Answer:
Too many people are unaware or indifferent to that.” Fines can cost up to $500 per truancy, due within 30 days unless a judge gives an extension. For many students and families, it's another debt they can't pay. And if fines aren't paid, they can convert into an arrest warrant when a student turns 17.
Explanation:
A bungee jumper jumps from a tall bridge that is 60m above a flat, dry creek bed. She is carrying an audio generator that emits sound at frequency 1200 Hz. The unstretched length of the bungee cord is 26m, and the spring constant is selected so that she stops just before hitting the creek bed. At what point in her fall is the beat frequency, determined by listening to the audio generator and the reflection from the creek bed, a maximum
Answer:
Explanation:
The original frequency of sound f₀
The apparent frequency of sound fa
For apparent frequency the formula is
fa = [tex]f_0\times\frac{V+v}{V-v }[/tex] , v is velocity of jumper which increases as he goes down .
Beat frequency
= fa - f₀
= [tex]f_0\times(\frac{V+v}{V-v }-1)[/tex]
= [tex]f_0\times(\frac{2v}{V-v })[/tex]
since v is very small in comparison to V , velocity of sound , in the denominator , v can be neglected.
beat frequency = [tex]f_0\times(\frac{2v}{V })[/tex]
v , the velocity of jumper will go on increasing as long as net force on the jumper is positive or
mg > kx where x is extension in the cord and k is its force constant . Below this point kx or restoring force becomes more than weight of the jumper and then net force on the jumper directs upwards. At this point beat frequency becomes maximum.
A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average force of 1.50 x 10^3 N. The hockey stick is in contact with the ball for 1.20 ms, and the ball leaves the stick with a velocity of 8.00 m/s. Let the direction of the force be the + x direction. Find the following (note: be careful with the sign/direction of the values):_______. 1. The final momentum of the ball 2. The impulse on the ball 3. The initial velocity of the ball
Answer:u=-22 m/s
Explanation:
Given
mass of puck [tex]m=0.06\ kg[/tex]
Average force [tex]f_{avg}=1.5\times 10^3\ N[/tex]
time of contact [tex]t=1.2ms=1.2\times 10^{-3}\ s[/tex]
puck leaves with a velocity of [tex]v=8\ m/s[/tex]
We know impulse is [tex]F_{avg}\Delta t[/tex][tex]=\text{change in momentum}[/tex]
therefore
[tex]1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i[/tex]
[tex]P_i=0.06\times 8-1.8[/tex]
[tex]P_i=0.48-1.8=-1.32\ kg-m/s[/tex]
Final momentum [tex]P_f=m\times v_f[/tex]
[tex]P_f=0.06\times 8[/tex]
[tex]P_f=0.48\ kg-m/s[/tex]
Impulse on the ball [tex]=F_{avg}\Delta t[/tex]
Impulse[tex]=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s[/tex]
Initial velocity is given by
[tex]u=\frac{P_i}{m}=\frac{-1.32}{0.06}[/tex]
[tex]u=-22\ m/s[/tex]
i.e. initially ball is moving towards -x-axis
What measurement stays the same at any point in a parallel circuit?
A. current
B. voltage
C. resistivity
D. power
Answer:Voltage
Explanation:
Voltage stays the same at any point in a parallel circuit
To understand how to use the principle of superposition in conjunction with the Biot-Savart (or Ampere's) law. From the Biot-Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by Bwire=μ0I2πd , where μ0 (=4π×10−7T⋅m/A) is the permeability constant, I is the current in the wire, and d is the distance from the wire to the location at which the magnitude of the magnetic field is being calculated. The same result can be obtained from Ampere's law as well.
The principle of superposition can be used in conjunction with the Biot-Savart (or Ampere's) law to calculate the magnetic field due to a combination of current-carrying wires. Ampere's law is a more general law that relates the magnetic field to the total current passing through a closed loop.
Explanation:The principle of superposition can be used in conjunction with the Biot-Savart (or Ampere's) law to calculate the magnetic field due to a combination of current-carrying wires. The Biot-Savart law allows us to calculate the magnetic field at any point due to an element of current in a wire. By integrating this law over the entire length of the wires and applying the principle of superposition, we can determine the total magnetic field produced by multiple wires.
For example, if we have two parallel wires carrying currents I1 and I2, the total magnetic field at a point due to both wires can be found by summing the individual magnetic fields produced by each wire using the Biot-Savart law.
It is important to note that Ampere's law can also be used to determine the magnetic field produced by current-carrying wires. It is a more general law that relates the magnetic field to the total current passing through a closed loop. The results obtained from Ampere's law are consistent with the Biot-Savart law.
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This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force wil At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, what is the direction of the force on the particle due to the magnetic field?
Answer:
force is -y direction
Explanation:
The magnetic force is given by the equation
F = q v x B
where bold letters indicate vectors.
