The amplitude of the pendulum at noon is approximately 0.78 m.
To determine the amplitude of the pendulum at noon, we need to consider the effect of damping on the oscillations. The amplitude of an oscillating system gradually decreases over time due to damping.
Given that the pendulum is started at 8:00 a.m. and the damping constant is 0.010 kg/s, we can use the formula for damped oscillations:
Amplitude = initial amplitude * e^(-damping constant * time)
The time elapsed from 8:00 a.m. to noon is 4 hours or 14400 seconds.
Using the given information, the initial amplitude of the pendulum is 1.2 m.
Amplitude at noon = 1.2 m * e^(-0.010 kg/s * 14400 s) = 0.78 m
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two locomotives approach each other on parallel tracks. each has a speed of 155 km/h with respect to the ground. if they are intially 8.5 km apart, how long will it be before they reach each other
The time the two locomotives will take to reach each other is 1.07 minutes.The speed of both the locomotives is 155 km/hr with respect to the ground.The distance between both the trains at initial point is 8.5 km
We have to calculate the time it will take for them to meet:Distance is equal to speed multiplied by time, so the distance between them (8.5 km) is equal to the relative speed between them multiplied by the time it takes them to meet.Let's calculate the relative speed:Relative speed = Speed of locomotive 1 + Speed of locomotive 2= 155 km/hr + 155 km/hr= 310 km/hrNow we can use the formula:Distance = Relative Speed × Time
We know the distance and the relative speed. Therefore,Time taken to meet = Distance / Relative speed= 8.5 km / 310 km/hr= 0.0274 hoursConvert hours to minutes:1 hour = 60 minutes0.0274 hours = 0.0274 × 60 minutes = 1.07 minutesSo, the time the two locomotives will take to reach each other is 1.07 minutes.
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Calculate the directivity of the helical antenna, HPBW and FNBW with the parameter: N = 7,F = 4GHz, C = 0.5A, s = 0.3 λ
The directivity of the helical antenna is 187,740, HPBW is 116 degrees and FNBW is 262 degrees.
To calculate the directivity of the helical antenna,
HPBW, and FNBW with the parameter:
N = 7, F = 4 GHz, C = 0.5A, s = 0.3 λ,
we need to use the following formulas:
Directivity = 15 * (N/D)^2HPBW = 58 * (λ/D)
FNBW = 131 * (λ/D)where,λ is the wavelength of the signal in metersD is the diameter of the helix in meters
We are given the following parameters:
N = 7F = 4 GHz
C = 0.5As = 0.3λ
λ = c/f = 3 x 10^8 / 4 x 10^9 = 0.075 m
D = C * λ = 0.5 * 0.075 = 0.0375 m
Directivity = 15 * (N/D)^2= 15 * (7/0.0375)^2= 15 * 12516= 187,740
HPBW = 58 * (λ/D)= 58 * (0.075/0.0375)= 116
FNBW = 131 * (λ/D)= 131 * (0.075/0.0375)= 262
Therefore, the directivity of the helical antenna is 187,740, HPBW is 116 degrees and FNBW is 262 degrees.
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a 2.50 kg blocl is pushed 2.20 m along a horizontal table by a constant force of 16.0 n directed at 25 degrees below the horizontal . if the coefficient of kinetic friction between the block ans the table is 0.213, what is the work done by the frictional force
To find the work done by the frictional force, we first need to calculate the net force acting on the block. Therefore, the work done by the frictional force is approximately 11.482 Joules.
The horizontal component of the applied force can be calculated as follows:
F[tex]_{horizontal }[/tex] = F[tex]_{applied}[/tex] × cos(25°)
F[tex]_{horizontal }[/tex] = 16.0 N × cos(25°)
F[tex]_{horizontal }[/tex] ≈ 14.495 N
Next, we need to calculate the force of kinetic friction:
F[tex]_{friction}[/tex] = coefficient of kinetic friction × normal force
The normal force can be calculated as the weight of the block:
Normal force = mass × gravitational acceleration
Normal force = 2.50 kg × 9.8 m/s²
Normal force ≈ 24.5 N
Now, we can calculate the force of kinetic friction:
F[tex]_{friction}[/tex] = 0.213 × 24.5 N
F[tex]_{friction}[/tex] ≈ 5.219 N
Since the block is being pushed horizontally, the work done by the frictional force is given by:
Work[tex]_{friction}[/tex] = F[tex]_{friction}[/tex] × displacement
Work[tex]_{friction}[/tex] = 5.219 N × 2.20 m
Work[tex]_{friction}[/tex] ≈ 11.482 J
Therefore, the work done by the frictional force is approximately 11.482 Joules.
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1)The dynamometer wattmeters are
a. More accurate on AC supply
b. More accurate on DC supply
C. None of these
d. Equally accurate on both AC and DC supply
2)In a dynamometer type wattmeter, the pressure coil connected across the load terminal is
a. Highly capacitive
b. Highly resistive
c. Highly inductive
d. Non inductive
1 ) d. Equally accurate on both AC and DC supply.
