Inside a freely-falling elevator, there would be no apparent weight for you.So option B is correct.
Inside a freely-falling elevator, there is still a gravitational force acting on you. However, since both you and the elevator are falling at the same rate, you would experience a sensation of weightlessness. Your apparent weight, which is the force exerted on a body due to gravity, would be zero. This is because there is no contact force between you and the elevator floor that would provide a normal force to counteract gravity. Therefore, the correct option is that there would be no apparent weight for you.
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3. When a real incompressible fluid flows through a circular pipe, energy is dissipated due to the viscosity of the fluid. The Moody diagram on page 9 represents this energy loss as a dimensionless friction factor (f) which is a function of the Reynolds number of the flow (Re) for both laminar and turbulent flow and also a function of the relative roughness (=/D) for turbulent flow. a) Explain this dependence of friction factor (f) upon the relative roughness (E/D) for turbulent flow and specifically why the friction factor increases with relative roughness at any given Reynolds number. Water with a density of 1000 kg/m³ and dynamic viscosity of 1.0 x 10³ Pa.s flows under gravity from a reservoir through a cast iron pipe of 75mm internal diameter and an equivalent roughness of 0.26mm at a flow rate of 600 litres per minute into the local atmosphere. The flow path comprises a sharp edged entrance from the reservoir into the pipe (loss factor (KL) of 0.5) and a 100m horizontal length of the cast iron pipe. There is no fitting or restriction at the outlet of the pipe into the local atmosphere and so no additional minor head loss. The liquid surface of the reservoir is exposed to the local atmosphere. b) Calculate the major head loss (hL) and minor head loss (hm) in the flow path and the height of water in the reservoir required above the sharp edged entrance into the pipe to achieve the required flow rate. c) If the height of water in the reservoir above the sharp edged entrance to the pipe and the pipe diameter and length are fixed, propose two other ways to increase the flow rate from the reservoir, evaluate their relative effectiveness and state which is the best option. Steady, uniform, and laminar flow of a fluid with dynamic viscosity (n) occurs between two horizontal, infinite, parallel plates separated by a distance (h) in the vertical direction (y). The lower plate (y=0) is stationary and the upper plate (y=h) moves with velocity (U) in the direction of flow (x). The vertical coordinate (y) where the maximum velocity (u) occurs, (y'), is given by below equation. Assume fluid of dynamic viscosity 0.5 Pa.s passes between the two plates which are 20mm apart with a pressure difference per unit length in the (x) direction of -500 Pa/m. h Undp hdx, 2 d) Calculate what happens to (y') as the upper plate velocity (U) increases from 0 (stationary) to 0.1 m/s and then to 0.2 m/s. With the aid of sketches, provide a physical explanation for this behaviour.
a) The friction factor increases with relative roughness at any given Reynolds number for turbulent flow because there is more resistance caused by the increased roughness. The rougher the pipe, the more it resists the flow, which results in a higher friction factor.
b) The following formulas can be used to calculate the major head loss (hL) and minor head loss (hm) in the flow path and the height of water in the reservoir required above the sharp-edged entrance into the pipe to achieve the required flow rate:
First, compute the velocity in the pipe:
[tex]v = Q/A = (600/1000) / [(pi/4)*(75/1000)^2] = 1.81 m/s[/tex]
where:
Q is the flow rate (l/min)
A is the cross-sectional area of the pipe (m²)
Compute the Reynolds number:
[tex]Re = (Dvρ) / μ = (75/1000)(1.81)(1000) / 1 x 10^-3 = 136,029[/tex]
Compute the friction factor:
Use the Moody chart to determine the friction factor:
From the chart, f = 0.03
Compute the major head loss:
[tex]hL = (fLv²) / (2gd) = (0.03)(100)(1.81²) / (2 x 9.81 x 100/1000) = 1.6 m[/tex]
where:
L is the pipe length (m)
g is the gravitational acceleration (9.81 m/s²)
Compute the minor head loss:
[tex]hm = KL(v²/2g) = 0.5(1.81²/2 x 9.81) = 0.17 m[/tex]
Compute the height of water:
Pump head = hL + hm = 1.6 + 0.17 = 1.77 m
c) Two ways to increase the flow rate from the reservoir are to increase the pipe diameter or decrease the pipe length. Increasing the pipe diameter is more effective than decreasing the pipe length because it has a greater impact on the flow rate. Doubling the pipe diameter, for example, would increase the flow rate by a factor of 16.
d) The value of y' decreases as the upper plate velocity U increases from 0 (stationary) to 0.1 m/s and then to 0.2 m/s. As the velocity of the upper plate increases, the flow rate and Reynolds number also increase. The increased flow rate pushes the maximum velocity point towards the lower plate.
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The phase difference between two identical sinusoidal waves propagating in the same direction is tt rad. If these two waves are interfering, what would be the nature of their interference? Operfectly destructive O partially constructive partially destructive O None of the listed choices. perfectly constructive
The phase difference between two identical sinusoidal waves propagating in the same direction is tt rad (where tt represents a specific angle in radians).
The nature of interference between these waves depends on the specific value of the phase difference. If the phase difference is an odd multiple of π (pi) radians (such as π, 3π, 5π, etc.), the interference is perfectly destructive. In this case, the peaks of one wave coincide with the troughs of the other wave, resulting in complete cancellation or destructive interference.
If the phase difference is an even multiple of π (pi) radians (such as 0, 2π, 4π, etc.), the interference is perfectly constructive. In this case, the peaks of one wave coincide with the peaks of the other wave, resulting in reinforcement or constructive interference. If the phase difference is any other value, the interference will be a combination of constructive and destructive interference, leading to partially constructive and partially destructive interference.
Therefore, the correct answer from the listed choices would be: Partially constructive, partially destructive.
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25. You hit a tuning fork that produces a 512HZ tone, and you hit another tuning fork with an unknown frequency. You hear a Waa-Waa sound (Beat frequency) of 5HZ. Explain what beat frequency is, and what value(s) frequency could have the other tuning fork been? 26. An ambulance is traveling away from you at 50.0 km/h has a siren that produces a 1,500.0 Hz sound. The outside temperature is 25.°C. What is the speed and frequency of the wave that you observe?
