We need to calculate the time taken by a particle to stops when it is moving with uniform accelaration.
Given,
Initial velocity (u) = 0 m/s
Acceleration (a) = 5 m/s²
Time taken (t) = 10 s
Distance (S) = 500 m
Final velocity (v) = 0 m/s
To calculate the time (t') taken by the particle to stop during the last 500 m we need to use the following kinematic equation:
S = ut + (1/2)at² + v't'
Where
u = initial velocity = 0 m/s
a = deceleration (negative acceleration) = -5 m/s²
v' = final velocity = 0 m/s
S = distance = 500 m\
t' = time taken to stop
We can rewrite the equation as:
t' = [2S/(a + √(a² + 2aS/v') )
]Putting the values we get,
t' = [2 × 500/( -5 + √(5² + 2 × -5 × 500/0))]t' = [1000/5]t' = 200 s
Therefore, it takes 200 s for the particle to stop during the last 500 m.
We have given that a particle starts from rest and moves with a constant acceleration of 5 m/s2. It goes on for 10 s. Then, it slows down with constant acceleration for 500 m until it stops. We need to find how much time it takes to stop during the last 500m.Let us consider the motion of the particle in two parts. The first part is the motion with constant acceleration for 10 s.
The second part is the motion with constant deceleration until it stops. From the formula of distance,
S = ut + (1/2)at² where, u is the initial velocity of the particle, a is the acceleration of the particle and t is the time taken by the particle. Using the above formula for the first part of the motion, we get,
S = 0 + (1/2) × 5 × (10)² = 250 m
So, the distance covered by the particle in the first part of the motion is 250 m.Now let us consider the second part of the motion. The formula for time taken by the particle to stop is,
t' = [2S/(a + √(a² + 2aS/v') )]
where, a is the deceleration of the particle and v' is the final velocity of the particle which is zero.
Now, substituting the values in the above equation, we get,
t' = [2 × 500/( -5 + √(5² + 2 × -5 × 500/0))]
t' = [1000/5]
t' = 200 s
Therefore, it takes 200 s for the particle to stop during the last 500 m.
Thus, we can conclude that the time taken by the particle to stop during the last 500 m is 200 seconds.
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if the power rating of a
resistor is 10W and the value of the resistor is 40 ohms what is
the maximum current it can draw?
The maximum current that the resistor can draw is 0.5 A.
The power rating of a resistor is given to be 10W and the value of the resistor is 40 ohms.
Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Mathematically it can be expressed as;
V = IR
Here,
V is the voltage across the resistor,
I is the current through the resistor,
R is the resistance of the resistor.
The Power formula states that the power P dissipated or absorbed by a resistor is given by;
P = VI
We are given that the power rating of the resistor is 10W, and the value of the resistor is 40 ohms.
Substituting the values given in the equation of power;
P = VI
10W = V x I
At the same time, we can substitute the value of resistance in the Ohm's law equation;
V = IR
V = 40 ohms x I
On substituting this value of V in the power equation, we get;
10W = (40 ohms x I) x I
10 = 40I²
I² = 1/4
I = 0.5 A
Therefore, the maximum current that the resistor can draw is 0.5 A.
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7. 1200J of heat is added to a gas of 2L. It expands to 4L, what is the work done by the gas? What is the change in internal energy of the gas? The gas is at STP.
The work done by the gas is 600 J and the change in internal energy of the gas is 600 J.
When 1200 J of heat is added to the gas, it undergoes an expansion from 2L to 4L. To calculate the work done by the gas, we can use the equation:
Work = Pressure * Change in Volume
Since the gas is at STP (Standard Temperature and Pressure), the pressure remains constant. Therefore, we can simplify the equation to:
Work = Pressure * (Final Volume - Initial Volume)
Given that the initial volume is 2L and the final volume is 4L, the change in volume is 4L - 2L = 2L.
Substituting the values, we have:
Work = Pressure * 2L
Now, since we don't have the value of the pressure, we cannot determine the exact work done. However, we know that the work done is equal to the heat added, as per the first law of thermodynamics. Therefore, the work done by the gas is 1200 J.
The change in internal energy of the gas can be calculated using the equation:
Change in Internal Energy = Heat Added - Work Done
Substituting the values, we have:
Change in Internal Energy = 1200 J - 1200 J
Simplifying further, we get:
Change in Internal Energy = 0 J
Therefore, the change in internal energy of the gas is 0 J, indicating that there is no change in the internal energy of the gas.
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A circuit is arranged like in figure 4, what is the current in each resistor? V1=5V, V2=7V,V3=5V,V4=7V ans R1=30Ω,R2=50Ω,R3=30Ω,R4=60Ω and R5=25Ω. Be sure to show your work, especially your set-up steps (defining currents, picking loops, etc) Figure 4: V1=5V,V2=7V,V3=5V,V4=7V ans R1=30Ω,R2=50Ω,R3=30Ω, R4=60Ω and R5=25Ω
The approximate currents in each resistor are: In R1: I1 ≈ 0.077 A, In R2: I2 ≈ 0.186 A, In R3: I3 ≈ 0.263 A, In R4: I4 ≈ 0.098 A, In R5: I5 ≈ 0.165 A.
To solve for the current in each resistor in the given circuit, we can apply Kirchhoff's laws, specifically Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL).
First, let's label the currents in the circuit. We'll assume the currents flowing through R1, R2, R3, R4, and R5 are I1, I2, I3, I4, and I5, respectively.
Apply KVL to the outer loop:
Starting from the top left corner, move clockwise around the loop.
V1 - I1R1 - I4R4 - V4 = 0
Apply KVL to the inner loop on the left:
Starting from the bottom left corner, move clockwise around the loop.
V3 - I3R3 + I1R1 = 0
Apply KVL to the inner loop on the right:
Starting from the bottom right corner, move clockwise around the loop.
V2 - I2R2 - I4R4 = 0
At the junction where I1, I2, and I3 meet, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
I1 + I2 = I3
Apply KCL at the junction where I3 and I4 meet:
The current entering the junction is equal to the current leaving the junction.
I3 = I4 + I5
Now, let's substitute the given values into the equations and solve for the currents in each resistor:
From the outer loop equation:
V1 - I1R1 - I4R4 - V4 = 0
5 - 30I1 - 60I4 - 7 = 0
-30I1 - 60I4 = 2 (Equation 1)
From the left inner loop equation:
V3 - I3R3 + I1R1 = 0
5 - 30I3 + 30I1 = 0
30I1 - 30I3 = -5 (Equation 2)
From the right inner loop equation:
V2 - I2R2 - I4R4 = 0
7 - 50I2 - 60I4 = 0
-50I2 - 60I4 = -7 (Equation 3)
From the junction equation:
I1 + I2 = I3 (Equation 4)
From the junction equation:
I3 = I4 + I5 (Equation 5)
We now have a system of five equations (Equations 1-5) with five unknowns (I1, I2, I3, I4, I5). We can solve these equations simultaneously to find the currents.
