\epsilon = \dfrac{NBA}{t}[/tex]
[tex]0.17= \dfrac{325\times B \times \pi (0.047)^2}{2.48\times 10^{-3}}[/tex]
[tex]B = 187 \times 10^{-6}\ T[/tex]
[tex]B = 187\ \mu T[/tex]
Magnetic field of the earth is equal to [tex]B = 187\ \mu T[/tex]
Consider a proton, a deuteron (nucleus of deuterium, i.e., Hydrogen-2), and an alpha particle (nucleus of Helium-4), all with the same speed. These particles enter a region of uniform magnetic field B, traveling perpendicular to B. What is the ratio of the deuteron's orbital radius to the proton's orbital radius?
Answer:
the ratio of the deuteron's orbital radius to the proton's orbital radius is 2 : 1
Explanation:
Detailed explanation and calculation is shown in the image below
A bucket of water of mass 16.0kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350m with mass 11.4kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.1m to the water. You can ignore the weight of the rope. a) What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be g = 9.80m/s^2 . b) With what speed does the bucket strike the water? Take the free fall acceleration to be g = 9.80m/s^2 . c) What is the time of fall? Take the free fall acceleration to be g = 9.80m/s^2 . d) While the bucket is falling, what is the force exerted on the cylinder by the axle? Take the free fall acceleration to be g = 9.80m/s^2 .
Answer:
Explanation:
We shall consider tension to be T.
For motion of bucket in downward direction
mg - T = ma , m is mass of bucket , a is linear acceleration of bucket
for rotational motion of wheel
Tortque by force Tr = I x α I is moment of inertia , α is angular acceleration , r is radius of the wheel.
Putting the values in the equation above
mg - I x α / r = ma
mg - 1/2 M r ²x a / r² = ma
mg - 1/2 M a = ma
mg = 1/2 M a + ma
a = mg / (m + .5 M)
= 16 X 9.8 / (16 +.5 x 11.4)
= 7.22 m /s²
mg - T = ma
T = m( g - a )
= 16 x ( 9.8 - 7.22 )
= 41.28 N.
b )
v² = u² + 2 a h , v is velocity when bucket falls by height h
= 0 + 2 x 7.22 x 10.1
= 145.84
v = 12.07
c )
v = u + at , t is time of fall
12.07 = 0 + 7.22 t
t = 1.67 s
d )
Force by axle on cylinder = T
= 41.28 N .
Three blocks are pushed along a rough surface by a force with magnitude P, as shown above. Fc is the magnitude of the contact force between blocks 2 and 3 and Ff, Fn, and Fg are the magnitudes of friction, normal, and gravitational forces on block 3, respectively. Please draw a FBD.
Answer:
See the FBD attached
Explanation:
Since Ff, Fn, and Fg are the magnitudes of friction, normal, and gravitational forces on block 3, you can build the FBD (free bdy diagram) for block 3.
The FBD of the block 3 must include all the forces acting on the block 3.
The FBD is shown on the graph attached.
Observe this:
Fc is the contact force that block 2 exerts on block 3. Assuming block 2 is to the left of block 3, the force is toward the right.Ff is the friction force exerted by the ground on the block 3. Its direction is opposite to the movement. Assuming the block is moving to the right, the direction of Ff is to the left. Also, assuming the block is accelerating to the right, the magnitude of Ff is less than the magnitude of Fc, and the arrow that represents Ff is shorter than the arrow that represents Fc.Fn is the normal force on block 3 due to the contact with the ground. It points upward, as it opposes the gravitaional force. Assuming the block is not moving vertically, the magnitude of Fn is equal to the magnitude of Fg, and both arrows have the same length.Fg is the gravitational force: the pulling force of the Earth on the block. It is equal to the weight of the block.A ball is thrown upward from a height of 256 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Find the attachments for complete solution
The function giving the height of the ball at time t is s(t) = 96t - 16t^2 + 256. The ball takes 6 seconds to reach the ground and achieves a maximum height of 400 feet.
Explanation:The physics problem you've posed can be approached by first understanding that the height of the ball (s(t)) at time t can be found by integrating the velocity function, due to v(t) being the derivative of s(t). So, integrating v(t)=96-32t with respect to t gives s(t) = 96t - 16t^2 + 256, where 256 is the constant of integration corresponding to the initial height from the ground.
To find out how long will the ball take to reach the ground, we solve the function s(t)=0, yielding t = 6 seconds as the answer, as that's when the ball hits the ground. Now, for the maximum height reached by the ball, it's where the ball has a velocity of zero before it begins its descent, i.e. v(t)=0. Solving this gives t=3 seconds, substituting this back into the s(t) equation gives the maximum reached height as 400 feet.
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Two particles of masses m1m1m_1 and m2m2m_2 (m1
Answer:
the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂
Explanation:
The center of mass, the point of a system where all external forces can be applied, is defined by
[tex]x_{cm}[/tex] = 1 /M ∑ [tex]x_{i} m_{i}[/tex]
where M is the total mass of the system, x_{i} m_{i} are the position and mass of each item in the system,
let's apply this equation to our case
the total mass is
M = m₁ + m₂
for the calculations we must fix a reference system, we will place it in the second mass (m₂)
x_{cm}= 1/M (m₁ d + m₂ 0)
where d is the distance between the two masses in this case d = 10 m
x_{cm} = m₁ / (m₁ + m₂) d
x_{cm} = m₁ / (m₁ + m₂) 10
as the mass m₁ <m₂
Let us analyze the answer if the masses sides were equal the center of mass would be x_{cm} = 5 m, but since m₁ <m₂ the center of mass must be closer to m₂.
Therefore the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