The direction of this force can be found by the right-hand rule, where the thumb points in the direction of speed, the other fingers extend in the direction of the magnetic field and the palm is in the direction of force for a positive charge
let's apply this to our case
the speed is in the direction + x the thumb
the magnetic field in the + z direction fingers extended
the palm is in the direction - y
force is -y direction
Final answer:
The force due to the magnetic field on a positively charged particle moving in the +x direction and perpendicular to the magnetic field will be in the +-y direction.
Explanation:
At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If q is positive, the direction of the force on the particle due to the magnetic field will be in the +-y direction.
The expression known as Froude's number is given as Upper F equals StartFraction v squared Over gl EndFraction . This number was used in shipbuilding and also in research on dinosaurs. In this example, l is given as well as Froude's number. Find the velocity, v (in meters per second) of a rhinoceros. It is known that gequals9.8 meters per second squared. Rhinoceros: lequals 1.2; Froude number equals 2.6
Answer:[tex]v=5.529\ m/s[/tex]
Explanation:
Given
number is given as [tex]=\frac{v^2}{gl}[/tex]
and [tex]number=2.6[/tex]
[tex]l=1.2\ m[/tex]
g=acceleration due to gravity[tex](9.8\ m/s^2)[/tex]
calculating for [tex]v[/tex]
[tex]v^2=gl(2.6)[/tex]
[tex]v=\sqrt{2.6\times 9.8\times 1.2}[/tex]
[tex]v=\sqrt{30.576}[/tex]
[tex]v=5.529\ m/s[/tex]
Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off?
The angular speed of the merry-go-round would increase when the boy jumps off due to the conservation of angular momentum. However, without specific values like the weights and distances involved, the precise speed cannot be calculated.
Explanation:The speed of the merry-go-round after the boy jumps off cannot be determined from the given information. However, we can use the principle of conservation of angular momentum to analyze the situation; this principle states that the initial angular momentum (while the boy was still on the merry-go-round) should be equal to the final angular momentum (after the boy jumps off). The angular momentum is the product of moment of inertia and angular speed. When the boy was sitting on the edge of the merry-go-round, he had a certain moment of inertia due to his mass and the distance from the center. Once he jumps off, the moment of inertia decreases, hence according to the conservation law, the angular speed must increase to keep the momentum constant.
However, the exact increase in angular speed could only be calculated if we knew the weight of the boy and of the merry-go-round, as well as the radius of the merry-go-round (to calculate moments of inertia), and the initial angular speed. Another simplification made in this situation is that the friction between the merry-go-round and its axle is negligible, which might not be true in real life scenarios.
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When the boy jumps off the merry-go-round, the total angular momentum of the system should conserve. As the boy's leaving reduces the moment of inertia, the angular speed of the merry-go-round must increase to preserve angular momentum. To calculate the exact increase, we would need specific values for the moment of inertia before and after the jump.
Explanation:The question asked is associated with angular velocity and moment of inertia in the context of a merry-go-round. As the boy jumps off the merry-go-round, the system's total angular momentum must be conserved, because there are no external torques acting on it.
When the boy was still on the merry-go-round, he was part of the system, contributing to the total angular momentum. But when he departs, he no longer contributes to the system's moment of inertia, effectively reducing it. As a result, the angular speed of the merry-go-round must increase to conserve angular momentum (since angular momentum = moment of inertia x angular speed).
Unfortunately, without specific values for the moment of inertia both before and after the boy jumps off, we can't calculate the exact increase in angular speed. However, the key concept is the conservation of angular momentum: when the moment of inertia decreases, the angular speed must increase to maintain constant total angular momentum.
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A plane circular loop of conducting wire of radius with possesses turn is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength Bis increased at a constant rate from to in a time interval of 10s. (a) What is the emf generated around the loop? (b) If the electrical resistance of the loop is , what current flows around the loop as the magnetic field is increased?
Complete question;
A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.
a) What is the emf generated around the loop?
b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?
Answer:
A) E.M.F generated around loop = 0.163 V
B)Current in loop; I = 0.011 A
Explanation:
A) We are given;
Initial magnetic field strength;B1 = 1T
Final magnetic field strength;B2 = 5T
Number of turns;N = 15 turns
Radius; r = 10cm = 0.1m
Angle;θ = 30°
Time interval; Δt = 10 s
Now, the formula for magnetic flux is: Φ = NABcosθ
Where;
N is number of turns
A is area = πr²
B is magnetic field strength
θ is angle
So, initial magnetic flux is;
Φ1 = NA(B1)cosθ
Plugging in the relevant values to obtain;
Φ1 = 15*(π*0.1²)(1)cos30
Φ1 = 0.4081 Wb
Similarly, final magnetic flux is;
Φ2 = NA(B2)cosθ
Plugging in the relevant values to obtain;
Φ2 = 15*(π*0.1²)(5)cos30
Φ2 = 2.0405 Wb
The time rate of change of the flux is;
dΦ_B/dt = (Φ2 - Φ1)/Δt
So, dΦ_B/dt = (2.0405 - 0.4081)/10
dΦ_B/dt = 0.163 Wb/s
Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.163 V
B) from Ohm's law, the current which flows around the loop in response to the emf is given as;
I = E/R
We are given R =15Ω
Thus; I = 0.1632/15
I = 0.011 A
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?