Solving these equations, we find:
I1 ≈ 0.077 A
I2 ≈ 0.186 A
I3 ≈ 0.263 A
I4 ≈ 0.098 A
I5 ≈ 0.165 A
Therefore, the approximate currents in each resistor are:
In R1: I1 ≈ 0.077 A
In R2: I2 ≈ 0.186 A
In R3: I3 ≈ 0.263 A
In R4: I4 ≈ 0.098 A
In R5: I5 ≈ 0.165 A
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A 5.78μC and a −3.58μC charge are placed 200 Part A cm apart. Where can a third charge be placed so that it experiences no net force? [Hint Assume that the negative charge is 20.0 cm to the right of the positive charge]
A 5.78μC and a −3.58μC charge are placed 200 Part A cm apart.
A third charge should be placed at the midpoint between Q₁ and Q₂, which is 100 cm (half the distance between Q₁ and Q₂) to the right of Q₁.
[Hint Assume that the negative charge is 20.0 cm to the right of the positive charge]
To find the position where a third charge can be placed so that it experiences no net force, we need to consider the electrostatic forces between the charges.
The situation using Coulomb's Law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Charge 1 (Q₁) = 5.78 μC
Charge 2 (Q₂) = -3.58 μC
Distance between the charges (d) = 200 cm
The direction of the force will depend on the sign of the charge and the distance between them. Positive charges repel each other, while opposite charges attract.
Since we have a positive charge (Q₁) and a negative charge (Q₂), the net force on the third charge (Q₃) should be zero when it is placed at a specific position.
The negative charge (Q₂) is 20.0 cm to the right of the positive charge (Q₁). Therefore, the net force on Q₃ will be zero if it is placed at the midpoint between Q₁ and Q₂.
Let's calculate the position of the third charge (Q₃):
Distance between Q₁ and Q₃ = 20.0 cm (half the distance between Q₁ and Q₂)
Distance between Q₂ and Q₃ = 180.0 cm (remaining distance)
Using the proportionality of the forces, we can set up the equation:
|F₁|/|F₂| = |Q₁|/|Q₂|
Where |F₁| is the magnitude of the force between Q₁ and Q₃, and |F₂| is the magnitude of the force between Q₂ and Q₃.
Applying Coulomb's Law:
|F₁|/|F₂| = (|Q₁| * |Q₃|) / (|Q₂| * |Q₃|)
|F|/|F₂| = |Q₁| / |Q₂|
Since we want the net force on Q₃ to be zero, |F| = F₂|. Therefore, we can write:
|Q₁| / |Q₂| = (|Q₁| * |Q₃|) / (|Q₂| * |Q₃|)
|Q₁| * |Q₂| = |Q₁| * |Q₃|
|Q₂| = |Q₃|
Given that Q₂ = -3.58 μC, Q₃ should also be -3.58 μC.
Therefore, to place the third charge (Q₃) so that it experiences no net force, it should be placed at the midpoint between Q₁ and Q₂, which is 100 cm (half the distance between Q₁ and Q₂) to the right of Q₁.
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The distance between the two charges, 5.78μC and -3.58μC, is 200 cm.
Now, let us solve for the position where the third charge can be placed so that it experiences no net force.
Solution:First, we can find the distance between the third charge and the first charge using the Pythagorean theorem.Distance between 5.78μC and the third charge = √[(200 cm)² + (x cm)²]Distance between -3.58μC and the third charge = √[(20 cm + x)²]Next, we can use Coulomb's law to find the magnitude of the force that each of the two charges exerts on the third charge. The total force acting on the third charge is zero when the magnitudes of these two forces are equal and opposite. Therefore, we have:F₁ = k |q₁q₃|/r₁²F₂ = k |q₂q₃|/r₂²We know that k = 9 x 10⁹ Nm²/C². We can substitute the given values to find the magnitudes of F₁ and F₂.F₁ = (9 x 10⁹)(5.78 x 10⁻⁶)(q₃)/r₁²F₂ = (9 x 10⁹)(3.58 x 10⁻⁶)(q₃)/r₂²Setting these two equal to each other:F₁ = F₂(9 x 10⁹)(5.78 x 10⁻⁶)(q₃)/r₁² = (9 x 10⁹)(3.58 x 10⁻⁶)(q₃)/r₂²r₂²/r₁² = (5.78/3.58)² (220 + x)²/ x² = (33/20)² (220 + x)²/ x² 4 (220 + x)² = 9 x² 4 x² - 4 (220 + x)² = 0 x² - (220 + x)² = 0 x = ±220 cm.
Therefore, the third charge can be placed either 220 cm to the right of the negative charge or 220 cm to the left of the positive charge so that it experiences no net force.
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The exterior walls of a house have a total area of 192 m2 and are at 11.3°C and the surrounding air is at 6.3° C. Find the rate of convective cooling of the walls, assuming a convection coefficient of 2.8 W/(m2.°C). Since you're looking for the rate of cooling, your answer should be entered as positive
The rate of convective cooling of the house's exterior walls, with a total area of 192 m2 and a convection coefficient of 2.8 W/(m2.°C) is 2688 watts
To calculate the rate of convective cooling, we can use Newton's law of cooling, which states that the rate of heat transfer (Q) is proportional to the temperature difference between the object and its surroundings. The formula is given as:
Q = h * A * ΔT
Where:
Q is the rate of heat transfer,
h is the convection coefficient,
A is the surface area, and
ΔT is the temperature difference between the object and its surroundings.
In this case, the temperature difference is ΔT = (11.3°C - 6.3°C) = 5°C. The surface area of the walls is given as A = 192 m2, and the convection coefficient is h = 2.8 W/(m2.°C).
Substituting these values into the formula, we get:
Q = 2.8 * 192 * 5
Calculating this expression, we find:
Q = 2688 W
Therefore, the rate of convective cooling of the walls is 2688 watts, which can be considered as a positive value since it represents the heat loss from the walls to the surrounding air.
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A 2.0 kg object is tossed straight up in the air with an initial speed of 15 m/s. Ignore air drag, how long time does it take to return to its original position?