25. Beat frequency refers to the phenomenon of interference between two sound waves with slightly different frequencies. When two sound waves of slightly different frequencies are played together, they create an oscillating sound pattern characterized by a periodic increase and decrease in amplitude, resulting in a "waa-waa" sound.
The beat frequency is equal to the difference between the frequencies of the two sound waves. In this case, the known tuning fork produces a tone of 512 Hz, and the beat frequency is 5 Hz. Therefore, the frequency of the unknown tuning fork can be either 517 Hz (512 Hz + 5 Hz) or 507 Hz (512 Hz - 5 Hz).
26. The observed frequency of a sound wave emitted by a moving source is affected by the motion of the source and the medium through which the sound wave travels. This effect is known as the Doppler effect.
In this scenario, the ambulance is traveling away from you at a speed of 50.0 km/h. The speed of sound in air at 25.°C is approximately 343 m/s. Using the formula for the Doppler effect, we can determine the observed frequency:
Observed frequency = Source frequency × (Speed of sound + Observer velocity) / (Speed of sound + Source velocity)
The source frequency is 1,500.0 Hz, and the observer velocity is 0 (assuming you are stationary). Plugging in the values, we find:
Observed frequency = 1,500.0 Hz × (343 m/s + 0) / (343 m/s + 50.0 km/h)
Simplifying the calculation, we find the observed frequency of the siren sound.
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The absolute pressure of an ideal gas in a bike tire is 1.5 atm Atre gauge is used to measare thim pressure in the tie What prestulf does hie gaine read A. 2.5 atm B. 5 atm 1.5 atm 3 atm E 0.5 atm
The pressure reading on the gauge would be 2.5 atm calculated by subtracting the atmospheric pressure from the absolute pressure. So, the correct answer is option A. 2.5 atm.
Explanation:
Gauge pressure is the pressure measured relative to atmospheric pressure. In this case, the absolute pressure inside the bike tire is given as 1.5 atm. Since the atmospheric pressure is typically around 1 atm, the gauge pressure can be calculated by subtracting the atmospheric pressure from the absolute pressure.
Absolute pressure = Gauge pressure + Atmospheric pressure
Absolute pressure = 1.5 atm + 1 atm
Absolute pressure = 2.5 atm
Therefore, the pressure reading on the gauge would be 2.5 atm.
So, the correct answer is option A. 2.5 atm.
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A point source emits sound waves isotropically. The intensity of the waves 3.20~\mathrm{m}3.20 m from the source is 1.76 \times 10^{-6}~\mathrm{W/m^2}1.76×10−6 W/m2. Let us assume that the energy of the waves is conserved. At what distance RR from the source, do sound waves have a sound level of 0db
At a distance of approximately 7.54 x [tex]10^{-4}[/tex]meters from the source (0.754 mm), the sound waves would have a sound level of 0 dB.
To determine the distance from the source at which sound waves have a sound level of 0 dB, we need to understand the relationship between sound intensity and sound level.
Sound intensity (I) is measured in watts per square meter (W/m²) and is related to sound level (L) in decibels (dB) through the following equation:
L = 10 log₁₀(I/I₀)
Where I₀ is the reference intensity, which corresponds to the threshold of hearing and is approximately 1.0 x [tex]10^{-12}[/tex]W/m².
In this case, the sound level is given as 0 dB, which means that the sound intensity is equal to the reference intensity:
L = 0 dB
I = I₀ = 1.0 x [tex]10^{-12}[/tex] W/m²
We are given the intensity at a distance of 3.20 m from the source, which is 1.76 x [tex]10^{-6}[/tex] W/m². To find the distance (R) at which the sound level is 0 dB, we need to find the point where the intensity decreases to the reference intensity.
Using the inverse square law for sound intensity, which states that sound intensity decreases with the square of the distance from the source:
I = I₀ / [tex]R^{2}[/tex]
Setting the two intensity values equal to each other:
1.76 x [tex]10^{-6}[/tex] W/m² = 1.0 x [tex]10^{-12}[/tex] W/m² / [tex]R^{2}[/tex]
[tex]R^{2}[/tex] = (1.0 x [tex]10^{-12}[/tex] W/m²) / (1.76 x [tex]10^{-6}[/tex] W/m²)
≈ 5.68 x [tex]10^{-7}[/tex] m²
Taking the square root of both sides:
[tex]R= \sqrt{5.68*10^{-7}m^{2} }[/tex]
≈ 7.54 x [tex]10^{-4}[/tex] m
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An astronaut orbiting the Earth is preparing to dock with a Westar Vi satelite. The satelise is in a circular erbit soo km above the Earth's surface, where the free fat acceieration is 8.29 ms
2
. Take the radur of the Earth as 6400 km. Determine the speed of the satelite. min Determine the time interval required to complete one orbit around the Earth, which is the period of the satelite. min
The speed of the satellite is approximately 7764 m/s, and the period of the satellite (time to complete one orbit) is approximately 86.3 minutes.
To determine the speed of the satellite, we can use the concept of centripetal acceleration. The centripetal acceleration is provided by the gravitational force between the satellite and the Earth. The formula for centripetal acceleration is:
a = v^2 / r,
where "a" is the centripetal acceleration, "v" is the velocity of the satellite, and "r" is the distance between the center of the Earth and the satellite's orbit (the radius of the Earth plus the altitude of the satellite).
Given that the free fall acceleration is 8.29 m/s^2 and the radius of the Earth is 6400 km, we can convert these values to meters and solve for the velocity:
8.29 m/s^2 = v^2 / (6400 km + 500 km),
where the altitude of the satellite above the Earth's surface is 500 km.
Simplifying the equation, we have:
v^2 = 8.29 m/s^2 * (6400 km + 500 km),
v^2 = 8.29 m/s^2 * (6900 km).
Now we can solve for the velocity:
v = √(8.29 m/s^2 * (6900 km)).
Calculating this expression, we find that the speed of the satellite is approximately 7764 m/s.
To determine the period of the satellite (the time interval required to complete one orbit), we can use the formula for the period of a circular orbit:
T = 2πr / v,
where "T" is the period, "r" is the distance between the center of the Earth and the satellite's orbit, and "v" is the velocity of the satellite.
Plugging in the values, we have:
T = 2π * (6400 km + 500 km) / 7764 m/s.
