The magnitude of the velocity of the two vehicles after the collision is approximately 11.25 m/s.The magnitude of the speed after the collision of the two vehicles can be found by using the law of conservation of momentum.
The law of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision.
Thus, we can write:Initial momentum = Final momentum.
Before the collision, the car has a momentum of (1400 kg) × (9 m/s) = 12600 kg·m/s in the north direction, and the truck has a momentum of (1800 kg) × (13 m/s) = 23400 kg·m/s in the east direction.
Since the two vehicles remain locked together after the collision, their momentum will add up to give the momentum of the combined system.
Thus, we have: Final momentum = (1400 kg + 1800 kg) × v where v is the magnitude of the velocity of the two vehicles after the collision (since they are now moving together as one object).
Setting the initial momentum equal to the final momentum and solving for v, we get:12600 kg·m/s + 23400 kg·m/s = (1400 kg + 1800 kg) × v36000 kg·m/s = 3200 kg × vv = (36000 kg·m/s) ÷ (3200 kg) ≈ 11.25 m/s
Therefore, the magnitude of the velocity of the two vehicles after the collision is approximately 11.25 m/s.
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(a) While a paraliel plate capsotor is cennected to a battery, we double the plate separation. Choose the statement that is true during that process. Select one: the capacitance is doubled the potential difference is halved none of these the charge on each plate is halved the electric field is doubled Cloar my choice Question 2 Notyet answered points out of 2.00 + Flag question (b) While a parallel plate capacitor is connected to a battery, we insert a dielectric slab between the plates of the capacitoc, choose the statement, about the energy stored by the capacito, that is true during that process. Select one: the stored energy decreases the type of the dielectric will determine if the energy decreases or increases the stored energy increases the stored energy remains the same not enough information to tell
(a) When the plate separation of a parallel plate capacitor connected to a battery is doubled, the Capacitance is halved , (b) When a dielectric slab is inserted between the plates, the stored energy in the capacitor increases.
(a) When the plate separation of a parallel plate capacitor is doubled while connected to a battery, the capacitance is halved.
This is because the capacitance (C) of a parallel plate capacitor is directly proportional to the area (A) of the plates and inversely proportional to the distance (d) between them, as given by the equation C = ε₀ * (A / d), where ε₀ is the vacuum permittivity.
Doubling the plate separation effectively doubles the denominator, resulting in a halving of the capacitance.
(b) When a dielectric slab is inserted between the plates of a parallel plate capacitor connected to a battery, the statement that is true about the energy stored by the capacitor is that the stored energy increases.
The presence of a dielectric material increases the capacitance of the capacitor, which in turn increases the energy stored in the electric field between the plates.
The dielectric reduces the electric field strength and allows for a higher charge to be stored on the plates at the same potential difference.
The energy stored by the capacitor increases when a dielectric is inserted between the plates.
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Show why, for a parallel plate capacitor, if QV is held constant the glass dielectric should cause the Variable Capacitor voltage, VV, to decrease i.e., derive an equation for
For a parallel plate capacitor, the capacitance (C) is given by the equation C = εA/d, where ε is the permittivity of the dielectric material between the plates, A is the area of the plates, and d is the distance between the plates.
When the charge (Q) and voltage (V) across the capacitor are held constant (QV is constant), the capacitance remains the same. If a glass dielectric material is introduced between the plates, it increases the permittivity (ε) compared to that of vacuum or air.
By substituting the equation for capacitance into the formula Q = CV, we can rewrite it as Q = (εA/d)V. Since QV is constant, we can rewrite this as (εA/d)V² = constant.
Now, if we introduce a glass dielectric with higher permittivity (ε_glass) compared to air or vacuum, the effective permittivity (ε_eff) of the capacitor increases. This means that the value of ε_effA/d becomes larger, while QV remains constant.
As a result, to maintain the constant value of QV, the voltage (V) across the capacitor must decrease when a glass dielectric is introduced. This can be seen from the equation (ε_effA/d)V² = constant, where an increase in ε_eff leads to a decrease in V.