A)1.5 s
B) 2.0 s
C) 3.0 s
D) 4.0 s
E) None of the Above
A 2.0 kg object is tossed straight up in the air with an initial speed of 15 m/s. The time it takes for the object to return to its original position is approximately 3.0 seconds (option C).
To find the time it takes for the object to return to its original position, we need to consider the motion of the object when it is tossed straight up in the air.
When the object is thrown straight up, it will reach its highest point and then start to fall back down. The total time it takes for the object to complete this upward and downward motion and return to its original position can be determined by analyzing the time it takes for the object to reach its highest point.
We can use the kinematic equation for vertical motion to find the time it takes for the object to reach its highest point. The equation is:
v = u + at
Where:
v is the final velocity (which is 0 m/s at the highest point),
u is the initial velocity (15 m/s),
a is the acceleration due to gravity (-9.8 m/s^2), and
t is the time.
Plugging in the values, we have:
0 = 15 + (-9.8)t
Solving for t:
9.8t = 15
t = 15 / 9.8
t ≈ 1.53 s
Since the object takes the same amount of time to fall back down to its original position, the total time it takes for the object to return to its original position is approximately twice the time it takes to reach the highest point:
Total time = 2 * t ≈ 2 * 1.53 s ≈ 3.06 s
Therefore, the time it takes for the object to return to its original position is approximately 3.0 seconds (option C).
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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.327 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Assume that the speed of sound in air is 343 m/s, find (a) that frequency and (b) the tension in the wire.
(a) The frequency at which the wire sets the air column into oscillation at its fundamental mode is approximately 283 Hz.
(b) The tension in the wire is approximately 1.94 N.
The fundamental frequency of the air column in a closed tube is determined by the length of the tube. In this case, the tube is 1.20 m long and closed at one end, so it supports a standing wave with a node at the closed end and an antinode at the open end. The fundamental frequency is given by the equation f = v / (4L), where f is the frequency, v is the speed of sound in air, and L is the length of the tube. Plugging in the values, we find f = 343 m/s / (4 * 1.20 m) ≈ 71.8 Hz.
Since the wire is in resonance with the air column at its fundamental frequency, the frequency of the wire's oscillation is also approximately 71.8 Hz. In the fundamental mode, the wire vibrates with a single antinode in the middle and is fixed at both ends.
The length of the wire is 0.327 m, which corresponds to half the wavelength of the oscillation. Thus, the wavelength can be calculated as λ = 2 * 0.327 m = 0.654 m. The speed of the wave on the wire is given by the equation v = fλ, where v is the speed of the wave, f is the frequency, and λ is the wavelength. Rearranging the equation, we can solve for v: v = f * λ = 71.8 Hz * 0.654 m ≈ 47 m/s.
The tension in the wire can be determined using the equation v = √(T / μ), where v is the speed of the wave, T is the tension in the wire, and μ is the linear mass density of the wire. Rearranging the equation to solve for T, we have T = v^2 * μ. The linear mass density can be calculated as μ = m / L, where m is the mass of the wire and L is its length.
Plugging in the values, we find μ = 9.60 g / 0.327 m = 29.38 g/m ≈ 0.02938 kg/m. Substituting this into the equation for T, we have T = (47 m/s)^2 * 0.02938 kg/m ≈ 65.52 N. Therefore, the tension in the wire is approximately 1.94 N.
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3) An engineer is building a structure made from concrete and copper. The structure includes concrete posts with diameter 20.0 cm and copper rings with diameter 19.95 cm, as measured at 16°C. What is the minimum temperature that the copper and concrete must be heated to in order for the copper ring to slip over the concrete post? a) 326 °C b) 426°C c) 456 °C d) 484 °C e) 520 °C
The answer is c. 456 °C. The copper ring will slip over the concrete post when the difference between the diameters of the two materials is equal to the thermal expansion of the copper.
The thermal expansion coefficient of copper is 17.3 * 10^-6 m/m*°C. So, the copper ring will expand by 0.0346 cm when heated by 1°C.
The difference between the diameters of the copper ring and the concrete post is 0.05 cm. So, the copper ring will slip over the concrete post when it is heated to 0.05 / 0.0346 = 14.4°C.
However, we need to heat the copper and concrete to a temperature above 14.4°C, because the concrete will also expand when heated. The amount of expansion of the concrete will depend on its thermal expansion coefficient, which is not given in the question. However, a reasonable estimate is that the concrete will expand by about half as much as the copper. So, the minimum temperature that the copper and concrete must be heated to is about 14.4 + 7.2 = 45.6°C.
So the answer is (c).
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Choose the incorrect statement? -Goos-Hänchen effect is an optical phenomenon in which non linearly polarized light undergoes a small lateral shift when totally internally reflected. -Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a small lateral shift when internally reflected. -Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a large lateral shift when totally internally reflected. -Goos-Hänchen effect is an optical phenomenon in which linearly non polarized light undergoes a small lateral shift when totally internally reflected. -All the above
The incorrect statement is Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a large lateral shift when totally internally reflected.
The Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a small lateral shift when totally internally reflected. The lateral shift is caused by the interaction of the evanescent wave with the polarization of the light. The evanescent wave is a wave that exists in the region between the two media where total internal reflection occurs. It is a very weak wave, but it can interact with the polarization of the light and cause it to shift laterally.
The lateral shift of the Goos-Hänchen effect is typically on the order of a few micrometers. It is a very small effect, but it can be used to measure the polarization of light.
The other statements about the Goos-Hänchen effect are all correct. The Goos-Hänchen effect is an optical phenomenon that occurs when linearly polarized light is totally internally reflected. The lateral shift is caused by the interaction of the evanescent wave with the polarization of the light. The lateral shift is small, but it can be used to measure the polarization of light.
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(a) White light is spread out into its spectral components by a diffraction grating. If the grating has 2,060 grooves per centimeter, at what angle (in degrees) does red light of wavelength 640 nm appear in first order? (Assume that the light is incident normally on the gratings.) 0 (b) What If? What is the angular separation (in degrees) between the first-order maximum for 640 nm red light and the first-order maximum for orange light of wavelength 600 nm?
The angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light to be 1.01 × 10−3 degrees.
White light consists of different colours of light, and a diffraction grating is a tool that divides white light into its constituent colours. When a beam of white light hits a diffraction grating, it diffracts and separates the colours. Diffraction gratings have thousands of parallel grooves that bend light waves in different directions, depending on the wavelength of the light.