2 ) c. Highly inductive.
Answer:
Explanation:
1- The correct answer is:
d. Equally accurate on both AC and DC supply
2- The correct answer is:
d. Non inductive
hope it helps!! :)
Design a series resonant circuit to generate 300 kV high
voltage. (HİGH VOLTAGE ENGİNEERİNG)
The inductance of 0.4776 H is needed in the series resonant circuit to generate 300 kV high voltage.
High voltage required = 300kV
Impedance of series resonant circuit,Z = R + jXLC
For a series resonant circuit at resonance, the impedance becomes purely resistive. So, Xl = Xc or L = 1/ωC, where ω is the resonant frequency. Hence,Z = R
For a series resonant circuit with R = 150, the impedance is 150 Ω at resonance.
Since voltage across capacitor and inductor are equal to each other and are equal to the applied voltage,
Therefore, voltage across inductor = voltage across capacitor = Vc= VL= V/2
Total voltage across capacitor and inductor = Vc + VL= V/2 + V/2= V∴ V = 300kVFor a series resonant circuit,V = I × Z or I = V/ZI = V/R = 300 × 10³ /150= 2000 A
Therefore, inductance of the series resonant circuit is given by L = 1/ωC = 1/ (2πfC)Inductance L = V/(2πfIL) = 300 × 10³ / (2π × 50 × 2000) = 0.4776 H
Thus, an inductance of 0.4776 H is needed in the series resonant circuit to generate 300 kV high voltage.
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a rocket is used to place a synchronous satellite in orbit about the earth. what is the speed of the satellite in orbit? 4070 m/s 2070 m/s 3070 m/s
The speed of the satellite in orbit is given by 3070 m/s.
We have given that a rocket is used to place a synchronous satellite in orbit about the earth.
Let's derive the equation for the speed of the satellite in orbit about the earth:
We know that the acceleration due to gravity (g) at a height (h) above the earth's surface is given by,
g = GM / (R + h)²Here,M = Mass of the earthR = Radius of the earthG = Gravitational constanth = Height above the surface of the earth
Now, the force of gravity acting on the satellite is given by,
F = m gwhere m is the mass of the satellite
As the satellite is in circular motion, there is a centripetal force that is given by,
F = m v² / R
where v is the speed of the satellite in orbit and R is the distance of the satellite from the center of the earth.
The above two equations are equal to each other,m g = m v² / Rg = v² / Rv = √(g R)
Now, substituting the values of R and g, we getv = √(GM / (R + h))
Putting values,G = 6.67 × 10⁻¹¹ N m² / kg²M = 5.97 × 10²⁴ kgR = 6371 km = 6371000 mh = 0 (as the synchronous satellite orbits the earth at the same angular rate as the earth rotates)
On substituting the above values, we getv = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (6371000))v = 3070 m/s
Therefore, the speed of the satellite in orbit is 3070 m/s.
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do the two cars ever have the same velocity at one instant of time? if so, between which two frames? check all t
Yes, the two cars can have the same velocity at one instant of time. The cars have the same velocity at one instant of time between dots 1 and 2.
What is Velocity?The speed and direction of an object's motion are measured by its velocity. In kinematics, the area of classical mechanics that deals with the motion of bodies, velocity is a fundamental idea.
A physical vector quantity called velocity must have both a magnitude and a direction in order to be defined.
What is instant of time?Accordingly, a time interval that is not zero must be the sum of time instants that are all equal to zero. However, even if you add many zeros, one should remain zero.
Yes, at one point in time, the two cars can have the same speed. Between dots 1 and 2, the speed of the cars is the same at that precise moment.
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Complete question is,
Do the two cars ever have the same velocity at one instant of time? If so, between which two frames? Check all that apply. Cars have the same velocity at one instant of time between dots 1 and 2. Cars have the same velocity at one instant of time between dots 2 and 3. Cars have the same velocity at one instant of time between dots 3 and 4. Cars have the same velocity at one instant of time between dots 4 and 5. Cars have the same velocity at one instant of time between dots 5 and 6. Cars never have the same velocity at one instant of time.
Four solutes are added to a solvent. all solutes have the same mass and solubility.
When four solutes with the same mass and solubility are added to a solvent, they are likely to dissolve to the same extent, resulting in a homogeneous mixture. The explanation lies in the nature of solubility and the interactions between solutes and solvents.
When solutes are added to a solvent, their solubility determines the extent to which they dissolve. If all four solutes have the same solubility, it means they have similar chemical properties and can form favorable interactions with the solvent molecules. As a result, they will dissolve to the same extent, leading to a homogeneous solution where the solutes are evenly distributed throughout the solvent.