Answer:
The volume is decreasing at 160 cm³/min
Explanation:
Given;
Boyle's law, PV = C
where;
P is pressure of the gas
V is volume of the gas
C is constant
Differentiate this equation using product rule:
[tex]V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}[/tex]
Given;
[tex]\frac{dP}{dt}[/tex] (increasing pressure rate of the gas) = 40 kPa/min
V (volume of the gas) = 600 cm³
P (pressure of the gas) = 150 kPa
Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( [tex]\frac{dv}{dt}[/tex]);
(600 x 40) + (150 x [tex]\frac{dv}{dt}[/tex]) = 0
[tex]\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min[/tex]
Therefore, the volume is decreasing at 160 cm³/min
Final answer:
Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C. To find the rate at which the volume is decreasing at a given instant, we differentiate the equation PV = C with respect to time and substitute the given values.
Explanation:
Boyle's law states that the volume of a gas is inversely proportional to its pressure when temperature and the amount of gas are constant. Mathematically, this can be expressed as PV = C, where P is the pressure, V is the volume, and C is a constant. In this scenario, the pressure is increasing at a rate of 40 kPa/min. To find the rate at which the volume is decreasing at this instant, we differentiate the equation PV = C with respect to time. Applying the chain rule, we get:
d(PV)/dt = d(C)/dt
P(dV/dt) + V(dP/dt) = 0
dV/dt = - (V/P)(dP/dt)
We substitute the given values where the volume V = 600 cm3 and pressure P = 150 kPa:
dV/dt = - (600 cm3 / 150 kPa) (40 kPa/min)
dV/dt = -160 cm3/min
Therefore, the volume is decreasing at a rate of 160 cm3/min at this instant.
If the temperature of this balloon were to decrease suddenly, how would the balloon change?
A. Its mass would increase.
B. Its mass would decrease.
C. Its volume would increase.
D. Its volume would decrease.
Final answer:
The volume of a balloon decreases when its temperature is suddenly lowered, in accordance with Charles's Law, because the gas particles inside the balloon move less vigorously due to a decrease in their average kinetic energy.
Explanation:
When the temperature of a balloon decreases suddenly, according to Charles's Law, the volume of the gas inside the balloon also decreases, assuming that all other conditions, such as pressure and the amount of gas, remain constant. This occurs because a decrease in temperature leads to a decrease in the average kinetic energy of the gas particles, meaning they move less vigorously and occupy less space. Therefore, the correct answer to the question would be D. Its volume would decrease.
g A simple pendulum (consisting of a point mass suspended by a massless string) on the surface of the earth has a period of 1.00 s. On a distant planet, a pendulum of identical length but twice the mass is found to have the same period. What is true about the acceleration due to gravity on the distant planet? [Assume both pendulums operate free of any friction or air resistance force]
Answer:
Explanation:
The formula for time period of a pendulum is given as follows :
T = 2π[tex]\sqrt{\frac{l}{g} }[/tex]
l is length of pendulum and g is acceleration due to gravity .
So time period of pendulum is not dependent on the mass of the pendulum . If time period is same and length is also the same then acceleration due to gravity will also be the same . Hence the acceleration due to gravity at distant planet will be same as that on the earth.
A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600c. The pursuit ship is traveling at a speed of 0.800c relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?
Answer:
a) toward the pursuit
b) 0.384c
Explanation:
a) The velocity of the cruiser relative to the pursuit should be toward the pursuit.
b) To find the speed of the cruiser relative to the pursuit ship you use the following formula:
[tex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/tex]
v: velocity of the cruiser as seen by Tatooine
u: velocity of the pursuit ship as seen by Tatooine
c: speed of light
By replacing the values of the parameters you obtain:
[tex]u'=\frac{0.800c-0.600c}{1-\frac{(0.800)(0.600)c^2}{c^2}}=0.384c[/tex]
When accounting for relativistic velocity addition, the velocity of the cruiser relative to the pursuit ship should be directed towards the pursuit ship for it to be 'caught up' with. The cruiser's relative speed is about -0.429c from the perspective of the pursuit ship.
Explanation:From the perspective of the observer on Tatooine, the cruiser is moving away from the planet at a speed of 0.600c and the pursuit ship is moving in the same direction at a speed of 0.800c. To figure out these velocities in a relative sense, we must use the theory of relativity, specifically the concept of relativistic velocity addition. In classical physics, velocities simply add or subtract, but this isn't the case when dealing with speeds close to the speed of light.
(a) For the pursuit ship to catch up with the cruiser, the velocity of the cruiser relative to the pursuit ship should be directed towards the pursuit ship.
(b) When solving for the relative speed of the cruiser from the perspective of the pursuit ship, we use the formula for relativistic velocity addition: V'=(v-u)/(1-(uv/c^2)). In this case, v is the speed of the cruiser (0.600c), u is the speed of the pursuit spacecraft (0.800c), and c is the speed of light. Plugging in the values, you will find that the speed of the cruiser relative to the pursuit ship is approximately -0.429c (where the minus sign indicates the cruiser is moving towards the pursuit ship relative to the pursuit ship's frame of reference).