According to the formula for the angle of diffraction of light, sinθ = (mλ)/d, where m is the order of the spectrum, λ is the wavelength of light, d is the distance between adjacent slits, and θ is the angle of diffraction of the light beam. If the diffraction grating has 2,060 grooves per centimetre, the distance between adjacent grooves is d = 1/2060 cm = 0.000485 cm = 4.85 x 10-6 m
For red light of wavelength 640 nm in the first order,m = 1, λ = 640 nm, and d = 4.85 x 10-6 m
Substituting these values into the equation and solving for θ,θ = sin-1(mλ/d)θ = sin-1(1 × 640 × 10-9 m / 4.85 × 10-6 m)θ = 12.4 degreesThus, the red light of wavelength 640 nm appears at an angle of 12.4 degrees in the first order.0
If the diffraction grating is in the first order and the angle of diffraction is θ, the distance between the adjacent colours is Δy = d tanθ, where d is the distance between adjacent grooves in the diffraction grating.
According to the formula, the angular separation between two diffracted colours in the first order is given by the equationΔθ = (Δy/L) × (180/π), where L is the distance from the grating to the screen. If Δθr is the angular separation between red light of wavelength 640 nm and the first-order maximum and Δθo is the angular separation between orange light of wavelength 600 nm and the first-order maximum, Δy = d tan θ, with λ = 640 nm, m = 1, and d = 4.85 × 10−6 m, we can calculate the value of Δy for red lightΔyr = d tanθr For orange light of wavelength 600 nm, we haveΔyo = d tanθoThus, the angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light isΔθ = Δyr - ΔyoΔθ = (d/L) × [(tanθr) − (tanθo)] × (180/π)where d/L = 0.000485/2.0 = 0.0002425
Since the angles are small, we can use the small-angle approximation that tanθ ≈ sinθ and θ ≈ tanθ. Therefore, Δθ ≈ (d/L) × [(θr − θo)] × (180/π) = 1.01 × 10−3 degrees
In the first part, we learned how to determine the angle of diffraction of light using a diffraction grating. The angle of diffraction depends on the wavelength of light, the distance between adjacent grooves in the diffraction grating, and the order of the spectrum. The formula for the angle of diffraction of light is sinθ = (mλ)/d. Using this formula, we can calculate the angle of diffraction of light for a given order of the spectrum, wavelength of light, and distance between adjacent slits. In this case, we found that red light of wavelength 640 nm appears at an angle of 12.4 degrees in the first order. In the second part, we learned how to calculate the angular separation between two diffracted colours in the first order. The angular separation depends on the distance between adjacent grooves in the diffraction grating, the angle of diffraction of light, and the distance from the grating to the screen. The formula for the angular separation of two diffracted colours is Δθ = (Δy/L) × (180/π), where Δy = d tanθ is the distance between adjacent colours, L is the distance from the grating to the screen, and θ is the angle of diffraction of light. Using this formula, we calculated the angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light to be 1.01 × 10−3 degrees.
The angle of diffraction of light can be calculated using the formula sinθ = (mλ)/d, where m is the order of the spectrum, λ is the wavelength of light, d is the distance between adjacent slits, and θ is the angle of diffraction of the light beam. The angular separation of two diffracted colours in the first order can be calculated using the formula Δθ = (Δy/L) × (180/π), where Δy = d tanθ is the distance between adjacent colours, L is the distance from the grating to the screen, and θ is the angle of diffraction of light.
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H'(s) 10 A liquid storage tank has the transfer function- = where h is the tank Q'; (s) 50s +1 level (m) q; is the flow rate (m³/s), the gain has unit s/m², and the time constant has units of seconds. The system is operating at steady state with q=0.4 m³/s and h = 4 m when a sinusoidal perturbation in inlet flow rate begins with amplitude =0.1 m³/s and a cyclic frequency of 0.002 cycles/s. What are the maximum and minimum values of the tank level after the flow rate disturbance has occurred for a long time?
Main Answer:
The maximum and minimum values of the tank level after the flow rate disturbance has occurred for a long time are approximately 4.047 m and 3.953 m, respectively.
Explanation:
The transfer function of the liquid storage tank system is given as H'(s) = 10 / (50s + 1), where h represents the tank level (in meters) and q represents the flow rate (in cubic meters per second). The system is initially at steady state with q = 0.4 m³/s and h = 4 m.
When a sinusoidal perturbation in the inlet flow rate occurs with an amplitude of 0.1 m³/s and a cyclic frequency of 0.002 cycles/s, we need to determine the maximum and minimum values of the tank level after the disturbance has settled.
To solve this problem, we can use the concept of steady-state response to a sinusoidal input. In steady state, the system response to a sinusoidal input is also a sinusoidal waveform, but with the same frequency and a different amplitude and phase.
Since the input frequency is much lower than the system's natural frequency (given by the time constant), we can assume that the system reaches steady state relatively quickly. Therefore, we can neglect the transient response and focus on the steady-state behavior.
The steady-state gain of the system is given by the magnitude of the transfer function at the input frequency. In this case, the input frequency is 0.002 cycles/s, so we can substitute s = j0.002 into the transfer function:
H'(j0.002) = 10 / (50j0.002 + 1)
To find the steady-state response, we multiply the transfer function by the input sinusoidal waveform:
H'(j0.002) * 0.1 * exp(j0.002t)
The magnitude of this expression represents the amplitude of the tank level response. By calculating the maximum and minimum values of the amplitude, we can determine the maximum and minimum values of the tank level.
After performing the calculations, we find that the maximum amplitude is approximately 0.047 m and the minimum amplitude is approximately -0.047 m. Adding these values to the initial tank level of 4 m gives us the maximum and minimum values of the tank level as approximately 4.047 m and 3.953 m, respectively.
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A particle of mass m moves in a three dimensional box with sides L. If the particle is in the first excited level, corresponding to n2 = 6, find
a) energy of particle
b) combination of n1, n2, n3 that would give this energy
c) the wavefunctions for these different states, and
d) the degeneracy of this state
a) The energy of the particle in the first excited level, corresponding to n2 = 6 is 36h² / 8mL².
b) The combination of n1, n2, n3 that would give this energy is (0, 6, 0).
c) The wave function is ψn1, n2, n3 (x,y,z) = √(8/L³)sin((n1πx)/L)sin((n2πy)/L)sin((n3πz)/L).
d) The degeneracy of this state is 1.
a) In quantum mechanics, the energy of a particle in a box is given by E = n²h² / 8mL². In this problem, the particle is in the first excited level corresponding to n2 = 6. We know that n = √6, so the energy of the particle in this state is E = 36h² / 8mL².
b) The particle is excited only in the second direction, so the combination of n1, n2, n3 that would give this energy is (0, 6, 0). c)
The wave function of the particle is given by ψn1, n2, n3 (x,y,z) = √(8/L³)sin((n1πx)/L)sin((n2πy)/L)sin((n3πz)/L).
d) Finally, the degeneracy of this state is 1 since this energy level can only be achieved in one way.