Simplifying and converting kilometers to meters, we find that the period of the satellite is approximately 5180 seconds or 86.3 minutes.
In summary, the speed of the satellite is approximately 7764 m/s, and the period of the satellite (time to complete one orbit) is approximately 86.3 minutes.
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explain why the electrical charge on an atom is zero
The electrical charge on an atom is zero due to presence of same number of protons and electrons.
An atom is the smallest entity comprising of three components. These are protons, neutrons and electrons. Protons and neutrons are centrally located forming the nucleus while electrons revolve around the nucleus. Protons are positively charged while electrons are negatively charged. Neutrons are neutral due to lack of charge.
The number of protons and electrons are same in an atom owing to balancing the overall charge in an atom. This makes the atom electrical neutral and hence the charge on an atom is zero.
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Electromagnetio radiation is emitied by accoleratng Part A charges. The rale at which ereegy is emined from an accelsating chargn that has charged and. acceleration a is given by d4x=6 crep 231a2 where c when fraction of tis energy does trad ale ber second? is the spend of light.
The rate at which energy is emitted from an accelerating charged particle is given by the formula dE/dt = (2/3) (e^2/4πε₀c³) a², where e is the charge of the particle and a is its acceleration. The expression (2/3) (e^2/4πε₀c³) represents a constant factor. The energy emitted per second is directly proportional to the square of the acceleration of the charged particle.
The rate at which energy is emitted from an accelerating charged particle can be derived from the theory of classical electrodynamics. The formula dE/dt = (2/3) (e^2/4πε₀c³) a² represents the power radiated by the charged particle. Here, e is the charge of the particle, a is its acceleration, ε₀ is the permittivity of free space, and c is the speed of light.
The expression (2/3) (e^2/4πε₀c³) represents a constant factor that depends on the properties of the particle and the medium in which it is accelerating. The energy emitted per second, or the power, is directly proportional to the square of the acceleration of the charged particle.
Therefore, the rate at which energy is emitted from an accelerating charged particle is determined by the square of its acceleration, and the constant factor (2/3) (e^2/4πε₀c³) represents the proportionality between the power and the acceleration.
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How many 40μF capacitors must be connected in parallel to store a charge of 1C with a potential of 100 V across the capacitors? 1. 1000 2. 625 3. 0500 4. 0400 5. 0250
The formula that relates capacitance (C), charge (Q), and potential difference (V) is Q = CV. Here, we need to find out how many 40μF capacitors must be connected in parallel to store a charge of 1C with a potential of 100 V across the capacitors.
We can find out the number of capacitors required using the formula:Q = CVQ = 1C, V = 100V, and C = 40μFThe formula is:
Q = CV=> C = Q/V=> 40μF = 1C/100V=> C = 0.01F
Now,
we can find the number of capacitors required using the formula:
N = Ceq/C, where Ceq is the equivalent capacitance.N = number of capacitors required C = capacitance of each capacitor Ceq = Q/VN = Ceq/C => N = (Q/V)/C => N = (1C/100V)/(40μF)=> N = 250Hence, 250 capacitors are needed to store a charge of 1C with a potential of 100 V across the capacitors. Therefore, the correct option is 5. 0250.
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What is the apparent weight of 125 cm3 of steel submerged in
water? ANS: 8.4 N
When an object is submerged in water, the apparent weight is less than its actual weight due to the buoyant force. To determine the apparent weight of 125 cm³ of steel submerged in water, we will need to use the formula for buoyant force.
Buoyant force = Weight of water displaced by the object
We know the volume of the steel is 125 cm³. Since 1 cm³ of water has a mass of 1 gram and the density of steel is 7.8 g/cm³, we can calculate the mass of the steel:
mass of steel = volume of steel × density of steel= 125 cm³ × 7.8 g/cm³= 975 g
To determine the weight of water displaced by the steel, we need to know the volume of water displaced.
This is equal to the volume of the steel:
volume of water displaced = volume of steel = 125 cm³
The weight of water displaced is equal to the weight of this volume of water, which we can calculate using the density of water and the volume of water displaced:
weight of water displaced = volume of water displaced × density of water= 125 cm³ × 1 g/cm³= 125 g
Now we can calculate the buoyant force acting on the steel:
Buoyant force = Weight of water displaced by the object= 125 g × 9.81 m/s²= 1.23 N
The apparent weight of the steel submerged in water is equal to the actual weight minus the buoyant force:
Apparent weight = Actual weight - Buoyant force
Actual weight = mass of steel × gravitational acceleration= 975 g × 9.81 m/s²= 9.57 N
Apparent weight = 9.57 N - 1.23 N = 8.34 N
Therefore, the apparent weight of 125 cm³ of steel submerged in water is 8.34 N (to two decimal places).
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"
(a) A one dimensional collision occurs between a cart of mass
10.0 kg moving to the right at 4.0 m/s and a block of mass 6.0 kg
moving to the left at 12.0 m/s. After the collision, the block
moves to
"
After the collision, the block moves to the right at 4.5 m/s. The velocity of the cart after the collision is approximately -5.9 m/s (to the left).
To solve this problem, we can apply the principles of conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision.
Given:
Mass of cart (m₁) = 10.0 kg
Initial velocity of cart (v₁i) = 4.0 m/s (to the right)
Mass of block (m₂) = 6.0 kg
Initial velocity of block (v₂i) = -12.0 m/s (to the left)
Final velocity of block (v₂f) = 4.5 m/s (to the right)
Let's denote the final velocity of the cart as v₁f.
Conservation of momentum equation:
m₁ v₁i + m₂ v₂i = m₁ v₁f + m₂ v₂f
Substituting the given values:
(10.0 kg * 4.0 m/s) + (6.0 kg * (-12.0 m/s)) = (10.0 kg * v₁f) + (6.0 kg * 4.5 m/s)
Simplifying the equation:
40.0 kg m/s - 72.0 kg m/s = 10.0 kg * v₁f + 27.0 kg m/s
Combining like terms:
-32.0 kg m/s = 10.0 kg * v₁f + 27.0 kg m/s
Rearranging the equation:
10.0 kg * v₁f = -32.0 kg m/s - 27.0 kg m/s
10.0 kg * v₁f = -59.0 kg m/s
Dividing both sides by 10.0 kg:
v₁f = (-59.0 kg m/s) / 10.0 kg
v₁f = -5.9 m/s
Therefore, the velocity of the cart after the collision is approximately -5.9 m/s (to the left).