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A hydrogen atom can be in the 1S state, whose energy we'll call 0, the 2S state, or any of 3 2P states. The 2S and 2P states have energies of 10.2 eV. There are other states with higher energy but we'll ignore them for simplicity. The 2P states have distinctive optical properties, so we're interested in how many are present even when it's a small fraction of the total. 1) What fraction of the H is in 2P states at T=5900 K, a typical Sun surface temperature?
Final answer:
The fraction of hydrogen atoms in the 2P states at T=5900 K is calculated using the Boltzmann distribution. The energy of the 2S and 2P states is 10.2 eV, and their degeneracies are factored into the Boltzmann factor to determine the relative populations of these states.
Explanation:
The fraction of hydrogen atoms in the 2P states at a temperature of 5900 K can be calculated using the Boltzmann distribution. According to quantum mechanics, the energy levels of a hydrogen atom involve principal quantum numbers, with the ground state being 1s and higher energy excited states being designated by higher quantum numbers and corresponding letters (s, p, d, f, etc.) for their angular momentum quantum numbers. Since we are given that the energy of 2S and 2P states are the same at 10.2 eV and the ground state (1S) has an energy we'll call 0, we can use the Boltzmann factor to find the relative populations of these states.
To calculate the fraction of hydrogen atoms in the 2P state, use the following Boltzmann factor equation: fraction = (g2P / gTotal) * exp(-E2/kT), where g2P is the degeneracy of the 2P state, gTotal is the total degeneracy of all states considered, E2 is the energy of the 2P state, k is the Boltzmann constant, and T is the temperature. The degeneracy of the 2P state is 3 (since there are three 2P states) and the only other state considered here is the 2S state, which has degeneracy 1, making gTotal = 4. Plugging in the energy of 10.2 eV for E2 and converting it to joules (multiply by 1.602 x 10-19 J/eV), using k = 1.38 x 10-23 J/K, and T = 5900 K, we can calculate the fraction of hydrogen in the 2P state.
Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 12 cm that is placed in a spatially uniform magnetic field of magnitude 0.30 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)
Given Information:
Number of turns = N = 52
Diameter of coil = d = 12 cm = 0.12 m
Time = t = 0.10 seconds
Magnetic field = B = 0.30 T
Required Information:
Induced electric field = E = ?
Answer:
Induced electric field = E = 4.68 V/m
Explanation:
The Maxwell's third equation can be used to find out the induced electric field,
∫E.dl = -dΦ/dt
Where E is the induced electric field, dl is the circumference of the loop and dΦ/dt is the rate of change of magnetic flux and is given by
Φ = NABcos(θ)
Where N is the number of turns, A is the area of coil and B is the magnetic field and cos(θ) = 1
Φ = NAB
∫E.dl = -dΦ/dt
E(2πr) = -d(NAB)/dt
E =1/(2πr)*-d(NAB)/dt
E =NA/(2πr)*-dB/dt
Area is given by
A = πr²
E =Nπr²/(2πr)*-dB/dt
E =Nr/2*-dB/dt
The magnetic field reduce from 0.30 to zero in 0.10 seconds
E =Nr/2*-(0.30 - 0)/(0 - 0.10)
E =Nr/2*-(0.30)/(-0.10)
E = Nr/2*-(-3)
The radius r is given by
r = d/2 = 0.12/2 = 0.06 m
E = (52*0.06)/2*(3)
E = 1.56*3
E = 1.56*3
E = 4.68 V/m
Therefore, the induced electric field in the coil is 4.68 V/m
The hydronium ion concentration in an aqueous solution is [H+] = 5.3 x10-4 M. What is the hydroxide ion concentration? Is this solution acidic or basic?
A.) [OH-] = 1.9 x10-11 M, acidic
B.) [OH-] = 4.7 x10-6 M, basic
C.) [OH-] = 1.1 x10-9 M, acidic
D.) [OH-] = 3.2 x10-8 M, acidic
Answer:a
Explanation:
Given
[tex][H^+]=5.3\times 10^{-4}\ M[/tex]
and [tex]pH+pOH=14[/tex]
Also [tex]pH=-\log [H^+][/tex]
therefore [tex]pH=-\log (5.3\times 10^{-3})[/tex]
[tex]pH=-(-4-\log (5.3))[/tex]
[tex]pH=3.275[/tex]
Thus [tex]pOH=14-3.275[/tex]
[tex]pOH=10.275[/tex]
and [tex]pOH=-\log [OH^{-}][/tex]
[tex][OH]^{-1}=10^{-10.25}[/tex]
[tex][OH]^{-1}=1.88\times 10^{-11}\ M[/tex]
As the pH is less than 7 therefore solution is acidic
What can you infer about the prefix "thermo-" from the
word "thermoregulation?"