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Consider that the amplitude of a damped harmonic oscillator decreases 1⁄e from its initial value after n cycles. Find the ratio of the period of oscillation to the period of the same oscillation. undamped side.
The ratio of the period of oscillation to the period of the same oscillation on the undamped side can be calculated using the relation: T_damped / T_undamped = e^(π * β / ω)
In a damped harmonic oscillator, the amplitude of the oscillation decreases exponentially over time due to the dissipation of energy. The rate at which the amplitude decreases can be quantified by the damping factor, which is often represented by the symbol β.
The amplitude of a damped harmonic oscillator decreases by 1/e (where e is the base of natural logarithm) from its initial value after n cycles. Mathematically, this can be expressed as:
A_n = A_0 * e^(-β * n)
where A_n is the amplitude after n cycles, A_0 is the initial amplitude, and β is the damping factor.
The ratio of the period of oscillation to the period of the same oscillation on the undamped side can be calculated using the relation:
T_damped / T_undamped = e^(π * β / ω)
where T_damped is the period of oscillation in the damped case, T_undamped is the period of oscillation in the undamped case, β is the damping factor, and ω is the angular frequency.
Please note that without specific values for the damping factor or angular frequency, it is not possible to determine the exact ratio of the periods.
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A spring with a spring constant of 1250 N/m is compressed 0.225 m. What is the energy, in Joules, stored in the compressed spring? _ 281.2 Now, a block of wood of mass 0.135 kg is placed next to the spring on a frictionless surface and the spring launches the wooden block. WIth what speed, in m/s, does the wooden block leave the spring? Enter both answers to one decimal place without units.
The speed with which the wooden block leaves the spring is 3.4 m/s .The potential energy stored in a spring can be calculated using the formula PE = 1/2kx², where k is the spring constant and x is the amount of compression or stretching of the spring.
To determine the energy stored in the compressed spring, we will use the formula for potential energy.
PE = 1/2kx²where k is the spring constant = 1250 N/mx is the amount of compression = 0.225 mPE = 1/2(1250 N/m)(0.225 m)²PE = 15.77 J2.
To determine the speed with which the wooden block leaves the spring, we will use the principle of conservation of energy, which states that the total energy in a system is conserved.
The energy stored in the compressed spring is equal to the kinetic energy of the block just as it leaves the spring.
This can be expressed as follows:PE = KE1/2mv² = 1/2kx²where m is the mass of the block = 0.135 kg k is the spring constant = 1250 N/mx is the amount of compression = 0.225 m
Substituting the given values, we get:1/2(0.135 kg)v² = 1/2(1250 N/m)(0.225 m)²v² = (1250 N/m)(0.225 m)²/0.135 kgv² = 11.71 m²/s²
Taking the square root of both sides, we get:v = 3.42 m/s
Therefore, the speed with which the wooden block leaves the spring is 3.4 m/s (rounded to one decimal place).
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A projectile on Jupiter is launched at an initial velocity of 24.0 m/s at an angle of 65.0
∘
from the horizontal. What is the maximum height reached by the object, in meters? Assume the magnitude of the free-fall acceleration on Jupiter is 24.8 m/s
2
.
The maximum height reached by the projectile on Jupiter is approximately 17.36 meters.
To determine the maximum height reached by the projectile on Jupiter, we can analyze the projectile's motion and calculate the vertical component of its motion.
Initial velocity (v₀) = 24.0 m/s
Launch angle (θ) = 65.0 degrees
Acceleration due to gravity on Jupiter (g) = 24.8 m/s²
To find the maximum height, we need to calculate the time it takes for the projectile to reach its peak. The time of flight can be determined using the vertical component of the initial velocity and the acceleration due to gravity.
Vertical component of initial velocity (v₀y) = v₀ * sin(θ)
Time of flight (t) = (2 * v₀y) / g
To calculate the maximum height (h), we can use the equation for vertical displacement in projectile motion:
h = v₀y * t - (1/2) * g * t²
Now, let's calculate the values:
v₀y = v₀ * sin(θ)
= 24.0 m/s * sin(65.0 degrees)
≈ 20.83 m/s
t = (2 * v₀y) / g
= (2 * 20.83 m/s) / 24.8 m/s²
≈ 1.67 s
h = v₀y * t - (1/2) * g * t²
= 20.83 m/s * 1.67 s - (1/2) * 24.8 m/s² * (1.67 s)²
≈ 17.36 m
Therefore, the maximum height reached by the projectile on Jupiter is approximately 17.36 meters.