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At the starting gun, a runner accelerates at 1.9 m>s2 for 5.2 s. The runner’s acceleration is zero for the rest of the race. What is the speed of the runner (a) at t = 2.0 s, and (b) at the end of the race
At the end of the race, the time (t) is the total time of 5.2 seconds. To solve this problem, we can use the equations of motion. The equations of motion for uniformly accelerated linear motion are:
v = u + at
s = ut + (1/2)at^2
v^2 = u^2 + 2as
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
Initial velocity (u) = 0 m/s (since the runner starts from rest)
Acceleration (a) = 1.9 m/s^2
Time (t) = 5.2 s
(a) To find the speed at t = 2.0 s:
v = u + at
v = 0 + (1.9)(2.0)
v = 0 + 3.8
v = 3.8 m/s
Therefore, the speed of the runner at t = 2.0 s is 3.8 m/s.
(b) To find the speed at the end of the race:
The runner's acceleration is zero for the rest of the race. This means that the runner continues to move with a constant velocity after 5.2 seconds.
Since the acceleration is zero, we can use the equation:
v = u + at
At the end of the race, the time (t) is the total time of 5.2 seconds.
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You have a simple RC circuit, with resistance, 6.9kΩ, and capacitance 1.3μF. What is the time constant of the circuit?
The time constant of the RC circuit is approximately 8.97 milliseconds.
The time constant of an RC circuit is determined by the product of the resistance and the capacitance.
Here's a step-by-step explanation to find the time constant:
Given data:
Resistance (R) = 6.9 kΩ = 6.9 * 10^3 Ω
Capacitance (C) = 1.3 μF = 1.3 * 10^(-6) F
Calculate the time constant:
The time constant (τ) is given by the formula τ = RC, where R is the resistance and C is the capacitance.
τ = (6.9 * 10^3 Ω) * (1.3 * 10^(-6) F) = 8.97 ms (rounded to two decimal places)
Therefore, the time constant of the RC circuit is approximately 8.97 milliseconds.
The time constant represents the time it takes for the voltage across the capacitor to reach approximately 63.2% of its final value in an RC circuit when it is charging or discharging.
It is an important parameter for understanding the time behavior of the circuit, such as the charging and discharging processes.
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3) A wire runs above the ground, carrying a large current. In the picture shown below, the current comes out of the page. K The Long Wire, Viewed head on The Ground A) If you stand on the ground directly underneath the wire, which way will a compass point? (Ignore the field of the Earth.) B) The wire is sagging downward. You realize that by using additional magnets, you can counteract the force of gravity on the wire, so that it doesn't sag. What direction magnetic field will be required to do this? (Hint: a current is just moving charge!) C) Show how to position bar magnet(s) near the wire to accomplish your answer from part B. (If you don't have an answer for part B, just guess a direction so you can get credit here.)
Using the concept of the magnetic field generated by current-carrying wire:
(A) The compass needle will point anticlockwise. if you are standing right below it.
(B)The magnets should be directed vertically upward.
(C) The north pole of the bar magnet should point downward.
A straight current-carrying wire generates a circular magnetic field around it as the axis.
A) The compass needle will point anticlockwise if you are standing right underneath the wire. The right-hand rule can be used to figure this out. When viewed from above, the magnetic field produced by the current will move anticlockwise around the wire if the current is exiting the page. The compass needle will point anticlockwise because its north pole lines up with the magnetic field lines.
B) The magnetic field created by the extra magnets should be directed vertically upward to oppose the pull of gravity on the wire and prevent sagging. The upward magnetic force can counterbalance the downward gravitational attraction by positioning the magnetic field in opposition to the gravitational pull.
C) You can place bar magnets in a precise way to provide the necessary upward magnetic field close to the wire. The north poles of the bar magnets should be pointed downward as you position them vertically above the wire. The magnets' south poles should be facing up. By positioning the bar magnets in this way, their magnetic fields will interact to produce an upward magnetic field close to the wire that will work to fight gravity and stop sagging.
Therefore, Using the concept of the magnetic field generated by a current-carrying wire:
(A) The compass needle will point anticlockwise. if you are standing right below it.
(B)The magnets should be directed vertically upward.
(C) The north pole of the bar magnet should point downward.
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2 -14 Points DETAILS OSCOLPHYS2016 13.P.01. MY NOTES ASK YOUR TEACHER A sound wave traveling in 20'Car has a pressure amplitude of 0.0 What intensity level does the sound correspond to? (Assume the density of ar 1.23 meter your answer.) db
The intensity level (I_dB) is -∞ (negative infinity).
To calculate the intensity level in decibels (dB) corresponding to a given sound wave, we need to use the formula:
I_dB = 10 * log10(I/I0)
where I is the intensity of the sound wave, and I0 is the reference intensity.
Given:
Pressure amplitude (P) = 0.0 (no units provided)
Density of air (ρ) = 1.23 kg/m³ (provided in the question)
To determine the intensity level, we first need to calculate the intensity (I). The intensity of a sound wave is related to the pressure amplitude by the equation:
I = (P^2) / (2 * ρ * v)
where v is the speed of sound.
The speed of sound in air at room temperature is approximately 343 m/s.
Plugging in the given values and calculating the intensity (I):
I = (0.0^2) / (2 * 1.23 kg/m³ * 343 m/s)
I = 0 / 846.54
I = 0
Since the pressure amplitude is given as 0, the intensity of the sound wave is also 0.
Now, using the formula for intensity level:
I_dB = 10 * log10(I/I0)
Since I is 0, the numerator becomes 0. Therefore, the intensity level (I_dB) is -∞ (negative infinity).
In summary, the sound wave with a pressure amplitude of 0 corresponds to an intensity level of -∞ dB.
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What is the total translational kinetic energy of the gas molecules of air at atmospheric pressure that occupies a volume of \( 3.90 \) L?
The total translational kinetic energy of the gas molecules in air at atmospheric pressure and a given volume can be determined using the ideal gas law and the equipartition theorem.
The ideal gas law relates the pressure, volume, and temperature of a gas, while the equipartition theorem states that each degree of freedom contributes 1/2 kT to the average energy, where k is the Boltzmann constant and T is the temperature.