A
It refers to the brain
reser to the brain
It refers to temperature
It refers to feedback
It refers to nerves and hormones
Answer:
it refers to temperature I did test with the same question and got it correct.
Explanation:
You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether you should buy smaller sizes of your clothing because you will be thinner on your trip due to length contraction. You also plan to save money by reserving a smaller cabin to sleep in because you will be shorter when you lie down. Should you (a) buy smaller sizes of clothing, (b) reserve a smaller cabin, (c) do neither of these things, or (d) do both of these things?
Answer:
Do neither of these things ( c )
Explanation:
For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula
L = (l) [tex]\sqrt{1 -\frac{v^{2} }{c^{2} } }[/tex]
where l = Measured distance from object at rest, L = contracted measured in relation to the observer , v = speed of clock , c = speed of light
you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .
A spring hangs from the ceiling with an unstretched length of x0=0.45 m . A m1=7.9 kg block is hung from the spring, causing the spring to stretch to a length x1=0.74 m.Find the length x2of the spring when a m2=3.3 kg block is hung from the spring. For both cases, all vibrations of the spring are allowed to settle down before any measurements are made.
To find the new length of the spring when a different mass is hung from it, calculate the spring constant using the initial mass, and then use Hooke's Law to solve for the extension caused by the new mass. Add this extension to the original unstretched length of the spring to find the stretched length.
Explanation:The student's question involves finding the new length of a spring x2 when a block with mass m2 of 3.3 kg is hung from it. Given an initial unstretched length (x0) and a stretched length (x1) with block m1, we can solve for the spring constant k using Hooke's Law, which states F = k * x, where F is the force, k is the spring constant, and x is the extension from the natural length of the spring.
First, we calculate the extension caused by m1: extension_m1 = x1 - x0 = 0.74 m - 0.45 m = 0.29 m. The force exerted by m1 is F1 = m1 * g, where g is the acceleration due to gravity (9.81 m/s2). Calculating the force, we get F1 = 7.9 kg * 9.81 m/s2. Then, the spring constant k can be found using k = F1 / extension_m1.
With the value of k, we can find the extension caused by m2: extension_m2 = F2 / k, where F2 = m2 * g. Finally, the length x2 of the spring with m2 hung from it is x2 = x0 + extension_m2.
A very long, straight solenoid with a cross-sectional area of 2.16 cm^2 is wound with 85.7 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)= (( 0.175 A/s^2 )t^2). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
1.74x10⁻⁵ V
Explanation:
n = 85.7 turns/cm => 8570 turns/metre
The field inside the long solenoid is given by B = μ₀ni
B = 4πx10⁻⁷ x 8570 x 0.175t² = 1.884x10⁻³ t²
dB/dt = 3.78x10⁻³ t
Cross-sectional Area'A'= 2.16 cm²=> 2.16 x [tex]10^{-4}[/tex] m²
Now, rate of change of flux linkage '|Emf|' is given by:
|Emf| = d(NAB)/dt = NA dB/dt
|Emf| = 5 x 2.16 x [tex]10^{-4}[/tex] x 3.78x10⁻³ t
|Emf| = 4.0824x10⁻⁶ t
Considering time 't' at which the current = 3.2A , we have
3.2 = 0.175T²
T² = 3.2/0.175
T = 4.28 s
|emf| = 4.0824x10⁻⁶ t => 4.0824x10⁻⁶ x4.28
|emf|= 1.74x10⁻⁵ V
Therefore,the magnitude of the emf induced in the secondary winding is 1.74x10⁻⁵ V
In aircraft design, the pressure coefficient Cp is usually measured during wind tunnel testing of an aircraft component to predict structural loads, as well as lift and drag forces. Let Cp=(p−p[infinity])/ (0.5*rhoU[infinity]^2)where U[infinity] is the airspeed of the test at pressure p[infinity] and using a fluid of density rho. The pressure difference p ‐ p[infinity] is measured directly where p is the pressure located at a point on the surface of the test component. Estimate the uncertainty in Cp under the following conditions: Δp = p − p[infinity] = 1,000 N/m2, uΔp = 15 N/m2; rho = 1.20 kg/m3, urho = 0.01 kg/m3; U[infinity] = 50.0 m/s, u_U[infinity] = 0.21 m/s. Assume 95% confidence in stated values.
Answer:
Check the explanation
Explanation:
From given data, it can be noted that 95% of given confidently data, means 5% of data is uncertain. According to the question, we have to calculate uncertainty in Cp .
Kindly check the attached image below for the step by step explanation to the question above.