Solubility is influenced by factors such as temperature, pressure, and the nature of the solute and solvent. When solutes have the same mass and solubility, it suggests that their molecular structures and properties are similar. This similarity allows them to interact with the solvent in a comparable manner, resulting in equal dissolution. It is important to note that solubility can vary for different solutes if their properties or the conditions of the solvent change. However, in the given scenario, where solutes have the same mass and solubility, they are expected to dissolve equally in the solvent.
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chegg A force is applied to a block to move it up a 30 degree incline. The incline is frictionless. If F
If a force F is applied to a block to move it up a frictionless incline at a 30-degree angle, the force required to overcome the gravitational component acting on the block is given by [tex]F_cos(30)[/tex], where F is the applied force.
When a block is placed on an inclined plane, the force of gravity can be divided into two components: one parallel to the incline and one perpendicular to it. The force parallel to the incline, also known as the gravitational component, is given by [tex]F_g = mgsin(30)[/tex], where m is the mass of the block and g is the acceleration due to gravity.
To move the block up the incline, an external force F must be applied in the opposite direction of the gravitational component. Since the incline is frictionless, the force required to overcome the gravitational component is equal to the applied force F. However, since the applied force is not acting directly against gravity but at an angle of 30 degrees, only the component of the applied force parallel to the incline contributes to overcoming gravity. This component is given by [tex]F_cos(30)[/tex].
Therefore, the force required to move the block up the frictionless incline is equal to [tex]F_cos(30)[/tex], where F is the applied force.
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2. Show that the D-T fusion reaction releases 17.6 MeV of energy. 3. In the D-T fusion reaction, the kinetic energies of 2H and H are small, compared with typical nuclear binding energies. (Why?) Find the kinetic energy of the emit- ted neutron.
The D-T fusion reaction releases 17.6 MeV of energy. This is so because the fusion reaction of deuterium and tritium produces a helium nucleus, a neutron, and energy. The D-T fusion reaction can be written as follows: 2H + 3H → 4He + n + 17.6 MeV. The energy released is in the form of kinetic energy of the helium nucleus and the neutron. The energy released is due to the difference in the mass of the initial particles and the mass of the products.Explanation:In the D-T fusion reaction,
the kinetic energies of 2H and H are small compared with typical nuclear binding energies. This is because the kinetic energies of 2H and H are not large enough to overcome the electrostatic repulsion between the positively charged nuclei. The energy required to bring the positively charged nuclei together is the Coulomb barrier. For the D-T reaction, the Coulomb barrier is about 0.1 MeV.
However, when the nuclei are brought together at very high temperatures and pressures, they can overcome the Coulomb barrier, and the fusion reaction occurs.The kinetic energy of the emitted neutron can be found using the law of conservation of energy. The energy released in the reaction is shared between the helium nucleus and the neutron. The helium nucleus carries most of the energy, and the neutron carries the rest. The kinetic energy of the emitted neutron can be calculated as follows:Kinetic energy of neutron = Energy released - Kinetic energy of helium nucleus- 17.6 MeV - 3.5 MeV (approximate kinetic energy of helium nucleus)= 14.1 MeVTherefore, the kinetic energy of the emitted neutron is 14.1 MeV.
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A body with a mass of m=Xkg starts sliding on an inclined plane and when it reaches the flat collides with a larger body with a mass of M=2m and move as a single body. After a while the resultant body hits a spring that has a stiffness coefficient of K=10Y N/m. Given that there is no friction during the movement and no damping during the compression of the spring calculate: 1. The velocity of the first mass just before the collision 2. The velocity of the resultant body 3. The maximum compression of the spring 4. The maximum height that the resultant body is to climb up the inclined plane after the decompression of the spring
The maximum height climbed by the resultant body is (3/2) times the initial height of the inclined plane. To solve the problem, we'll apply the principles of conservation of momentum and conservation of mechanical energy.
Velocity of the first mass just before the collision: Before the collision, the first mass has a mass m and an initial velocity v. Since there is no friction, the only force acting on it is due to gravity. We can calculate its velocity using the equation of motion: mgh = (1/2)mv^2 where h is the vertical height of the inclined plane. Since it starts from rest, we have: gh = (1/2)v^2 v = √(2gh) Velocity of the resultant body: After the collision, the two bodies stick together and move as a single body. The mass of the resultant body is M + m = 3m. Since there is no external force acting on the system, the momentum is conserved. Therefore: (mv) + (Mv') = (3m)V where v' is the velocity of the resultant body. Since the first mass is moving in the opposite direction of the resultant body, its velocity is negative. Rearranging the equation: v' = (mv) / (3m + M) v' = v / (3 + 2) = v / 5 Maximum compression of the spring: When the resultant body hits the spring, the energy is conserved. The initial kinetic energy of the system is given by: (1/2)(3m)V^2 This energy is stored as potential energy in the compressed spring (1/2)kx^2 where k is the stiffness coefficient of the spring and x is the maximum compression of the spring. Equating the two energies: (1/2)(3m)V^2 = (1/2)kx^2 x^2 = (3mV^2) / k x = √((3mV^2) / k) Maximum height climbed by the resultant body: After the decompression of the spring, the resultant body starts to move up the inclined plane. The mechanical energy is conserved, so the potential energy at the maximum height is equal to the initial potential energy stored in the compressed spring: mgh' = (1/2)kx^2 where h' is the maximum height. Rearranging the equation: h' = (kx^2) / (2mg) Substituting the expression for x^2 from step 3: h' = (k / (2mg)) * ((3mV^2) / k) h' = (3mV^2) / (2mg) Therefore, the maximum height climbed by the resultant body is (3/2) times the initial height of the inclined plane.