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The complete question is:
A one dimensional collision occurs between a cart of mass 10.0 kg moving to the right at 4.0 m/s and a block of mass 6.0 kg moving to the left at 12.0 m/s. After the collision, the block moves to the right at 4.5 m/s. What is the velocity of the cart after the collision?
An object has a circular path with radius 8.00 cm. The angular velocity of the object is 150
rad/s. Determine (a) tangential velocity and (b) centripetal force.
Therefore, the tangential velocity of the object is 12 m/s and the centripetal force acting on the object is 22500 N
To determine the tangential velocity and centripetal force of an object moving in a circular path, we can use the following formulas:
(a) Tangential velocity (v):
v = r * ω
where r is the radius of the circular path and ω is the angular velocity.
(b) Centripetal force (F):
F = m * a = m * ([tex]v^2[/tex] / r)
where m is the mass of the object, v is the tangential velocity, and a is the centripetal acceleration.
Radius, r = 8.00 cm = 0.08 m
Angular velocity, ω = 150 rad/s
(a) Tangential velocity:
v = r * ω
v = 0.08 m * 150 rad/s
Calculate the value:
v = 12 m/s
(b) Centripetal force:
F = m * ([tex]v^2[/tex] / r)
F = m * (12 [tex]m/s)^2[/tex] / 0.08 m
Simplify the equation and substitute the appropriate values:
F = m * 1800 [tex]m^2/s^2[/tex] / 0.08 m
Calculate the value:
F = m * 22500 N.
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Two point charges are located a distance of 2 m apart. Charge one is +2C and charge two is −3C. What is the potential energy for that configuration? [K=9
∗
10
∧
9Nm
∧
2/C
∧
2] −27
∗
10
∧
9 J −9
∗
10
∧
9 J −13.5
∗
10
∧
9 J −14
∗
10
∧
9 J
Let us first calculate the electrostatic force experienced by the point charges due to each other.
The force experienced by charge 1 due to charge 2 is:
[tex]$$\begin{aligned} F_{1,2} &=\frac{1}{4\pi\varepsilon_0}\frac{Q_1Q_2}{r^2}\\ &=\frac{1}{4\pi(9\times10^9)}\frac{2\times(-3)}{2^2}\\ &=\frac{-3}{4\pi(9\times10^9)}\\ &= -1.25\times10^{-10}N\end{aligned}$$[/tex]
Where
r = 2m
is the distance between the two-point charges, and
Q1 = 2C and Q2 = -3C
are the magnitudes of the two-point charges.
Now, the potential energy of the two-point charges is given by:
[tex]$$U_{1,2}=K_e\frac{Q_1Q_2}{r}$$$$\begin{aligned} U_{1,2} &= (9\times10^9)\frac{(2)(-3)}{2}\\ &=(-27\times10^9)J\\ &= -2.7\times10^{10}J\end{aligned}$$[/tex]
the potential energy for the configuration is -2.7×10¹⁰J, which is represented by option D.
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Use physical standards used to develop the Celsius and Fahrenheit temperature scales. Now, come up with a new temperature scale that is based on different physical standards. Be as imaginative as possible.
The Celsius and Fahrenheit temperature scales were both established using the properties of substances under specific conditions.
One of the physical standards that was used to develop the Celsius temperature scale is the melting point of ice (0°C) and boiling point of water (100°C) under atmospheric pressure.
On the other hand, the Fahrenheit temperature scale was established using a mixture of water, salt, and ice that resulted in a temperature of 0°F, and the human body temperature was used as a reference point for 98.6°F.
Now, let's create a new temperature scale based on different physical standards. We can call it the Quantum temperature scale, which uses the properties of an atom as a reference point.
The idea is to make use of the atomic resonance frequency, which is the frequency at which an atom will absorb a photon of light. Each atom has a unique resonance frequency that corresponds to a specific temperature.
Let's use the hydrogen atom as an example. The hydrogen atom has a resonance frequency of 1.42 GHz at a temperature of 0K (Kelvin).
The Quantum temperature scale would use this frequency as its reference point. As the temperature increases, the resonance frequency of the hydrogen atom will shift, and the scale would be calibrated accordingly.
For example, at 100K, the resonance frequency of the hydrogen atom would be 1.44 GHz. Therefore, 100K would be equivalent to 1.44 GHz on the Quantum temperature scale.
The Quantum temperature scale would be an imaginative and precise way of measuring temperature, as it would not be based on human reference points or the properties of substances but rather the unique properties of atoms.
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A flat, square surface with side iengen 4.80 cm is in tha xy-plane at z=0. Calculate the magnifude of the flux through this surface produced by a magnetic field
H
3
=(0.150 T)i+(0.250 T)j+(0.475 T)k. Fxpress your answer in webers.
Given data:A flat, square surface with side iengen 4.80 cm is in tha xy-plane at z=0.
The magnetic field,
H3 = (0.150 T)i + (0.250 T)j + (0.475 T)k.
To calculate:The magnitude of the flux through this surface produced by a magnetic field.
First, let's calculate the area of the given square surface.
A = side2= (4.80 cm)2= 23.04 cm2 = 0.002304 m2
The flux is calculated by the formula,
φ = B .
Awhere B is the magnetic field and A is the area of the surface. As we need to calculate the magnitude of flux through the given surface. Therefore, we use the formula as,
φ = ∣B∣. ∣A∣. cos θ
As the surface is in the xy-plane, so its normal vector n is in the direction of z-axis and makes an angle of 90° with the direction of magnetic field vector,
H3.cosθ = cos90° = 0So,φ = ∣B∣. ∣A∣. cos θ= ∣B∣. ∣A∣ × 0= 0
Weber (Wb)Hence, the magnitude of the flux through this surface produced by the given magnetic field is 0 Weber (Wb).
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what is the fractional decrease in amplitude per cycle?
Fractional decrease in amplitude per cycle is the percentage decrease of amplitude per cycle.