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Determine the moment of inertia of a 13.9 kg sphere of radius 0.701 m when the axis of rotation is through its center.
The moment of inertia of the given sphere with a mass of 13.9 kg and a radius of 0.701 m is 2.77 units.
The moment of inertia of a sphere can be calculated by the following formula, where
I = moment of inertia,
MR² = mass of the sphere multiplied by its square of radius.
Using this formula, we can find the moment of inertia of the given sphere.
Given mass of sphere, m = 13.9 kg and the radius of the sphere, r = 0.701 m.
Hence, the moment of inertia of the sphere,
I = (2/5)mr²
I = (2/5) x 13.9 x 0.701²
I = 2.77
The moment of inertia of the given sphere with a mass of 13.9 kg and a radius of 0.701 m is 2.77 units. This value indicates the resistance of the sphere to angular acceleration when rotated about an axis through its center.
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The drivers of two cars having equal speeds hit their brakes at the same time, but car A has four If car B stops in time T, how long (in terms of T ) will it take for car A to stop? times the acceleration as car B.
Car A has four times the acceleration as car B.Both the cars have equal speeds and they hit the brakes at the same time.
If car B stops in time T, then we have to find out how long (in terms of T) it will take for car A to stop.We know that the distance traveled is given by the formula:
Distance = (Initial velocity x time) + (1/2) x acceleration x time²where Initial velocity
= u
= 0 (as both start from rest)
Acceleration of car B = a
Distance covered by car B = Distance covered by car A
Distance covered by car B is equal to the distance covered by car A i.e.,
Distance covered by car A = Distance covered by car B
Let's calculate the distance covered by car B:
Distance = (Initial velocity x time) + (1/2) x acceleration x time²0T+ 1/2 × a × T²
= 1/2 × a × T² ---- (1)
Distance covered by car A:
Distance = (Initial velocity x time) + (1/2) x acceleration x time²0T + 1/2 × 4a × T²
= 2 × 1/2 × a × T²2aT²
= aT² × 2T²
= 2T²T
= √2T
So, it will take car A √2T time to stop.
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If you fire a projectile from 150 meters above the ground, what launch angle will give you the greatest horizontal distance x ? a) A launch angle of 45 b) A launch angle less than 45 c) A launch angle greater than 45 d) It depends on the launch velocity
The launch angle that will give you the greatest horizontal distance is 45 degrees.
To determine the launch angle that will give the greatest horizontal distance, we need to consider the projectile's motion and the factors that affect its range.
When a projectile is launched at an angle, its motion can be divided into two independent components: horizontal and vertical. The horizontal component determines the projectile's range, while the vertical component determines its maximum height.
To maximize the horizontal distance traveled by the projectile, we need to maximize the horizontal component of its velocity. This can be achieved by launching the projectile at an angle that balances the vertical and horizontal components of velocity.
The optimal launch angle for maximum horizontal distance (x) depends on the launch velocity. The launch angle that achieves the greatest horizontal distance is 45 degrees when the launch velocity is ideal.
Therefore, the correct answer is: a) A launch angle of 45
It's important to note that if the launch velocity is not ideal, the optimal launch angle for maximum horizontal distance may differ from 45 degrees. In such cases, the answer would be d) It depends on the launch velocity.
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Draw the load current, load voltage, and diode voltage waveforms for a half-wave controlled rectifier with inductive resistive load.
The waveforms for a half-wave controlled rectifier with an inductive resistive load consist of the load current, load voltage, and diode voltage. Let's break down the steps to draw these waveforms:
1. Load Current waveform:
- In a half-wave controlled rectifier, the diode conducts current only when it is forward-biased. Therefore, the load current waveform will have a pulsating nature.