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a 5.00 kg object has a moment of inertia of 1.20 kg m2. what torque is needed to give the object an angular acceleration of 2.0 rad/s2?
The amount of torque needed to give the object an angular acceleration of 2.0 rad/s² is 2.40 N m.
To calculate the torque needed to give an object an angular acceleration, you can use the following formula:
Torque (τ) = Moment of Inertia (I) × Angular Acceleration (α)
In this case, the moment of inertia (I) is given as 1.20 kg m², and the angular acceleration (α) is given as 2.0 rad/s². We can substitute these values into the formula to find the torque:
τ = 1.20 kg m² × 2.0 rad/s²
Calculating this expression:
τ = 2.40 N m
Therefore, the torque needed to give the 5.00 kg object an angular acceleration of 2.0 rad/s² is 2.40 N m.
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An object starts from rest to 20 m/s in 40 s with a constant acceleration. What is its acceleration in m/s^2
An object starts from rest to 20 m/s in 40 s with a constant acceleration.. The acceleration of the object is 0.5 m/s^2.
To find the acceleration of the object, we can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the object starts from rest (u = 0 m/s) and reaches a final velocity of 20 m/s (v = 20 m/s) in 40 seconds (t = 40 s), we can substitute these values into the equation and solve for acceleration. 20 = 0 + a * 40
Simplifying the equation, we have: 20 = 40a Dividing both sides of the equation by 40, we get: a = 0.5 m/s^2
Therefore, the acceleration of the object is 0.5 m/s^2. This means that the object's velocity increases by 0.5 m/s every second, leading to a final velocity of 20 m/s after 40 seconds of constant acceleration.
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The latent heat of vaporization for water at room temperature is 2430 J/g . Consider one particular molecule at the surface of a glass of liquid water, moving upward with sufficiently high speed that it will be the next molecule to join the vapor.(d) Why are you not burned by water evaporating from a vessel at room temperature?
Evaporation occurs at room temperature because individual water molecules can gain enough energy to overcome the attractive forces between them and escape into the air. However, you are not burned by water evaporating from a vessel at room temperature because the energy required for evaporation is taken from the surrounding environment, which includes the glass and the surrounding air.
When a water molecule at the surface of a glass of liquid water gains enough energy, it can break free from the liquid phase and enter the gas phase, becoming vapor. This process is called evaporation. However, for a molecule to gain sufficient energy, it must absorb heat from its surroundings. In this case, the heat energy needed for evaporation is taken from the glass, the surrounding air, and potentially your skin if it comes into contact with the evaporating water.
As the water molecules gain energy and evaporate, they cool down the surrounding environment. This cooling effect is the reason why evaporating water feels cold. The energy absorbed from the environment is used to break the intermolecular bonds within the liquid and convert the water molecules into vapor.
Therefore, while the process of evaporation requires energy, it is the surrounding environment that provides this energy. As a result, you are not burned by water evaporating from a vessel at room temperature because the necessary heat is taken from the environment rather than being released onto your skin.
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An angle-modulated signal is given by s(t) = 20 cos [2740(10°)t +5 sin(274000t)] a. If this is a PM signal with k, = 10, what is the message signal? P b. Plot message signal and PM signal using MATLAB c. If this is a FM signal with k, = 4000 Hz/V. What is the message signal? d. Plot message signal and FM signal using MATLAB
In the given expression s(t) = 20 cos [2740(10°)t + 5 sin(274000t)], the term "5 sin(274000t)" represents the message signal. It is a sinusoidal signal with a frequency of 274000 Hz and an amplitude of 5 units.
In the context of angle modulation, the message signal refers to the original baseband signal that carries the information or data to be transmitted. It is also known as the modulating signal. The message signal can be any continuous waveform that represents the desired information, such as an audio signal in the case of broadcasting or a data signal in the case of digital communication.
a. To find the message signal for the PM (Phase Modulation) signal, we need to extract the term that represents the variation in phase. In this case, the message signal can be obtained from the term "5 sin(274000t)".
b. To plot the message signal and PM signal using MATLAB, you can use the following code:
t = 0:0.0001:0.02; % Time vector
message_signal = 5*sin(274000*t); % Message signal
pm_signal = 20*cos(2740*10*pi*t + message_signal); % PM signal
figure;
subplot(2,1,1);
plot(t, message_signal);
xlabel('Time (s)');
ylabel('Amplitude');
title('Message Signal');
subplot(2,1,2);
plot(t, pm_signal);
xlabel('Time (s)');
ylabel('Amplitude');
title('PM Signal');
c. For the FM (Frequency Modulation) signal with k_f = 4000 Hz/V, the message signal can be obtained from the term "5 sin(274000t)".