What is amplitude?The amplitude of a wave refers to the maximum displacement of a point on a wave from its resting position. In other words, it is the height of a wave, or how far it deviates from its undisturbed position.What is fractional decrease?The fractional decrease of a wave's amplitude is the percentage decrease in amplitude from the original value. It is also known as the damping ratio and is denoted by ζ. The formula for calculating the fractional decrease in amplitude per cycle is as follows:ζ= (a - b) / a,
Where a is the initial amplitude and b is the amplitude after a cycle.
For example, if a wave has an initial amplitude of 10 cm and a final amplitude of 8 cm after one cycle, then the fractional decrease in amplitude is:ζ= (10 - 8) / 10= 0.2 or 20%Therefore, the fractional decrease in amplitude per cycle is 20%.
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What is the wavelength of the photon with energy E=3.3×10
−18
J. Use nm (nanometer) for the unit of the wavelength. Question 10 1pts Free electrons that are ejected from a filament by thermionic emission is accelerated by 6.4kV of electrical potential difference. What is the kinetic energy of an electron after the acceleration? Answer in the unit of eV.
To calculate the wavelength of a photon given its energy, you can use the following formula: E = hc/λ
λ = hc/E
Substituting the given values:
λ = (6.626 × 10^-34 J·s × 3 × 10^8 m/s) / (3.3 × 10^-18 J)
Simplifying the expression:
λ = (6.626 × 3) / 3.3 × 10^(-34 + 8 + 18)
λ ≈ 6.03 × 10^-7 m
To convert this to nanometers, we multiply by 10^9:
λ ≈ 6.03 × 10^(-7 + 9) nm
λ ≈ 603 nm
Therefore, the wavelength of the photon with energy E = 3.3 × 10^-18 J is approximately 603 nm. Moving on to the second question, to calculate the kinetic energy of an electron accelerated by an electrical potential difference.
Kinetic energy (K.E.) = qV
Substituting the given values:
K.E. = (1.6 × 10^-19 C) × (6.4 × 10^3 V)
Simplifying the expression:
K.E. = 10.24 × 10^(-13) eV
K.E. ≈ 10.24 × 10^(-13) eV
Therefore, the kinetic energy of an electron after acceleration by 6.4 kV of electrical potential difference is approximately 10.24 × 10^(-13) eV.
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Two identical traveling waves of amplitude 0.7 m, moving in the same direction, are out of phase by Pi/6rad. Find the amplitude of the resultant wave.
The amplitude of the resultant wave is 1.4 m.
To find the amplitude of the resultant wave, we need to consider the interference of the two traveling waves. Given that the waves are identical in amplitude (0.7 m) and are out of phase by π/6 radians, we can use the principle of superposition to determine the resultant amplitude.
When two waves interfere constructively, their amplitudes add up, and when they interfere destructively, their amplitudes cancel out. In this case, since the waves are out of phase, they will interfere constructively.
To determine the amplitude of the resultant wave, we can use the formula:
Resultant amplitude = √(Amplitude1^2 + Amplitude2^2 + 2 * Amplitude1 * Amplitude2 * cos(Δφ))
Where Amplitude1 and Amplitude2 are the amplitudes of the two waves, and Δφ is the phase difference between them.
Plugging in the given values, we have:
Resultant amplitude = √((0.7 m)^2 + (0.7 m)^2 + 2 * (0.7 m) * (0.7 m) * cos(π/6))
Simplifying the expression, we find:
Resultant amplitude ≈ √(0.49 m^2 + 0.49 m^2 + 2 * 0.49 m^2 * cos(π/6))
Resultant amplitude ≈ √(1.96 m^2 + 0.98 m^2)
Resultant amplitude ≈ √(2.94 m^2)
Resultant amplitude ≈ 1.4 m
Therefore, the amplitude of the resultant wave is 1.4 m.
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A 789 kg car starts from rest and accelerates uniformly over a time of 5 seconds for a distance of 450 km. Determine the force exerted by the car.
The force exerted by the car is approximately 28,404,000 Newtons. This force is responsible for the acceleration of the car during the 5-second time interval and the distance traveled.
To determine the force exerted by the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:
Force = mass * acceleration
Given that the car has a mass of 789 kg, we need to find the acceleration it undergoes. To calculate the acceleration, we can use the equation of motion:
distance = (1/2) * acceleration * time^2
In this case, the distance is 450 km, which is 450,000 meters, and the time is 5 seconds. Rearranging the equation, we can solve for acceleration:
acceleration = (2 * distance) / (time^2)
Substituting the given values:
acceleration = (2 * 450,000 m) / (5 s)^2
= 36,000 m/s^2
Now that we have the acceleration, we can calculate the force exerted by the car:
Force = mass * acceleration
= 789 kg * 36,000 m/s^2
= 28,404,000 N
Therefore, the force exerted by the car is approximately 28,404,000 Newtons. This force is responsible for the acceleration of the car during the 5-second time interval and the distance traveled.
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Starting from rest, a motorboat travels around a circular path of r = 50 m at a speed that increases with time, v = 0.2 t^2 m/s. what is the magnitude of its total acceleration at t = 3 s? 8 m/s^2 O 1.2 m/s^2 O 6.2 m/s^2 O 5.02 m/s^2
the magnitude of the total acceleration of the motorboat at t = 3 s is approximately 1.27 m/s². Therefore, the correct option is 1.2 m/s².
Substituting the given velocity function and radius into the centripetal acceleration formula:
ac = (0.2t²)² / 50 = 0.04t⁴ / 50 m/s²
At t = 3 s, we can calculate the tangential acceleration (at) and the centripetal acceleration (ac):
at = 0.4(3) = 1.2 m/s²
ac = 0.04(3)⁴ / 50 ≈ 0.432 m/s²
To find the total acceleration (a), we can use the Pythagorean theorem:
a = √((at)² + (ac)²)
= √(1.2² + 0.432²)
≈ √(1.44 + 0.186624)
≈ √1.626624
≈ 1.27 m/s²
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A car is driving back and forth along the r axis. The position of the car is x(t)=t
2
−t
3
. (a) What is the instantaneous velocity of the car v(t) ? (b) What is the instantancous ncccleration of the car a(t)? (c) Between t=0 and t=2, when is tlo position of the car at ita maximum? (d) What is the displacement of the car from t=0 to t=2? (c) What is the average velocity of the car from f=0(v)=2 ? (f) What is the average acceleration of the car from t=0+0t=2 ? (g) Sketch the function x(t) from t=0 to t=2.