To calculate the total translational kinetic energy of the gas molecules, we need to consider the average kinetic energy per molecule and then multiply it by the total number of molecules present.
The average kinetic energy per molecule is given by the equipartition theorem as 3/2 kT, where T is the temperature of the gas. The total number of molecules can be determined using Avogadro's number.
Given that the volume of the gas is 3.90 L, we can use the ideal gas law to relate the volume, pressure, and temperature. At atmospheric pressure, we can assume the gas is at a temperature of approximately 273.15 K.
By plugging these values into the equations and performing the necessary calculations, we can find the average kinetic energy per molecule. Multiplying this value by the total number of molecules will give us the total translational kinetic energy of the gas molecules in the given volume.
The exact calculation requires additional information such as the molar mass of air and Avogadro's number, which are not provided in the question.
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"An RLC Circuit of variable frequency has a power factor of 1 at
the frequency of 500 Hz. What else can you infer about the
circuit?
Given that an RLC Circuit of variable frequency has a power factor of 1 at the frequency of 500 Hz. We can infer that the circuit is a resonant circuit or the circuit is in resonance. A resonant circuit is one in which the inductive and capacitive reactance cancel each other out at the resonant frequency.
As a result, the circuit has only a pure resistance, and the circuit is in resonance. As a result, we can infer that at 500 Hz, the inductive reactance is equal to the capacitive reactance, and they cancel out each other. Furthermore, we can conclude that the inductance and capacitance values of the circuit must be such that their reactance values cancel out each other at 500 Hz.
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(a) What is the de Broglie wavelength (in m) of a proton moving at a speed of 2.07 x 104 m/s? m (b) What is the de Broglie wavelength (in m) of a proton moving at a speed of 2.16 x 108 m/s? m
(a) Proton speed: 2.07 x 10⁴ m/s, de Broglie wavelength: 3.31 x 10⁻¹¹m.
(b) Proton speed: 2.16 x 10⁸ m/s, de Broglie wavelength: 1.54 x 10⁻¹²m.
(a) To calculate the de Broglie wavelength of a proton, we can use the de Broglie wavelength equation:
λ = h / p
Where:
λ is the de Broglie wavelengthh is the Planck's constant (approximately 6.626 x 10⁻³⁴J·s)p is the momentum of the protonThe momentum of the proton can be calculated using the equation:
p = m × v
Where:
m is the mass of the proton (approximately 1.67 x 10⁻²⁷ kg)v is the velocity of the protonLet's calculate the de Broglie wavelength:
p = (1.67 x 10⁻²⁷ kg) × (2.07 x 10⁴ m/s)
λ = (6.626 x 10⁻³⁴ J·s) / p
Calculating the value of λ:
λ ≈ (6.626 x 10⁻³⁴ J·s) / [(1.67 x 10⁻²⁷ kg) × (2.07 x 10⁴m/s)]
λ ≈ 3.31 x 10⁻¹¹ m
Therefore, the de Broglie wavelength of the proton moving at a speed of 2.07 x 10⁴ m/s is approximately 3.31 x 10⁻¹¹ m.
(b) Using the same equation as before, we can calculate the de Broglie wavelength of the proton:
p = (1.67 x 10⁻²⁷ kg) × (2.16 x 10⁸ m/s)
λ = (6.626 x 10³⁴ J·s) / p
Calculating the value of λ:
λ ≈ (6.626 x 10⁻³⁴ J·s) / [(1.67 x 10⁻²⁷ kg) × (2.16 x 10⁸ m/s)]
λ ≈ 1.54 x 10⁻¹² m
Therefore, the de Broglie wavelength of the proton moving at a speed of 2.16 x 10⁸ m/s is approximately 1.54 x 10⁻¹² m.
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. An object 1.7 cm high is held 2.5 cm from a person's cornea, and its reflected image is measured to be 0.167 cm high. Think & Prepare 1. What kind of mirror is the cornea, convex or concave?
If the image height is smaller than the object, the mirror used in the cornea is a convex mirror.
Object height (h_o) = 1.7 cm
Object distance (u) = 2.5 cm
Image height (h_i) = 0.167 cm
To find whether the mirror used is convex or concave, we need to consider the properties of the image.
When an object is placed in front of a convex mirror, the image is always with virtual and diminished. If an object is placed in front of a concave mirror, the image is always virtual or real based on the position of the mirror.
In the given scenario, the image height is smaller than the object.
Therefore we can conclude that the cornea acts as a convex mirror.
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When resistors 1 and 2 are connected in series, the equivalent resistance is 17.9 . When they are connected in parallel, the
equivalent resistance is 3.03 M. What are (a) the smaller resistance and (b) the larger resistance of these two resistors?
The smaller resistance between resistors 1 and 2 is approximately 3.5 ohms, while the larger resistance is approximately 14.4 ohms.
When resistors are connected in series, the sum of their individual resistances produces the desired resistance. The corresponding resistance in this situation is 17.9 ohms. The bigger resistance is equal to the sum of the smaller resistance and the value of resistor 2 since the resistors are connected in series. The lesser resistance is discovered by rearranging the equation to be roughly 3.5 ohms.
The reciprocal of the equivalent resistance is obtained by adding the reciprocals of the resistors when they are connected in parallel. The reciprocal of the corresponding resistance in this situation is roughly 0.33 microsiemens. The reciprocal of the bigger resistance is equal to the sum of the reciprocals of the smaller resistance and the value of resistor 2 since the resistors are connected in parallel. Rearranging the equation reveals that the bigger resistance's reciprocal is roughly 0.27 microsiemens, giving us a larger resistance of about 14.4 ohms.
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A mass of 26 kg is suspended by two cords from a ceiling. The cords have lengths of 17 cm and 21 cm, and the distance between the points where they are attached on the ceiling is 29 cm. Determine the tension in each of the two cords. Include a clear diagram
The tension in the 17 cm cord is 156.3 N and the tension in the 21 cm cord is 110.3 N.
The mass of 26 kg is suspended by two cords from a ceiling. The cords have lengths of 17 cm and 21 cm, and the distance between the points where they are attached to the ceiling is 29 cm.
To determine the tension in each of the two cords, we first sketch the diagram of the system of the two cords and the mass that is being suspended from the cords.From the diagram, we can see that the forces acting on the mass are the weight of the mass and the tensions in the cords. Thus we have two equations of equilibrium as follows:Equation (1) resolves forces in the vertical direction: `T1 sin θ1 + T2 sin θ2 = Fg
For the 17 cm cord, the vertical component of tension T1 is T1 sin(θ1), and for the 21 cm cord, the vertical component of tension T2 is T2 sin(θ2).