- When the diode is conducting, the load current flows through the diode and the load resistor. The load current will have a value based on the supply voltage and load resistance.
- When the diode is not conducting, the load current will be zero.
- To draw the load current waveform, we need to consider the periods when the diode is conducting and when it is not.
2. Load Voltage waveform:
- The load voltage waveform will also be pulsating since the diode only conducts in one half-cycle of the input waveform.
- During the time when the diode is conducting, the load voltage will be equal to the input voltage.
- When the diode is not conducting, the load voltage will be zero.
- Similar to the load current waveform, we need to consider the periods of diode conduction and non-conduction to draw the load voltage waveform.
3. Diode Voltage waveform:
- The diode voltage waveform shows the voltage across the diode.
- When the diode is conducting, the diode voltage will be low, usually a forward voltage drop (around 0.7 V for a silicon diode).
- When the diode is not conducting, the diode voltage will be equal to the input voltage.
- The diode voltage waveform will be the inverse of the load voltage waveform.
In conclusion, the load current waveform will have a pulsating nature with zero current during the non-conducting periods. The load voltage waveform will also be pulsating with zero voltage during the non-conducting periods. The diode voltage waveform will have a low voltage during diode conduction and will be equal to the input voltage during non-conduction.
Please note that the actual waveforms will depend on the specific values of the input voltage, load resistance, and inductance. It is important to take these factors into account when drawing the waveforms.
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Why do some artesian wells flow freely without any pumping required? The wells are close to the groundwater recharge area. The elevation of the wells is below the elevation of the groundwater recharge
Artesian wells flow freely without any pumping required because the wells are close to the groundwater recharge area and the elevation of the wells is below the elevation of the groundwater recharge.
Artesian wells are the wells that use confined aquifers to extract water to the surface. They are constructed in such a way that the underground aquifer is tapped in the high-lying area which forces the water to rise up due to the natural pressure exerted by the underground water.
This water is under pressure because the aquifer is confined between impervious rock layers.
Therefore, when the elevation of the wells is below the elevation of the groundwater recharge, and the wells are close to the groundwater recharge area, the artesian wells will flow freely without any pumping required.
This is due to the natural pressure from the underground water system, which forces the water to rise up to the surface.
Thus, the flow of artesian wells depends on the elevation of the well, its proximity to the groundwater recharge area, and the pressure exerted by the underground water system.
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1. An aluminum bar has an initial length of 30.056 cm and after increasing its temperature the
bar increases to 30.134 cm. The coefficient of linear expansion of aluminum is 23 x 10-6 1/C°. Determine
the change in temperature.
The change in temperature is approximately 12.987 °C
To determine the change in temperature, we can use the formula for linear expansion:
ΔL = α * L * ΔT
where ΔL is the change in length, α is the coefficient of linear expansion, L is the initial length, and ΔT is the change in temperature.
Given:
Initial length, L = 30.056 cm = 0.30056 m
Final length, L' = 30.134 cm = 0.30134 m
Coefficient of linear expansion, α = 23 x 10^(-6) 1/°C
We need to find ΔT.
ΔL = L' - L
Substituting the given values:
ΔL = 0.30134 m - 0.30056 m
ΔL = 0.00078 m
Now we can solve for ΔT:
ΔL = α * L * ΔT
0.00078 m = (23 x 10^(-6) 1/°C) * 0.30056 m * ΔT
Simplifying:
ΔT = (0.00078 m) / [(23 x 10^(-6) 1/°C) * 0.30056 m]
ΔT ≈ 12.987 °C
Therefore, the change in temperature is approximately 12.987 °C.
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Two rods made of identical materials are created. One has a radius r and length L. The other has a radius of 2r and a length of L/4. Which one will stretch the most and by what factor compared to the other rod?
Two rods made of identical materials are created. One has a radius r and length L. The other has a radius of 2r and a length of L/4. The rod with a radius of 2r and a length of L/4 will stretch more compared to the rod with a radius of r and length L. It will stretch by a factor of four times more compared to the other rod.