(a) The instantaneous velocity of the car, v(t) is given by the derivative of its position with respect to time, that isv(t) = dx(t)/dt= 2t - 3t². Thus, the instantaneous velocity of the car is 2t - 3t².
(b) The instantaneous acceleration of the car, a(t) is given by the derivative of its velocity with respect to time, that is,a(t) = dv(t)/dt= d/dt(2t - 3t²) = 2 - 6tThus, the instantaneous acceleration of the car is 2 - 6t.
(c) The position of the car is maximum when the velocity is equal to zero. Thus, 2t - 3t² = 0 or t = 0 or t = 2/3. Since the velocity is increasing from negative to positive values, this means that the position of the car is maximum at t = 2/3.
(d) The displacement of the car from t = 0 to t = 2 is given by the definite integral of its velocity over that interval, that is,Δx = ∫(v(t) dt) between 0 and 2.Δx = ∫(2t - 3t² dt) between 0 and 2Δx = [t² - t³] between 0 and 2Δx = 4 - 8/3 = 4/3.
(e) The average velocity of the car from t = 0 to t = 2 is given by the ratio of the displacement to the time interval, that is,v(avg) = Δx/Δt = (4/3)/(2 - 0) = 2/3.
(f) The average acceleration of the car from t = 0 to t = 2 is given by the ratio of the change in velocity to the time interval, that is,a(avg) =[tex]Δv/Δt = (v(2) - v(0))/(2 - 0)a(avg) = (2(2) - 3(2)² - 2(0) + 3(0)²)/(2 - 0)a(avg) = -4/2 = -2.[/tex]
(g) The function x(t) from t = 0 to t = 2 is shown below.
The axis on the left is the y-axis and the axis on the right is the x-axis.
The function is x(t) = t² - t³.
The maximum point on the graph is at t = 2/3 and x = 4/27.
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Three strain gauges were arranged in the form of a rectangular rosette and positioned on a test surface. The measured strains were as follows: & 1 = 200x 106 &2 = 100 x 106 &3 = 50 x 106 Determine a) the principal strains and the principle stresses b) the direction of the greater principal strain relative to gauge 1 and sketch the Mohr strain circle. Take the Young Modulus of Elasticity value to be E = 200 GN/m² and Poisson's ratio u = 0.28.
a) The correct values for the principal strains are:
ε₁ = 261.803 x 10⁻⁶ε₂ = 38.197 x 10⁻⁶The correct values for the principal stresses are:
σ₁ = 1197.674 MPaσ₂ = -697.674 MPab) The correct direction of the greater principal strain relative to gauge 1 is approximately 7.03 degrees.
Please note that the values provided earlier in the answer were incorrect, and these revised values are the accurate ones based on the calculations.
To find the principal strains, we use the equation:
ε = [(ε1 + ε2)/2] ± √[(ε1 - ε2)/2]² + ε3²
Where ε1, ε2, and ε3 are the strains measured by the gauges. Substituting the values, we get:
ε = [(200 x 106 + 100 x 106)/2] ± √[(200 x 106 - 100 x 106)/2]² + (50 x 106)²
ε = 150 x 106 ± 111.803 x 106
Therefore, the principal strains are 261.803 x 106 and 38.197 x 106.
To find the principal stresses, we use the equation:
σ = (E/[(1+u)(1-2u)]) x [(ε1 + ε2) ± √[(ε1 - ε2)² + 4ε3²]]
Substituting the values, we get:
σ = (200 x 109/[(1+0.28)(1-2(0.28))]) x [(200 x 106 + 100 x 106) ± √[(200 x 106 - 100 x 106)² + 4(50 x 106)²]]
σ = 1197.674 MPa and -697.674 MPa
Therefore, the principal stresses are 1197.674 MPa and -697.674 MPa.
To find the direction of the greater principal strain relative to gauge 1, we use the equation:
tan(2θ) = [(2ε1 - ε2 - ε3)/√[(ε1 - ε2)² + 4ε3²]]
Substituting the values, we get:
tan(2θ) = [(2(200 x 106) - 100 x 106 - 50 x 106)/√[(200 x 106 - 100 x 106)² + 4(50 x 106)²]]
tan(2θ) = 0.2679
Therefore, 2θ = 14.06° and θ = 7.03°.
To sketch the Mohr strain circle, we plot the principal strains on the x and y axes and the corresponding principal stresses on the vertical axis. We then draw a circle with radius equal to half the difference between the principal stresses. The circle intersects the vertical axis at the average of the principal stresses. The point on the circle corresponding to the greater principal strain gives the direction of the maximum shear stress.
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Which of the following takes place when a transverse pulse wave traveling down a string is reflected off of a fixed end of a string? harmonics overtones phase reversal no phase reversal
When a transverse pulse wave traveling down a string is reflected off of a fixed end of a string, a phase reversal occurs. The reflected wave is inverted when it comes back.
This means that the crests of the wave become troughs and the troughs become crests.
A transverse wave on a string is where the particles of the medium (string) vibrate perpendicular to the direction the wave is traveling. The reflection of a wave can occur when a wave encounters a new medium and changes direction, such as when light reflects off a mirror.
When a wave reflects off of a fixed end of a string, the wave is reversed and reflected back along the same string. This is called a fixed boundary condition.
There are two different types of boundary conditions.
A fixed boundary is when the string is anchored at both ends, and the ends of the string can’t move up and down.
When the pulse wave hits this fixed boundary, it will bounce back with a phase reversal, meaning that the wave will be inverted and will return to its original direction of travel with a reflected wave.
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A phase reversal occurs when a transverse pulse wave reflects off a fixed end of a string, causing the wave to reflect back along the string in opposite direction while inverting its wave disturbance pattern.
Explanation:When a transverse pulse wave traveling down a string reflects off a fixed end, a phase reversal takes place. This is a 180° change in phase with respect to the incident wave, as opposed to no phase change occurring when reflecting off a free end. During a phase reversal, the incident pulse or wave that travels down the string reflects back along the string in the opposite direction, with an inversion in its wave disturbance pattern. Nodes, where the wave disturbance is zero, appear at the fixed ends where the string is immobile. This phenomenon, where standing waves are created due to reflections of waves from the ends of the string, is common in stringed musical instruments, where the wave reflection is regulated by the boundary conditions of the system.