d. To plot the message signal and FM signal using MATLAB, you can use the following code:
t = 0:0.0001:0.02; % Time vector
message_signal = 5*sin(274000*t); % Message signal
fm_signal = cos(2740*10*pi*t + 4000*integrate(message_signal)); % FM signal
figure;
subplot(2,1,1);
plot(t, message_signal);
xlabel('Time (s)');
ylabel('Amplitude');
title('Message Signal');
subplot(2,1,2);
plot(t, fm_signal);
xlabel('Time (s)');
ylabel('Amplitude');
title('FM Signal');
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a pet lamb grows rapidly, with its mass proportional to the cube of its length. when the lamb's length changes by 14.4%, its mass increases by 15.0 kg. find the lamb's mass at the end of this process.
The lamb's mass at the end of the process is approximately 74.32 kg.
Let's assume the initial length of the lamb is L and its corresponding mass is M. According to the given information, the mass of the lamb is proportional to the cube of its length. Therefore, we can write the equation as:
M = kL^3
where k is the constant of proportionality.
When the lamb's length changes by 14.4%, its new length becomes L + 0.144L = 1.144L. As a result, its new mass becomes M + 15.0 kg.
Substituting the new length and mass values into the equation, we get:
M + 15.0 = k(1.144L)^3
Now, let's divide this equation by the original equation to eliminate the constant k:
(M + 15.0)/M = [(1.144L)^3]/(L^3)
Simplifying the equation, we have:
1 + 15.0/M = 1.144^3
Now, we can solve for M:
15.0/M = 1.144^3 - 1
M = 15.0/(1.144^3 - 1)
Calculating this expression, the lamb's mass at the end of the process is approximately 74.32 kg.
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two projectiles are launched at 100 m/s, the angle of elevation for the first being 30° and for the second 60°. which of the following statements is false?
Given data: Two projectiles are launched at 100 m/s, the angle of elevation for the first being 30° and for the second 60°.To find which of the following statements is false. Solution: Firstly, let's write the formulas of motion along the x-axis and y-axis separately along with the given data of each projectile and calculate the horizontal and vertical components of their velocity and acceleration of each projectile along the x-axis and y-axis as follows:
For projectile 1:Initial velocity, u = 100 m/s Angle of projection, θ = 30°Horizontal component of initial velocity, u cos θ = 100 × cos 30° = 100 × √3 / 2 = 50√3 m/s Vertical component of initial velocity, u sin θ = 100 × sin 30° = 100 × 1 / 2 = 50 m/s Acceleration due to gravity, a = -9.8 m/s² (downward)Here, the negative sign indicates that the direction of the acceleration due to gravity is opposite to that of the vertical velocity along the upward direction as per the chosen coordinate axis.
For projectile 2:Initial velocity, u = 100 m/s Angle of projection, θ = 60°Horizontal component of initial velocity, u cos θ = 100 × cos 60° = 100 × 1 / 2 = 50 m/s Vertical component of initial velocity, u sin θ = 100 × sin 60° = 100 × √3 / 2 = 50√3 m/s Acceleration due to gravity, a = -9.8 m/s² (downward)Here, the negative sign indicates that the direction of the acceleration due to gravity is opposite to that of the vertical velocity along the upward direction as per the chosen coordinate axis.
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a roller coaster is moving at 25 m/s at the bottom of a hill. three seconds later it reaches the top of the hill moving at 10m/s.
The acceleration of the roller coaster is -5 m/s².
The roller coaster's initial velocity at the bottom of the hill is 25 m/s and it takes three seconds to reach the top of the hill, where its velocity is 10 m/s.
To find the acceleration of the roller coaster, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Using the given values, we have:
acceleration = (10 m/s - 25 m/s) / 3 s
= (-15 m/s) / 3 s
= -5 m/s²
Therefore, the acceleration of the roller coaster is -5 m/s².
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A pipe is 0.90 m long and is open at one end but closed at the other end. If it resonates with a tone whose wavelength is 0.72 m, what is the wavelength of the next higher overtone in this pipe?
Answer
0.40 m
0.51 m
0.36 m
0.45 m
0.58 m
If the pipe resonates with a tone whose wavelength is 0.72 m, the wavelength of the next higher overtone in this pipe is 0.36 m.
Given data:
Length of the pipe = L = 0.90 m
Length of the wave resonates with the tone = λ₁ = 0.72 m
We know that, in a closed-open pipe the frequency of the sound wave that resonates in the tube is given by:
f = nv/4L ---(1)
where v = velocity of sound
n = harmonic number that the pipe resonates within = 1 for fundamental frequency and so on
To calculate the wavelength of the next higher overtone, we can use the below formula:
λ₂ = λ₁/n ---(2)
where n is the harmonic number of the required overtone.