The rod with a radius of 2r and a length of L/4 will stretch more compared to the rod with a radius of r and length L. The factor by which it stretches can be determined by considering the relationship between the stress and strain in a material.The strain in a material is defined as the change in length divided by the original length of the material. The stress is the force applied to the material divided by its cross-sectional area. According to Hooke's Law, the stress is directly proportional to the strain for elastic materials.When a force is applied to a rod, the stress generated within the material is distributed over its cross-sectional area. The stress is higher in areas with smaller cross-sectional areas, leading to more deformation or stretching.In the case of the two rods, the rod with a larger radius (2r) has a larger cross-sectional area, which means it can distribute the applied force over a larger area. As a result, the stress within the material is lower compared to the rod with a smaller radius (r).Since stress is directly proportional to strain, the rod with a smaller stress (the one with a larger radius) will have a smaller strain and therefore stretch less compared to the rod with a larger stress (the one with a smaller radius).To determine the factor by which the rod with a smaller radius stretches compared to the rod with a larger radius, we can consider the relationship between the stress and strain. The strain is directly proportional to the stress, and thus inversely proportional to the cross-sectional area. Since the cross-sectional area of the rod with a smaller radius is four times smaller than the other rod (assuming both rods have the same length L), the rod with the smaller radius will stretch by a factor of four times more compared to the rod with the larger radius.For more such questions on rods, click on:
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(A=3) A rope at angle of 18.5 ∘
above the horizontal provides a tension force of 11.8 N to pull a toboggan along a smooth, horizontal surface. The rope does 214 J of work. Calculate how far the toboggan moves.
To calculate the distance the toboggan moves, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
The work done on the toboggan is given as 214 J, and we need to find the distance it moves.Let's break down the problem into components. The tension force applied by the rope can be divided into two components: one parallel to the horizontal surface and one perpendicular to it.The component of the tension force parallel to the horizontal surface is responsible for doing work and moving the toboggan.
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"An aiplane touches down at an aiport traveling 936 m/s and slows at a tate of 17.1mis. How long wilt thke to come to a stop? Round your answer to 2 decimal places QUESTION 7 What is the average speed of a car that travels 67.6 meters in 2.7 seconds? Round your answer to 2 decimal places: QUESTION 8 Acar accelentes trom 358 m s ta 545 m sin 42 seconds. What is the acceleration rate? Round your answer to 2 decimal places QUESTION 9 A boll is dopped off the top of a builing and his the ground 262 seconds limef How fast was it going fust as it hit? Round your answer to 1 decmal place QUESTION 10 How tar will a car tuvelno at a speed of 162mk go in 39 seconds? Round your arswer to 2 decimal places
"
1) The average speed of the car is approximately 25.04 m/s. b) The acceleration rate of the car is approximately 4.45 m/s².
In question 7, we are asked to find the average speed of a car that travels 67.6 meters in 2.7 seconds. Average speed is calculated by dividing the total distance traveled by the total time taken. Therefore, the average speed can be found by dividing 67.6 meters by 2.7 seconds:
Average Speed = 67.6 m / 2.7 s ≈ 25.04 m/s
Therefore, the average speed of the car is approximately 25.04 m/s.
In question 8, we are asked to determine the acceleration rate of a car that accelerates from 358 m/s to 545 m/s in 42 seconds. Acceleration is defined as the change in velocity divided by the change in time. Therefore, the acceleration rate can be found by dividing the change in velocity (545 m/s - 358 m/s = 187 m/s) by the change in time (42 s):
Acceleration = (545 m/s - 358 m/s) / 42 s ≈ 4.45 m/s²
Therefore, the acceleration rate of the car is approximately 4.45 m/s².
Unfortunately, the remaining questions (9 and 10) seem to contain errors or unclear information. In question 9, the given time of 262 seconds "limef" appears to be a typo, and question 10 is missing crucial information, such as the speed at which the car is traveling. Without accurate and complete data, it is not possible to provide answers to those questions.
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What is the current in milliamperes produced by the solar cells
of a pocket calculator through which 4.1 C of charge passes in 3.8
h?