A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively. (a) Find the energy stored in the inductor when the current reaches its maximum value. J (b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch
Answer:
a) 20.81 J
b) 8.29 J
Explanation:
V = iR + L di/dt
where
i = a(1-e^-kt)
for large t
i = V/R
i = 24 / 9.4
i = 2.55 A
so
i = 2.55(1-e^-kt)
di/dt = 2.55 k e^-kt
24 = 24-24e^-kt + 6.4(2.55)k e^-kt
24 = 6.4(2.55) k
k = 24 / (6.4 * 2.55)
k = 24 / 16.32
k = 1.47 = R/L
so
i = 2.55(1-e^-(Rt/L))
current is maximum at great t
i max = 2.55 - 0
energy = (1/2) L i^2
E = (1/2)(6.4)2.55^2
E = 20.81 Joules
one time constant T = L/R and e^-(Rt/L) = 1/e = .368
i = 2.55 (1 - 0.368)
i = 2.55 * 0.632
i = 1.61 amps
energy = (1/2)(6.4)1.61^2
E = 8.29 Joules
The energy stored in the inductor when the current reaches its maximum value is 20.80 J. After one time constant, the energy stored is approximately 8.25 J.
we need to determine the energy stored in the inductor at two different points in time.
a) The energy stored in an inductor is given by the formula:
U = (1/2) × L × I²
Maximum Current
In an RL circuit, the maximum current is achieved when the circuit reaches a steady state. The maximum current is given by:
I_max = V/R, where V is the voltage and R is the resistance.
Given V = 24.0 V and R = 9.40 Ω, we have:
I_max = 24.0 V / 9.40 Ω = 2.55 A
Now, the energy stored in the inductor at the maximum current is:
U_max = (1/2) × L × I_max²
Given L = 6.40 H, we get:
U_max = (1/2) × 6.40 H × (2.55 A)² = 20.80 J
b) Energy After One Time Constant
The time constant for an RL circuit is defined as:
τ = L/R
Given L = 6.40 H and R = 9.40 Ω, we calculate:
τ = 6.40 H / 9.40 Ω ≈ 0.68 s
At one time constant, the current has reached approximately 63.2% of its maximum value:
I_τ ≈ 0.632 × I_max = 0.632 × 2.55 A ≈ 1.61 A
The energy stored in the inductor at this time is:
U_τ = (1/2) × L × I_τ² = (1/2) × 6.40 H × (1.61 A)² ≈ 8.25 J
complete question:
A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively.
(a) Find the energy stored in the inductor when the current reaches its maximum value. J
(b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch is closed. j
The three small spheres are welded to the light rigid frame which is rotating in a horizontal plane about a vertical axis through O with an angular velocity w= 20 rad/s. If a couple MO= 30 N.m is applied to the frame for 5 seconds, compute the new angular velocity.
The new angular velocity is computed using the kinematic expression w² = wo² + 2a0, where 'wo' is the original angular velocity 'a' is the angular acceleration and 'w' is the new angular velocity. The calculation is guided by principles of physics primarily involving angular momentum and acceleration.
Explanation:The calculation for the new angular velocity can be gleaned from the kinematic expression w² = wo² + 2a0. The formula described represents the angular motion of the frame. The angular velocity originally is 20 rad/s and a couple MO of 30 N.m is applied to the frame for 5 seconds. The development of this concept involves principals of angular momentum as well as angular acceleration.
The angular acceleration can be calculated using the relation a = nett
Bearing in mind the initial angular velocity, the applied couple MO, and the duration of its application, the new angular velocity can be computed using the given formula. Note that the laws of physics, specifically the law of conservation of angular momentum, play an essential part in this calculation.
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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 75.0-g piece of copper at 60.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C.
Determine the specific heat of the unknown sample.
Answer:
The specific heat of the unknown sample is 1822.14 J/kg.k
Explanation:
Given;
mass of aluminum calorimeter, [tex]M_c[/tex] = 100 g
mass of water, [tex]M_w[/tex] = 250 g
stabilizing temperature of water-calorimeter, ΔT = 10.0°C
mass of copper, [tex]M_c_u[/tex] = 75 g
initial temperature of copper, [tex]T_{cu}[/tex] = 60.0°C
mass of unknown sample, [tex]M_u[/tex] = 70.0 g
initial temperature of unknown sample, [tex]T_u[/tex] = 100°C.
The final temperature of the entire system, t = 20.0°C
Apply the principle of conservation of energy;
energy used to heat water and calorimeter is equal to energy released by copper and unknown sample.