Calculation:
We know that the frequency of sound in the tube, f₁ is given by:
f₁ = nv/4Lf₁ = v/4L * nf₁ = (343/4*0.9) * 1f₁ = 95.3 Hz.
The speed of sound in air is given by v = 343 m/s. So, from (2), we haveλ₂ = λ₁/2λ₂ = 0.72/2λ₂ = 0.36 m. Therefore, the wavelength of the next higher overtone in this pipe is 0.36 m.
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Both power supplies in the circuit network shown below has 0.5 12 internal resistance. E = 18 V (0.512) R2 2.5 2 R a b 6.0 22 R3 1.5 12 Ez = 45 V (0.52) a) Find the electric currents passing through the resistors R1, R2, and R3 b) What is the total energy supplied by the two batteries during a period of 60 s? c) What is the total energy disscipated through Ri, R2, and R3 during this time? d) What is the total energy dissipated in the batteries during this time? Hint: Find from the lecture, how the internal resistance of a battery affects a circuit. Draw a new circuit including this effect, before attempting to find the currents.
Previous question
In the circuit network shown, there are two power supplies with internal resistances of 0.5 Ω. The voltage of one supply is 18 V and the other is 45 V. We need to find the electric currents passing through resistors R1, R2, and R3, as well as calculate the total energy supplied by the batteries, the total energy dissipated through the resistors, and the total energy dissipated in the batteries over a period of 60 seconds.
To find the electric currents passing through resistors R1, R2, and R3, we need to analyze the circuit taking into account the internal resistances of the power supplies. By applying Kirchhoff's voltage law and Ohm's law, we can calculate the currents.
To calculate the total energy supplied by the batteries over a period of 60 seconds, we need to multiply the total power supplied by the time. The power supplied by each battery is given by the product of its voltage and the current passing through it.
The total energy dissipated through resistors R1, R2, and R3 can be calculated by multiplying the power dissipated by each resistor by the time.
The total energy dissipated in the batteries can be calculated by subtracting the total energy dissipated through the resistors from the total energy supplied by the batteries.
To take into account the effect of the internal resistance of the batteries, we need to draw a new circuit that includes this resistance. This will affect the voltage drops across the resistors and the currents flowing through the circuit.
By solving the circuit equations and performing the necessary calculations, we can find the values of the electric currents, the total energy supplied by the batteries, the total energy dissipated through the resistors, and the total energy dissipated in the batteries over the given time period of 60 seconds.
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Which of the following lines exists in a p-V diagram for water? O all of the mentioned O saturated liquid lines O saturated vapor line saturated solid line
In a p-V (pressure-volume) diagram for water, the line that exists is the saturated liquid line. This line represents the boundary between the liquid and vapor phases of water at equilibrium. It indicates the conditions at which water exists as a saturated liquid.
The saturated vapor line, on the other hand, represents the boundary between the liquid and vapor phases of water when it exists as a saturated vapor. The saturated solid line is not applicable in a p-V diagram for water, as water does not have a stable solid phase at standard atmospheric conditions.
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A 28 kg suitcase is being pulled with constant speed by a handle that is at an angle of 25 ∘ above the horizontal.
Part A
If the normal force exerted on the suitcase is 180 N , what is the force F applied to the handle? F=____________N
The exact force applied to the handle (F) is approximately 320.36 N.
To find the force applied to the handle (F), we need to analyze the forces acting on the suitcase.
Given information:
Mass of the suitcase (m) = 28 kg
Angle above the horizontal (θ) = 25°
Normal force (N) = 180 N
We can break down the forces acting on the suitcase into horizontal and vertical components. The force applied to the handle (F) will have both horizontal and vertical components.
The vertical component of the force (F_y) will counteract the gravitational force acting on the suitcase and is given by:
F_y = mg,
where m is the mass of the suitcase (28 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
F_y = (28 kg)(9.8 m/s²) = 274.4 N.
Since the suitcase is being pulled with a constant speed, the net force in the horizontal direction is zero. The horizontal component of the force (F_x) is responsible for canceling out the frictional force.
Now, we can find the horizontal component of the force (F_x) using the angle (θ) and the normal force (N):
F_x = N × cos(θ).
F_x = 180 N × cos(25°) ≈ 162.85 N.
Therefore, the force applied to the handle (F) is the vector sum of the horizontal and vertical components:
F = √(F_x² + F_y²).
F = √(162.85² + 274.4²) ≈ 320.36 N.
So, the force F applied to the handle is approximately 320.36 N.
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. a resident of the above mentioned building was peering out of her window at the time the water balloon was dropped. if it took 0.15 s for the water balloon to travel across the 3.45 m long window, what floor does the resident live on?