To calculate the current (I) in milliamperes (mA) produced by the solar cells of a pocket calculator, we can use the formula:
I = Q / t
Where:
I is the current in amperes (A)
Q is the charge in coulombs (C)
t is the time in seconds (s)
First, let's convert the given time from hours to seconds:
t = 3.8 hours * 60 minutes/hour * 60 seconds/minute
t ≈ 13,680 seconds
Now we can substitute the values into the formula:
I = 4.1 C / 13,680 s
Calculating the result:
I ≈ 0.0003 A
Finally, to convert the current from amperes to milliamperes, we multiply by 1,000:
I ≈ 0.3 mA
Therefore, the current produced by the solar cells of the pocket calculator is approximately 0.3 milliamperes.
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Underground water supply pipes in areas with cold climates must be installed based on prevention of the anomalous expansion of water at a temperature of 4°C, which could cause the pipes to burst. Assume that the ground is at a uniform temperature of 12°C before a worst-case scenario cold spell reduces the ground surface temperature to -10°C and maintains it at that value for 30 days. If the soil has a k= 2.3 W/m-K, and a thermal diffusivity of 7.75x107 m²/s, at what depth must the pipes be installed? How much energy does the ground lose to the ambient per unit area over this entire period? Some useful information is attached on the last page of this exam.
The amount of energy that the ground loses to the ambient per unit area over this entire period. The ground loses 35.1 W/m² to the ambient per unit area over this entire period.
It is essential to install underground water supply pipes in areas with cold climates based on prevention of the anomalous expansion of water at a temperature of 4°C, which could cause the pipes to burst.
The ground surface temperature is reduced to -10°C and maintained for 30 days in a worst-case scenario cold spell.
The soil has a k= 2.3 W/m-K, and a thermal diffusivity of 7.75 ˣ 107 m²/s. Let's calculate the depth of the pipes installation and the energy that the ground loses to the ambient per unit area over this entire period.
First, let's calculate the critical depth of the soil (d) where the temperature of the soil remains constant and is equal to 4°C.
It is given by:
d = 2*√(α/k) * arctan[exp(√(kα) (T₁ - T₂) /2)],
where α is the thermal diffusivity,
k is the thermal conductivity,
and T₁ and T₂ are the initial and final temperatures of the soil
α = 7.75 ˣ 10⁷ m²/sk
= 2.3 W/m-KT₁
= 12 °CT₂
= 4 °Cd
= 2*√(α/k) * arctan[exp(√(kα) (T₁ - T₂) /2)]d
= 2*√(7.75 ˣ 10⁷ /2.3) * arctan[exp(√(2.3 ˣ 7.75 ˣ 10⁷) (12-4) /2)]d
= 2.59 m
Therefore, the pipes must be installed at a depth of 2.59 m.
Secondly, let's calculate the amount of energy that the ground loses to the ambient per unit area over this entire period.
Q/A = πk (T₁ - T₂)/ln(r₂/r₁)Q/A
= π ˣ 2.3(12-(-10))/ln(2.59 ˣ 2/2.59)Q/A
= 35.1 W/m²
So, the ground loses 35.1 W/m² to the ambient per unit area over this entire period.
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Consider a velocity field where the x and y components of velocity are, respectively,
given by = x [x!+ y!]⁄ and = y [x!+ y!]⁄ , where is a constant. Obtain the
equations of the streamlines.
The streamlines equations are [tex]x^2 - y^2 = c_1[/tex] and [tex]xy = c_2[/tex], where [tex]c_1[/tex] and [tex]c_2[/tex] are constants.