[tex]Q = M_{cu}C_{cu} \delta T_{cu} + M__{u}C{_u} \delta T_u[/tex]
where;
Q is energy used to heat water and calorimeter
[tex]C_c_u[/tex] is the specific heat capacity of copper
[tex]C_u[/tex] is the specific heat capacity of unknown sample
Make [tex]C_u[/tex] subject of the formula;
[tex]C_u = \frac{Q-M_c_u C_c_u \delta T_c_u}{M_u \delta T_u} \\\\C_u = \frac{(C_wM_w +C_cM_c)\delta T-M_c_u C_c_u \delta T_c_u}{M_u \delta T _u} \\\\C_u = \frac{(4186*0.25 +900*0.1)10-0.075* 387 *40}{0.07* 80} \\\\C_u = \frac{11365 -1161}{5.6} \\\\C_u = 1822.14 \ J/kg.k[/tex]
Therefore, the specific heat of the unknown sample is 1822.14 J/kg.k
The specific heat of the unknown sample is;
c_u = 1823 J/Kg.k
We are given;
Mass of Aluminum calorimeter; m_c = 100 g = 0.1 kg
Mass of water; m_w = 250 g = 0.25 kg
Initial temperature of Calorimeter and water; T_c = T_w = 10°C = 283 K
Mass of Copper; m_cu = 75 g = 0.075 kg
Initial temperature of Copper; T_cu = 60°C = 333 K
Mass of unknown sample; m_u = 70 g = 0.07 kg
Initial temperature of unknown substance = 100°C = 373 K
Final temperature of system; T_f = 20°C = 293 K
Formula for quantity of heat is;
Q = mcΔt
where;
m is mass
c is specific heat capacity
Δt is change in temperature;
For the calorimeter and water , we have;
Q_cw = (m_w*c_w + m_c*c_c)Δt
Specific heat capacity of water is; c_w = 4186 J/Kg.K
Specific heat capacity of aluminium is 900 J/Kg.K
Thus;
Q_cw = ((0.25 * 4186) + (0.1 * 900))(293 - 283)
Q_cw = 11365 J
For the unknown sample and the piece of copper;
Q_cu,u = (m_cu*c_cu*Δt) + (m_u*c_u*Δt)
specific heat capacity of copper; c_cu = 385 J/Kg.K
Thus;
Q_cu,u = (0.075*385*(333 - 293)) + (0.07*c_u*(373 - 293))
Q_cu,u = 1155 + 5.6c_u
From conservation of energy principle;
Q_cw = Q_cu,u
Thus;
11365 = 1155 + 4.2c_u
c_u = (11365 - 1155)/5.6
c_u = 1823 J/Kg.k
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A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight WWW is sitting a distance LLL to the left of the pivot, at what distance L1L1L_1 must she place her second child of weight www on the right side of the pivot to balance the seesaw? Express your answer in terms of LLL, WWW, and www.
Answer:
L1L1L_1 = ( LLL × WWW)/ www.
Explanation:
From the question we are given that the children's weight are not equal, thus, there will be a need to balance the seesaw "so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground''.
Also, From the question, we are gven that the weight of the heavier child = WWW, the heavier child's is sitting a distance = LLL, the weight of the second child= www.
Therefore, in order to balance the seesaw, she will need to place her second child of weight www on the right side of the pivot to balance the seesaw at a distance of;
L1L1L_1 = ( LLL × WWW)/ www.
In a shipping company distribution center, an open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. You can ignore friction between the cart and the floor. A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal. The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds. The package comes to a stop in the cart after 4 seconds. What is:
Answer:
(1). 9.35 m/s.
(2).V2 = 3.22 m/s to the left.
Explanation:
So, we are given the following data or parameters or infomation in the question above as;
(1). "open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. "
(2). "A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal."
(3). "The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds."
(4). "The package comes to a stop in the cart after 4 seconds."
(a). So, in order to solve this question, we will be making use of the equations below;
(1). Gravitational potential energy = mass × acceleration due to gravity × height.
(2). Kinetic energy = 1/2 × mass × (velocity)^2.
(3). Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
So, starting from equation (1) above;
Gravitational potential energy = mass × acceleration due to gravity × height.
Gravitational potential energy = 15 × 9.8 × 4 = 588 J.
Kinetic energy = 1/2 × mass × (velocity)^2. = 1/2 × 15 × (3)^2.= 67.5 J.
kinetic energy (final) = 1/2 × 15 × v^2. = 7.5v^2.
Hence, Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
[potential energy (final) = 0].
67.5 + 588 = 7.5v^2.
v = 9.35 m/s
(b). -3 cos (27°) = -2.67 m/s.
15 × -2.67 + 50 × 5 =( 15 + 50) × v2.
=> 209.5 = 65v2.
=> V2 = 3.22 m/s to the left.
(1) The speed of the package before it lands on the cart is 9.35 m/s.
(2) The final speed of the cart is 3.22 m/s
Conservation of energy and momentum:The gravitational potential energy of a system is given by
PE = mgh
where m is the mass
g is the acceleration due to gravity
and h is the height.
The kinetic energy is given by:
KE = (1/2) mv²
According to the law of conservation of energy:
Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
The final potential energy will be zero since taking the cart as the reference as ground.
PE(initial) = 15 × 9.8 × 4 = 588 J
KE(initial) = (1/2) mv² = 1/2 × 15 × 3² = 67.5 J
KE(final) = 1/2 × 15 × v² = 7.5v²
So,
67.5 + 588 = 7.5v²
v = 9.35 m/s
(b) The package leave at an angle of 27° down the horizontal with velocity 3 m/s, so the horizontal speed is:
v ' = -3 cos (27°) = -2.67 m/s.
Negative sign, as the package is moving opposite to the cart
After the package lands on the cart, both move together, so from conservation of momentum:
15 × (-2.67) + 50 × 5 =( 15 + 50) × V
where V is the speed of cart and package together
209.5 = 65 V
V = 3.22 m/s to the left.