The resident lives on the floor numbered as follows:Floor = height above ground level / height of each floor= (0.109575 / h) / h= 0.109575 / h2
Given that a resident of the above mentioned building was peering out of her window at the time the water balloon was dropped and it took 0.15 s for the water balloon to travel across the 3.45 m long window. We are required to find what floor does the resident live on?We can make use of the formula:$$d = v_0 t + \frac{1}{2} at^2$$Where, d is distance traveledv0 is the initial velocityt is timea is accelerationWe know that the balloon is moving horizontally and that there is no air resistance acting on it. Thus, its horizontal velocity is constant and given by the equation v0 = d/t.As there is no vertical force acting on the balloon except for gravity (ignoring air resistance), its vertical acceleration is equal to acceleration due to gravity, i.e., a = -9.81 m/s2Now, the time taken by the water balloon to travel across the window is 0.15 s.Thus, the horizontal velocity is given by:v0 = d/t = 3.45/0.15 = 23 m/sNow, the vertical velocity is given by the formula:v = v0 + atInitially, the balloon is at rest, thus, v0 = 0.v = at = -9.81 × 0.15 = -1.4715 m/sThe negative sign indicates that the balloon is moving downwards.Hence, we can use the formula to find the distance traveled by the balloon from the window of the resident:$$d = v_0 t + \frac{1}{2} at^2$$Substituting the known values, we get:d = 23 × 0.15 + 0.5 × (-9.81) × (0.15)2 = 0.254 mThe distance traveled by the balloon from the window of the resident is 0.254 m.Now, let's suppose the height of each floor of the building is h m, and the resident lives at a height of hF above the ground level.The time taken by the water balloon to fall from a height of hF is given by the formula:t = sqrt(2hF / g)Where, g is the acceleration due to gravity, which is equal to 9.81 m/s2.Substituting the known values, we get:t = sqrt(2hF / g) = sqrt(2hF / 9.81)The time taken by the water balloon to travel across the 3.45 m long window is the same as the time taken by it to fall from a height of hF, i.e.,0.15 = sqrt(2hF / 9.81)Squaring both sides of the equation, we get:0.0225 = 2hF / 9.81hF = 0.0225 × 9.81 / 2Hence, the resident lives at a height of 0.109575 m above the ground level, which is the same as 0.109575 / h meters above the ground level, where h is the height of each floor.
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a track star in the broad jump goes into the jump at 12 m/s and launches himself at 20° above the horizontal. how long is he in the air before returning to earth? (g = 9.8 m/s2)
The track star is in the air for approximately 1.9 seconds before returning to the ground.
To determine the time the track star spends in the air, we can use the kinematic equation for vertical motion:
y = v0y * t + (1/2) * g * t^2
Where:
y is the vertical displacement (0 since he returns to the same height),
v0y is the initial vertical velocity (v0 * sinθ),
t is the time in the air, and
g is the acceleration due to gravity (9.8 m/s^2).
Since the track star launches himself at an angle of 20° above the horizontal, we can break down the initial velocity into its vertical and horizontal components. The vertical component is given by v0y = v0 * sinθ, where v0 is the initial velocity (12 m/s) and θ is the launch angle (20°).
Plugging in the values, we have:
0 = (12 * sin20°) * t + (1/2) * 9.8 * t^2
Simplifying the equation:
4.8t - 4.9t^2 = 0
Factoring out t:
t(4.8 - 4.9t) = 0
This equation gives us two possible solutions: t = 0 (which is the starting point) and t = 4.8/4.9. Since we're interested in the time spent in the air, we discard the t = 0 solution.
Therefore, the track star is in the air for approximately 4.8/4.9 = 0.98 seconds, or rounded to one decimal place, 1.9 seconds.
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A three-phase overhead transmission line is supported on 4-disc suspensio n insulators. The voltages across the second and third discs are 13.2KV an d 18KV respectively. Calculate the line voltage and string efficiency
The line voltage across the insulators is 22.88 kV and the string efficiency is 10.81%
Given data:
The voltages across the second and third discs are 13.2 kV and 18 kV respectively.
Formula:
Line voltage = 3V1 = √3V2
V1 = 13.2 kV
V2 = 18 kV
To calculate the line voltage across the insulators, let's use the given formula.
Line voltage = 3V1 = √3V2
= √3 x 13.2 kV
= 22.88 kV
Therefore, the line voltage across the insulators is 22.88 kV.
The formula for string efficiency is:
String efficiency = (Voltage across all insulators) / (Total voltage of the line) × 100
The total voltage of the line is V1 + V2 + V3 = 13.2 kV + 13.2 kV + 18 kV = 44.4 kV
The voltage across all insulators is V3 - V2 = 18 kV - 13.2 kV = 4.8 kV
Now, let's calculate the string efficiency:
String efficiency = (Voltage across all insulators) / (Total voltage of the line) × 100
= (4.8 kV / 44.4 kV) × 100
= 10.81%
Therefore, the string efficiency is 10.81%.
Hence, the line voltage across the insulators is 22.88 kV and the string efficiency is 10.81%.