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John always paddles his canoe at constant speed v with respect to the still water of a river. One day, the river current was due west and was moving at a constant speed that was a little less than v with respect to that of still water. John decided to see whether making a round trip across the river and back, a north-south trip (in which he paddles in the north/south direction, but doesn't actually travel in either the north or south direction, respectively), would be faster than making a round trip an equal distance east-west. What was the result of John's test? The time for the north-south trip was greater than the time for the east-west trip. One cannot tell because the exact speed of the river with respect to still water is not given. The time for the north-south trip was equal to the time for the east-west trip. The time for the north-south trip was less than the time for the east-west trip.
John always paddles his canoe at a constant speed v with respect to the still water of a river. One day, the river current was due west and was moving at a constant speed that was a little less than v with respect to that of still water.
John decided to see whether making a round trip across the river and back, a north-south trip (in which he paddles in the north/south direction, but doesn't actually travel in either the north or south direction, respectively), would be faster than making a round trip an equal distance east-west. We have to find out the result of John's test.The time for the north-south trip was equal to the time for the east-west trip is the result of John's test.What we can infer from the given problem is that John paddles his canoe at a constant speed v with respect to the still water of a river.
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A wave along a string has the following equation where x is in metres and t is in seconds. y = 0.16 sin(34 t - 4.4 x) mm Find (a) The amplitude of the wave Number: __________ Units: __________ (b) The frequency of the wave Number: __________ Units: __________ (c) The wavelength of the wave Number: __________ Units: __________ (d) The speed of the wave Number: __________ Units: __________
(a) The amplitude of the wave is 0.16 mm. Amplitude is the maximum displacement of a particle from its position of rest, in simple harmonic motion. Here, it is the maximum value of y, which is 0.16 mm.
(b) The frequency of the wave is 17 Hz. The general equation of a wave is y = A sin(ωt - kx + φ) .Comparing this with the given equation, we can see that ω = 34, which is the angular frequency. The frequency f is given by the relation f = ω / 2π = 34 / (2 × π) ≈ 5.41 Hz.
But note that the value of the argument of the sine function, 34 t - 4.4 x, must be in radians.
Hence, we can convert 5.41 Hz to its radian measure by multiplying it by 2π. This gives us the frequency of the wave in rad/s, which is approximately 34 rad/s.
(c) The wavelength of the wave is 0.72 m. Wavelength λ is given by the formula λ = 2π / k, where k is the wave number. Comparing the given equation with the general equation of a wave, we can see that k = 4.4.
Hence, we have λ = 2π / k = 2π / 4.4 ≈ 1.44 m. But note that the wavelength is given in metres, not millimetres. So, the wavelength of the wave is 1.44 m.
(d) The speed of the wave is 24.48 m/s. The speed v of a wave is given by the relation v = ω / k.
We have already calculated the values of ω and k in parts (b) and (c).
So, we can substitute these values to get the speed of the wave: v = ω / k = 34 / 4.4 ≈ 7.73 m/s.
However, note that the units of v are m/s, not mm/s.
Hence, we need to convert 7.73 m/s to mm/s by multiplying it by 1000. This gives us the speed of the wave in mm/s, which is approximately 7730 mm/s.
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Problem 12: An electron moves in the positive
x-direction at 3x106 m/s measured within precision of 0.10%.
Find
uncertainty in measuring its position assuming its going in a
straight fashion.
The electron is moving in the positive x-direction at a velocity of 3 × 106 m/s. The precision is 0.10%. To find: Uncertainty in measuring its position.
Uncertainty principle: The product of uncertainty in position and the uncertainty in momentum of a particle is always greater than or equal to Planck's constant.Δx.Δp ≥ h / 4π.
The momentum of the electron can be calculated using its mass and velocity as follows:p = mv where,m = mass of the electron = 9.1 × 10-31 kgv = velocity of the electron = 3 × 106 m/s.
Therefore,p = (9.1 × 10-31 kg) × (3 × 106 m/s)p = 27.3 × 10-25 kg m/s.
The uncertainty in momentum can be calculated as follows:Δp = (0.10 / 100) × pΔp = 0.10% of 27.3 × 10-25 kg m/sΔp = (0.10 / 100) × 27.3 × 10-25 kg m/sΔp = 0.0273 × 10-25 kg m/sΔp = 2.73 × 10-27 kg m/s.
Now, substituting the values of h and Δp in the uncertainty principle formula:
Δx.Δp ≥ h / 4πΔx ≥ h / 4πΔpΔx ≥ (6.626 × 10-34 J s) / 4π(2.73 × 10-27 kg m/s)Δx ≥ 6.626 × 10-34 J s / 4π(2.73 × 10-27 kg m/s)Δx ≥ 6.05 × 10-7 m.
Therefore, the uncertainty in measuring the position of the electron is 6.05 × 10-7 m.
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ou have a resistor of resistance 200 Ω , an inductor of inductance 0.400 H, a capacitor of capacitance 6.00 μF and a voltage source that has a voltage amplitude of 33.0 V and an angular frequency of 240 rad/s. The resistor, inductor, capacitor, and voltage source are connected to form an L-R-C series circuit.
part a.What is the impedance of the circuit?
part b.What is the current amplitude?
part c.What is the phase angle of the source voltage with respect to the current?
part d.
Does the source voltage lag or lead the current?
part e.
What is the voltage amplitude across the resistor?
part f.
What is the voltage amplitude across the inductor?
part g.
What is the voltage amplitudes across the capacitor?
part h. Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.
part g.
a) The impedance of the L-R-C series circuit can be calculated using the formula:
�
=
�
2
+
(
�
�
−
�
�
)
2
Z=
R
2
+(X
L
−X
C
)
2
Where:
�
Z is the impedance of the circuit.
�
R is the resistance of the resistor.
�
�
X
L
is the reactance of the inductor.
�
�
X
C
is the reactance of the capacitor.