The question was incomplete. After searching it, it is likely that it was as given below:
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 27∘ from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what is (1) the speed of the package just before it lands in the cart and (2) the final speed of the cart?
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In deserts the temperatures can reach extreme values in deep of summer and winter. A steel bridge is being constructed and it is supposed to be 1234.567 m long at 233.15 K in the coldest possible winter. What will be the possible length of this bridge when the temperature rises to a possible extreme of +140.0°F in hottest part of summer? The average coefficient of linear expansion of the steel is 11.0123 × 10-6 K-1.
Answer:
The length of the bridge during the hottest part of summer is [tex]L_s = 1235.925 m[/tex]
Explanation:
From the question we are told that
The length of the steel bridge is [tex]L = 1234.567m[/tex]
The temperature for this length is [tex]T_1 = 233.15K[/tex]
The temperature at summer [tex]T_2 = + 140.0F = \frac{140 - 32}{180} *100 + 273= 333 K[/tex]
The coefficient of linear expansion is [tex]\alpha = 11.0123*10^{-6} K^{-1}[/tex]
Generally the change in length of the steel bridge is mathematically represented as
[tex]\Delta L = \alpha L \Delta T[/tex]
Substituting value
[tex]\Delta L = 11.0123*10^{-6} * 1234.567 (333-233.15)[/tex]
[tex]\Delta L = 1.3575 \ m[/tex]
The length of the bridge in summer is mathematically evaluated as
[tex]L_s = L + \Delta L[/tex]
Substituting values
[tex]L_s = 1234.567 + 1.3575[/tex]
[tex]L_s = 1235.925 m[/tex]
Compared to the current in the 10.-ohm resistance in the circuit shown at the right, the current in the 5.0-ohm resistance is
a. one-half as great
b. one-fourth as great
c. the same
d. twice as great
Answer:
The answer is d. Twice as great.
Explanation:
The German physicist and mathematician Georg Simon Ohm says in his basic law of electrical circuits or Ohm's law that the potential difference V that is applied at the ends of a conductor is proportional to the intensity I of the current that circulates and that the electrical resistance R is the ratio factor between I and V.
The equation would be I = V / R
If by any chance R is reduced because it is inversely proportional, the current would increase.
So if the current goes from 10 to half 5, its current would double.
how much work is done to move 3c of charge through a potential difference of 1.5v
Answer:4.5 joules
Explanation:
charge=3c
Potential difference=1.5v
Work=potential difference x charge
Work=1.5 x 3
Work=4.5
Work=4.5 joules
A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20°. Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it. Group of answer choices 0.27 N ⋅ m 0.41 N ⋅ m 0.38 N ⋅ m 0.77 N ⋅ m 0.14 N ⋅ m
The torque on a single circular loop in an external magnetic field can be calculated using the formula T = NIAB sin 0. We need to find the area of the loop using the given circumference and then substitute the values of the current, magnetic field strength, area, and angle into the formula to find the torque.
Explanation:To calculate the magnitude of the torque exerted on the circular loop by the magnetic forces, we need to use the formula for torque on a current-carrying loop in a uniform magnetic field, which is T = NIAB sin 0, where N represents the number of turns in the loop, I represents the current, A represents the area of the loop, B represents the magnetic field strength, and 0 represents the angle between the field and the plane of the loop.
In this case, since there is only a single loop, N is equal to 1. The current I is given as 4.0 A. The magnetic field strength B is given as 2.0 T. The angle 0 is 20°. The area of the loop A can be calculated from the circumference given as 80 cm or 0.8 m. Recall that the circumference of a circle is given by the formula 2πr. If the circumference C is given by 80 cm or 0.8 m, the radius r can be found by dividing the circumference by 2π. Once you've found the radius r, the area of the circle is πr².
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The correct answer is "0.14 N.m". The final torque value is 0.14 N·m on the loop.
To determine the torque exerted on the loop, we use the formula:
Torque (τ) = nIBA sin(θ)
First, find the area (A) of the loop. The circumference (C) of the loop is given by 80 cm, which is 0.80 m. From the circumference, we can find the radius (r):
C = 2πr -> r = C / 2π = 0.80 m / 2π ≈ 0.127 m
Now, calculate the area (A) of the loop:
A = πr² = π(0.127 m)² ≈ 0.0507 m²
Next, calculate the torque:
τ = nIBA sin(θ) = 1 × 4.0 A × 2.0 T × 0.0507 m² × sin(20°)
Using the value of sin(20°) ≈ 0.342:
τ ≈ 1 × 4.0 A × 2.0 T × 0.0507 m² × 0.342 ≈ 0.14 N · m
Therefore, the magnitude of the torque exerted on the loop is 0.14 N·m.
the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet
Answer:
The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.
The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.
Explanation:
Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.
Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.
Answer:
The force that is exerted when a shopping cart is pushed:
-Contact
The force that causes a metal ball to move toward a magnet:
-Noncontact