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A horse exerts 500 n of force on a heavy wagon. the wagon pulls back on the horse with an equal force. the wagon still accelerates because:__________
The wagon still accelerates despite exerting an equal force on the horse because of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Newton's third law of motion states that when an object exerts a force on another object, the second object exerts an equal and opposite force back on the first object. In this scenario, the horse exerts a force of 500 N on the wagon, and as a reaction, the wagon pulls back on the horse with an equal force of 500 N.
Despite the equal and opposite forces, the wagon still accelerates. This is because the forces act on different objects. The 500 N force exerted by the horse on the wagon causes the wagon to accelerate forward.
At the same time, the 500 N force exerted by the wagon on the horse does not affect the horse's motion significantly, as the horse is generally much larger and more massive than the wagon.
As a result, the net force acting on the wagon is not zero, leading to its acceleration. The wagon experiences a forward force from the horse and a negligible backward force from the wagon itself, allowing it to accelerate in the direction of the applied force.
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A reconnaissance plane flies 416km away from its base at 414 m/s then flies back to its base at 621 m/s . whats its average speed? answer in units of m/s
A reconnaissance plane flies 416 km away from its base at 414 m/sThen flies back to its base at 621 m/s. We need to calculate the average speed of the plane.
The average speed of the reconnaissance plane can be calculated by using the formula for the average speed which is;(average speed) = (total distance traveled) / (total time taken)In this problem, the plane travels 416 km away from its base and then flies back to its base, which means that the total distance traveled is;Total distance traveled = (416 km) + (416 km)Total distance traveled = (832 km) = (832000 m)We can calculate the total time taken by the reconnaissance plane by using the formula for time which is;time = (distance traveled) / (speed)The time taken to fly 416 km at 414 m/s can be calculated as;time = (416 km) / (414 m/s)time = (1008240 m) / (414 m/s)time = 2436.23 sThe time taken to fly back to its base at 621 m/s can be calculated as;time = (416 km) / (621 m/s)time = (1008240 m) / (621 m/s)time = 1623.96 sThe total time taken by the reconnaissance plane can be calculated as;total time taken = (2436.23 s) + (1623.96 s)total time taken = (4058.19 s)Now, we can substitute the calculated values in the formula for the average speed to calculate the average speed of the plane as follows;average speed = (832000 m) / (4058.19 s)average speed = 204.81 m/sTherefore, the average speed of the reconnaissance plane is 204.81 m/s (rounded to two decimal places).
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a projectile is fired from an airless world with a speed =escape. what is the total energy (tot =ke−/) of the projectile?
The total energy of the projectile is zero: tot = KE - PE = KE - mgh = 0.
The total energy (tot =ke−/) of the projectile when fired from an airless world with a speed equal to escape velocity is zero.
This is because the projectile has just enough kinetic energy to escape the gravitational field of the airless world, so there is no potential energy associated with its position relative to the world's gravitational field.
The escape velocity of a planet is given by:ve2=2GM/R
where ve is the escape velocity of the planet, G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
In this case, since the projectile is fired with a speed equal to escape velocity, its kinetic energy is equal to its potential energy: KE = PE = mgh
where m is the mass of the projectile, g is the acceleration due to gravity, and h is the height above the surface of the planet.
Since there is no gravitational field to do work on the projectile, its kinetic energy is conserved, and there is no potential energy associated with its position.
Therefore, the total energy of the projectile is zero: tot = KE - PE = KE - mgh = 0.
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Calculate the resistivity of rainwater with a conductivity of
100 µS/cm
The task is to calculate the resistivity of rainwater with a given conductivity of 100 µS/cm.
Resistivity is the inverse of conductivity and is a measure of a material's resistance to the flow of electric current. To calculate the resistivity of rainwater with a conductivity of 100 µS/cm, we can use the formula: Resistivity = 1 / Conductivity.
In this case, the given conductivity of rainwater is 100 µS/cm. By substituting this value into the formula, we can calculate the resistivity of rainwater. The resistivity will be expressed in units of ohm-cm (Ω·cm).
Resistivity is a fundamental property that characterizes the electrical behavior of a material. It represents the intrinsic resistance of the material to the flow of electric current. In the context of rainwater, the conductivity value indicates its ability to conduct electricity. By calculating the resistivity from the given conductivity, we can determine the inverse of this conductivity, which gives us a measure of the rainwater's resistance to electric current flow.
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justify your answer about which car if either completes one trip around the track in less tame quuantitatively with appropriate equations
To determine which car completes one trip around the track in less time, we can analyze their respective velocities and the track distance.
The car with the higher average velocity will complete the track in less time. Let's denote the velocity of Car A as VA and the velocity of Car B as VB. The track distance is given as d.
We can use the equation:
Time = Distance / Velocity
For Car A:
Time_A = d / VA
For Car B:
Time_B = d / VB
To compare the times quantitatively, we need more information about the velocities of the cars.
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