In this case,
�
=
200
R=200 Ω,
�
�
=
�
�
=
(
240
rad/s
)
(
0.400
H
)
X
L
=ωL=(240rad/s)(0.400H), and
�
�
=
1
�
�
=
1
(
240
rad/s
)
(
6.00
×
1
0
−
6
F
)
X
C
=
ωC
1
=
(240rad/s)(6.00×10
−6
F)
1
. By substituting these values into the formula, you can calculate the impedance of the circuit.
b) The current amplitude can be calculated using Ohm's Law, which states that
�
=
�
�
I=
Z
V
, where
�
I is the current amplitude,
�
V is the voltage amplitude of the source, and
�
Z is the impedance of the circuit.
c) The phase angle of the source voltage with respect to the current can be calculated using the formula:
�
=
arctan
(
�
�
−
�
�
�
)
θ=arctan(
R
X
L
−X
C
)
d) If the phase angle (
�
θ) is positive, it means that the source voltage leads the current. If
�
θ is negative, it means that the source voltage lags the current.
e) The voltage amplitude across the resistor (
�
�
V
R
) can be calculated using Ohm's Law:
�
�
=
�
⋅
�
V
R
=I⋅R.
f) The voltage amplitude across the inductor (
�
�
V
L
) can be calculated using the formula:
�
�
=
�
⋅
�
�
V
L
=I⋅X
L
.
g) The voltage amplitude across the capacitor (
�
�
V
C
) can be calculated using the formula:
�
�
=
�
⋅
�
�
V
C
=I⋅X
C
h) The voltage amplitude across the capacitor can be greater than the voltage amplitude across the source in a series L-R-C circuit because the capacitor's reactance (
�
�
X
C
) can be larger than the reactance of the inductor (
�
�
X
L
). This can result in a higher voltage drop across the capacitor compared to the source voltage. Additionally, the impedance of the circuit depends on the individual values of the resistor, inductor, and capacitor, which can contribute to different voltage amplitudes across the components.
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If a 221.7-kg weight attached to a paddle wheel in oil falls from rest to 3.000 m/s and the work of the falling weight is transferred to the water [use water's specific heat =4182 J/(kgK) ] with nearly no loss to other forms of energy, how many kelvin of temperature does the work done by the fall raise 1.5 kg of water?
The work done by the fall raises the temperature of 1.5 kg of water by approximately 0.15 K.
To determine the temperature increase caused by the work done by the falling weight on the water, we need to calculate the amount of thermal energy transferred to the water. The thermal energy transferred can be calculated using the equation:
Q = mcΔT
where Q is the thermal energy transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change.
Given:
Mass of water (m) = 1.5 kg
Specific heat capacity of water (c) = 4182 J/(kg·K)
To calculate the thermal energy transferred, we need to determine the work done by the falling weight. The work done is given by the equation:
W = ΔKE
where W is the work done, and ΔKE is the change in kinetic energy of the weight.
The change in kinetic energy can be calculated using the equation:
ΔKE = 0.5m[tex]v^{2}[/tex]
where m is the mass of the weight and v is its velocity.
Given:
Mass of weight (m) = 221.7 kg
Initial velocity (v₁) = 0 m/s
Final velocity (v₂) = 3.000 m/s
Calculating the change in kinetic energy:
ΔKE = 0.5 * 221.7 kg * (3.000 m/[tex]s^{2}[/tex])
Calculating the result:
ΔKE = 997.65 J
Now, we can calculate the thermal energy transferred to the water:
Q = mcΔT
Rearranging the equation to solve for ΔT:
ΔT = Q / (mc)
Substituting the known values:
ΔT = 997.65 J / (1.5 kg * 4182 J/(kg·K))
Calculating the result:
ΔT ≈ 0.15 K
Therefore, the work done by the fall raises the temperature of 1.5 kg of water by approximately 0.15 K.
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A particle moves in a straight line with a constant acceleration of 4.88 m/s2 in the positive x direction.
a) If its initial velocity is 2.19 m/s in the positive x direction, then how long, in seconds, does it take to move 6.23 m?
b) What is the particle’s final velocity in m/s?
The time taken to move 6.23 m is approximately 1.19 seconds. The particle's final velocity is 8.06 m/s.
Initial velocity (u) = 2.19 m/s
Acceleration (a) = 4.88 m/s²
Distance (s) = 6.23 m
To find:Time taken (t) = ?
We know, v² = u² + 2as
Where,v = final velocity = ?
u = initial velocity = 2.19 m/s
a = acceleration = 4.88 m/s²
s = distance = 6.23 m
Let's find the final velocity,v² = u² + 2as
v² = (2.19)² + 2(4.88)(6.23)
v² = 4.7961 + 60.3248
v² = 65.1209
v = √65.1209
v ≈ 8.06 m/s
So, the final velocity of the particle is approximately 8.06 m/s.
a) Now, let's find the time taken,t = (v - u) / at
t = (8.06 - 2.19) / (4.88)
t ≈ 1.19 s
Therefore, the time taken to move 6.23 m is approximately 1.19 seconds.
b) The particle's final velocity is 8.06 m/s.
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A boat that travels at 4.0 m/s in still water sets out across a river that is flowing at 2.5 m/s. a) What heading would the boat need to take in order to travel straight across the river? ( 2 pts) b) If the river is 50.0 m wide where will the boat land if it aims straight across the river?
What heading would the boat need to take in order to travel straight across the river?
To travel straight across the river, the boat must aim directly perpendicular to the current because the boat's heading will be equal to the angle that the boat forms with the current plus 90°.
Let h be the heading the boat needs to take to travel straight across the river.
Since the sine of an angle is the opposite side over the hypotenuse, we can determine h as follows:
[tex]$$\sin h=\frac{2.5}{4}$$ $$h=\sin^{-1} (\frac{2.5}{4})$$ $$h = 38.66^{\circ}$$[/tex]
the boat must head 38.66° upstream to travel straight across the river.
If the river is 50.0 m wide where will the boat land if it aims straight across the river?
The boat's velocity relative to the river is the difference between its velocity in still water and the velocity of the river.
To determine how long it takes the boat to cross the river, we first need to determine the boat's velocity relative to the river.
[tex]$$v_{BR} = v_{BW} - v_R$$[/tex]
where [tex]$v_{BR}$[/tex] is the velocity of the boat relative to the river,
[tex]$v_{BW}$[/tex] is the velocity of the boat in still water, and[tex]$v_R$[/tex]is the velocity of the river.
[tex]$$v_{BR} = 4 - 2.5 = 1.5 m/s$$[/tex]
We can now calculate how long it will take the boat to cross the river.
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