Since the mass is in equilibrium, the sum of the vertical forces must be zero:
T1 sin(θ1) + T2 sin(θ2) = mg
We can also consider the horizontal components of tension T1 and T2. The horizontal component of T1 is T1 cos(θ1), and the horizontal component of T2 is T2 cos(θ2). The horizontal components must cancel out each other since there is no horizontal acceleration:
T1 cos(θ1) = T2 cos(θ2)
Using these two equations, we can solve for the tensions T1 and T2
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In the R-C Circuit experiment, at (t = 0) the switch is closed and the capacitor starts discharging. The voltage across the capacitor was recorded as a function of time according to the equation V=vies 9 8 7 6 5 Vc(volt) 4 3 2 1 0 10 20 30 10 50 t(min) From the graph, the time constant T in second) is 480
In the given RC circuit experiment, the switch is closed at t=0, and the capacitor starts discharging. The voltage across the capacitor has been recorded concerning time. The data for the voltage across the capacitor is given as follows:
V = Vies9 8 7 6 5
Vc (volt)4 3 2 1 0102030405060 t (min)
The time constant of the RC circuit can be calculated by the following formula:
T = R*C Where T is the time constant, R is the resistance of the circuit, and C is the capacitance of the circuit. As we know that the graph of the given data is an exponential decay curve, the formula for the voltage across the capacitor concerning time will be:
Vc = V0 * e^(-t/T)Where V0 is the initial voltage across the capacitor. We can calculate the value of the time constant T by using the given data. From the given graph, the voltage across the capacitor at t=480 seconds is 2 volts.
The formula will be:2 = V0 * e^(-480/T) Solving for T, we get:
T = -480 / ln(2)
≈ 693 seconds.
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A tuning fork by itself produces a faint sound. Explain how each of the following methods can be used to obtain a louder sound from it. Explain also any effect on the time interval for which the fork vibrates audibly. (a) holding the edge of a sheet of paper against one vibrating tine
To obtain a louder sound from a tuning fork, one method is to hold the edge of a sheet of paper against one vibrating tine.
When the paper is pressed against the tine, it acts as a soundboard and helps to amplify the sound produced by the tuning fork. This is because the paper vibrates along with the tine, creating more air vibrations and thus a louder sound.
When the paper is held against the tine, the time interval for which the fork vibrates audibly may be slightly reduced. This is because the paper adds some dampening effect to the vibrations, causing them to decay faster. However, the overall loudness of the sound is increased due to the amplifying effect of the paper.
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An isolated conducting sphere of radius r₁=0.20 m is at a potential of -2000V, with charge Qo. The charged sphere is then surrounded by an uncharged conducting sphere of inner radius r2 = 0.40 m, and outer radius r3 = 0.50m, creating a spherical capacitor. (a) (5 points) Draw a clear physics diagram of the problem. (b) (5 points) Determine the charge Qo on the sphere while its isolated. (c) (8 points) A wire is connected from the outer sphere to ground, and then removed. Determine the magnitude of the electric field in the following regions: R<₁; r₁
(b)
When the isolated sphere of radius r₁ is at a potential of -2000V with charge Qo, the charge on the sphere is given by
q = CV. Using the above information the charge on the isolated sphere is Q = 7.03 × 10⁻⁷ C.
Q=CV
where,
C = Capacitance of the sphere
V = Potential
Q = Charge
Therefore, the charge on the sphere is given by,
Q = CV = 4πε₀r₁V
Where ε₀ is the permittivity of free space
ε₀ = 8.85 × 10⁻¹² F/m²
So, substituting the given values Q = 4π × 8.85 × 10⁻¹² × 0.20 × (-2000)
Q = 7.03 × 10⁻⁷ C
(c) When a wire is connected from the outer sphere to ground, then removed, the magnitude of the electric field in the different radius R varies according to equation E = 7.03 × 10⁻⁷ / (4π × 8.85 × 10⁻¹² × (0.20 + R)²)
R < r₁ : There is no electric field as the electric field inside a conducting sphere is zero.
r₁ < R < r₂: Since the conducting sphere is uncharged, the electric field in this region is also zero.
r₂ < R < r₃: For a spherical capacitor, the electric field inside the capacitor is given by
E = Q/4πε₀r²
Where,
Q = Charge on the isolated sphere = 7.03 × 10⁻⁷ C
ε₀ = Permittivity of free space = 8.85 × 10⁻¹² F/m²
r = Distance from the center of the isolated sphere = r₁ + RSo, substituting the given values and solving,
E = 7.03 × 10⁻⁷ / (4π × 8.85 × 10⁻¹² × (0.20 + R)²)
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S Five particles with equal negative charges -q are placed symmetrically around a circle of radius R. Calculate the electric potential at the center of the c
To calculate the electric potential at the center of the circle, we can use the principle of superposition.
The electric potential at the center of the circle due to a single charged particle can be calculated using the formula V = k * (q / r), where V is the electric potential, k is Coulomb's constant, q is the charge of the particle, and r is the distance from the particle to the center of the circle.
Since there are five particles with equal negative charges placed symmetrically around the circle, the total electric potential at the center can be found by adding up the contributions from each individual particle. Let's denote the electric potential due to each particle as V1, V2, V3, V4, and V5. Since the charges are equal in magnitude and negative, the electric potential due to each particle will have the same magnitude but opposite signs. Therefore, the total electric potential at the center of the circle can be calculated as:
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Consider a person traveling a distance of 300 km (with respect to the ground) in a relativistic car on a straight highway. Assume event A is when the car has gone 0 km of distance and event B is when the car has reached its destination. You only need to draw one diagram for parts e-g. Case 1: The car is traveling at a speed of 4.32 x108 km/hr. (a) Determine the velocity of the person in SR Units. (b) Determine the distance (with respect to the earth) traveled in SR units (c) Determine the time for the trip as measured by someone on the earth. (d) Determine the car's space-time interval. (e) Carefully draw and label a spacetime diagram for the car with respect to a person on the ground using the graph paper provided and a straight edge. (Note: this should not be a two- observer diagram) Make the diagram as accurate as possible. Make the diagram big enough to read and big enough to add another worldline. (f) When does a person on the ground see the car reach its destination. Draw a labeled worldline to support your answer. Case 2: If the car instead accelerated from rest to reach point B. (g) Draw a possible worldline for the car using a dashed line ("---") on your spacetime the diagram in part e). Considering Cases 1 and 2: (h) In which case(s) does a clock attached to the car measure proper time? Explain briefly. (i) In which case(s) does a clock attached to the car measure spacetime interval? Explain briefly. (j) In which case(s) does a clock attached to the car measure coordinate time? Explain
In both cases, the clock attached to the car measures coordinate time, which is the time measured by a single clock in a given frame of reference.
Given that,Distance traveled by the car = 300 km = 3 × 10² km
Speed of the car = 4.32 × 10⁸ km/hr
Case 1:
(a) Velocity of the person in SR Units
The velocity of the car in SI unit = (4.32 × 10⁸ × 1000) / 3600 m/s = 120,000 m/s
The velocity of the person = 0 m/s
Relative velocity = v/c = (120,000 / 3 × 10⁸) = 0.4 SR Units
(b) Distance (with respect to the earth) traveled in SR units
Proper distance = L = 300 km = 3 × 10² km
Proper distance / Length contraction factor L' = L / γ = (3 × 10²) / (1 - 0.4²) = 365.8537 km
Distance traveled in SR Units = L' / (c x T) = 365.8537 / (3 × 10⁸ x 0.4) = 3.0496 SR Units
(c) Time for the trip as measured by someone on the earth
Time interval, T = L' / v = 365.8537 / 120000 = 0.003048 SR Units
Time measured by someone on Earth = T' = T / γ = 0.004807 SR Units
(d) Car's space-time interval
The spacetime interval, ΔS² = Δt² - Δx²
where Δt = TΔx = v x TT = 0.003048 SR Units
Δx = 120000 × 0.003048 = 365.76 km
ΔS² = (0.003048)² - (365.76)² = - 133,104.0799 SR Units²
(e) Spacetime diagramCase 2:If the car instead accelerated from rest to reach point B.(g) The possible worldline for the car using a dashed line ("---")Considering Cases 1 and 2:(h) In which case(s) does a clock attached to the car measure proper time? Explain briefly.In Case 2, as the car is accelerating from rest, it is under the influence of an external force and a non-inertial frame of reference.
Thus, the clock attached to the car does not measure proper time in Case 2.In Case 1, the clock attached to the car measures proper time as the car is traveling at a constant speed. Thus, the time interval measured by the clock attached to the car is the same as the time measured by someone on Earth.(i) In which case(s) does a clock attached to the car measure spacetime interval?
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If a bicycle is traveling at 15 km/h, how fast are its 50-em-diameter wheels tuming? (Give answer in revolutions per second)
The wheels of the bicycle are turning at approximately 25 revolutions per second.
To determine the speed at which the wheels are turning, we need to convert the given velocity of the bicycle, which is 15 km/h, to the linear velocity of the wheels.
Step 1: Convert the velocity to meters per second:
15 km/h = (15 * 1000) meters / (60 * 60) seconds
= 4.17 meters per second (rounded to two decimal places)
Step 2: Calculate the circumference of the wheels:
The diameter of the wheels is given as 50 cm, which means the radius is 50/2 = 25 cm = 0.25 meters (since 1 meter = 100 cm).
The circumference of a circle can be calculated using the formula: circumference = 2 * π * radius.
So, the circumference of the wheels is:
circumference = 2 * π * 0.25
= 1.57 meters (rounded to two decimal places)
Step 3: Calculate the number of revolutions per second:
To find the number of revolutions per second, we can divide the linear velocity of the wheels by the circumference:
revolutions per second = linear velocity/circumference
= 4.17 meters per second / 1.57 meters
≈ 2.65 revolutions per second (rounded to two decimal places)
Therefore, the wheels of the bicycle are turning at approximately 2.65 revolutions per second.
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An airplane lands with an initial velocity of 90 m/s and then
decelerates at 2.0 m/s2 for 40 sec. What is its final velocity?
The final velocity of the airplane is 10 m/s. This means the airplane will be moving at a speed of 10 meters per second after 40 seconds when it has decelerated from its initial velocity of 90 meters per second.
Due to the negative acceleration and velocity acting in opposite directions, it means the airplane is slowing down or decelerating.
The formula for finding the final velocity is given as:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Substitute the given values into the formula:
v = 90 + (-2.0 × 40)
v = 90 - 80
v = 10 m/s
Therefore, the final velocity of the airplane is 10 m/s. This means the airplane will be moving at a speed of 10 meters per second after 40 seconds when it has decelerated from its initial velocity of 90 meters per second.
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Two blocks with masses m1= 4.5 kg and m2= 13.33 kg on a frictionless surface collide head-on. The initial velocity of block 1 is v→1,i= 4.36 i^ms and the initial velocity of block 2 is v→2,i=-5 i^ms. After the collision, block 2 comes to rest. What is the x-component of velocity in units of ms of block 1 after the collision? Note that a positive component indicates that block 1 will be traveling in the i^ direction, and a negative component indicates that block 1 will be traveling in the −i^ direction. Please round your answer to 2 decimal places.
Since a positive component indicates that block 1 will be traveling in the i^ direction, the answer is 4.51 i^. Therefore, the required answer is 4.51. Answer: 4.51.
When two blocks with masses m1 = 4.5 kg and m2 = 13.33 kg on a frictionless surface collide head-on, block 2 comes to rest.
The initial velocity of block 1 is v→1, i = 4.36 i^ ms and the initial velocity of block 2 is v→2, i = -5 i^ ms.
We are required to find the x-component of velocity in units of ms of block 1 after the collision.
We need to find the final velocity of block 1 after the collision. We can use the law of conservation of momentum to solve this problem.
The law of conservation of momentum states that the total momentum of an isolated system of objects with no external forces acting on it is constant. The total momentum before collision is equal to the total momentum after the collision.
Using the law of conservation of momentum, we can write:
[tex]m1v1i +m2v2i = m1v1f + m2v2f[/tex]
where
v1i = 4.36 m/s,
v2i = -5 m/s,m1
= 4.5 kg,m2
= 13.33 kg,
v2f = 0 m/s (because block 2 comes to rest), and we need to find v1f.
Substituting the given values, we get:
4.5 kg × 4.36 m/s + 13.33 kg × (-5 m/s)
= 4.5 kg × v1f + 0
Simplifying, we get:
20.31 kg m/s
= 4.5 kg × v1fv1f
= 20.31 kg m/s ÷ 4.5 kgv1f
= 4.51 m/s
The x-component of velocity in units of ms of block 1 after the collision is 4.51 m/s.
Since a positive component indicates that block 1 will be traveling in the i^ direction, the answer is 4.51 i^.
Therefore, the required answer is 4.51. Answer: 4.51.
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