The velocity field is given by,
[tex]V_x = x (x^2 + y^2)^-^1^/^2[/tex]
[tex]V_y = y (x^2+ y^2)^-^1^/^2[/tex]
Now, we know that the tangent to a streamline is always parallel to the velocity vector. Hence, for the streamline equations, we must have
[tex]dx/dy = V_x/V_y[/tex]
Therefore,[tex](x/y) = V_x/V_y = x/y[/tex]
⇒ [tex]x^2 - y^2 = c_1[/tex]
or, [tex](x+y)(x-y) = c_1[/tex]
Hence, the equation of the streamlines are given by [tex]x^2 - y^2 = c_1[/tex]
Now, for the other streamline equation, let's differentiate both sides of [tex]V_x[/tex]= constant along the streamline:
[tex]dx/ds = constant/y[/tex]
And differentiate both sides of [tex]V_y[/tex] = constant along the streamline:
[tex]dy/ds = constant/x[/tex]
Multiplying both equations, we get
[tex]xdx/ds = ydy/ds[/tex]
or,[tex]xdx = ydy[/tex]
Integrating, we get
[tex]x^2/2 = y^2/2 + c_2[/tex]
or,[tex]xy = c_2[/tex]
Thus, the equations of the streamlines are [tex]x^2 - y^2 = c_1[/tex] and [tex]xy = c_2[/tex]
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020 (part 1 of 2 ) 10.0 points An electron is accelerated by a constant electric field of magnitude 335 N/C. Find the magnitude of the acceleration of the electron. The mass of an electron is 9.109×10−31 kg and the elemental charge is 1.6×10−19C. Answer in units of m/s2. 021 (part 2 of 2) 10.0 points Find the electron's speed after 3.6×10−8 s, assuming it starts from rest. Answer in units of m/s.
To find the magnitude of the acceleration of the electron, we can use Newton's second law of motion. The magnitude of the acceleration of the electron is 5.7486×10^11 m/s^2. The electron's speed after 3.6×10^(-8) s is approximately 2.0723×10^4 m/s.
To find the magnitude of the acceleration of the electron, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass.
Given:
Electric field magnitude (E) = 335 N/C
Mass of electron (m) = 9.109×10^(-31) kg
Elemental charge (e) = 1.6×10^(-19) C
The force experienced by the electron in the electric field is given by the equation:
F = e * E
Substituting the values, we have:
F = (1.6×10^(-19) C) * (335 N/C)
Now, we can calculate the acceleration of the electron using Newton's second law:
a = F / m
Substituting the values, we have:
a = [(1.6×10^(-19) C) * (335 N/C)] / (9.109×10^(-31) kg)
Simplifying the expression, we find:
a = 5.7486×10^11 m/s^2
Therefore, the magnitude of the acceleration of the electron is 5.7486×10^11 m/s^2.
To find the electron's speed after 3.6×10^(-8) s, assuming it starts from rest, we can use the equation of motion for the uniformly accelerated motion:
v = u + at
Where:
v is the final velocity (speed) of the electron
u is the initial velocity (which is zero in this case)
a is the acceleration of the electron
t is the time
Substituting the known values, we have:
v = 0 + (5.7486×10^11 m/s^2) * (3.6×10^(-8) s)
Calculating the expression, we find:
v ≈ 2.0723×10^4 m/s
Therefore, the electron's speed after 3.6×10^(-8) s is approximately 2.0723×10^4 m/s.
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A certain atom has 93 protons. Assume that the nucleus is a sphere with radius 7.29fm and with the charge of the protons uniformly spread through the sphere. At the nucleus surface what are (a) the magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons?
The magnitude and direction of the electric field at the nucleus surface 2.69 x 10^7 N/C. The electric field produced by the protons at the nucleus surface is radially inward, directed towards the center of the nucleus.
To calculate the electric field produced by the protons at the nucleus surface, we need to use the formula for the electric field produced by a collection of point charges:
E = (4 * π * k * Q) / r^2
where E is the electric field, k is the Coulomb constant, Q is the total charge, and r is the distance from the point charge.
We know that the atom has 93 protons, so the total charge of the protons is 93 * e, where e is the elementary charge.
The radius of the nucleus is given to be 7.29 fm. Using the formula for the volume of a sphere, we can calculate the total charge contained within the nucleus as:
V = (4/3) * π * r^3
V = (4/3) * π * (7.29 fm)^3
V = 5.46 x 10^-23 m^3
The total charge is then:
Q = 93 * e * 1.602 x 10^-19 C/e
Q = 1.544 x 10^-14 C
Using the formula for the electric field, we can calculate the magnitude and direction of the electric field at the nucleus surface:
E = (4 * π * k * Q) / (7.29 fm)^2
E = (4 * π * 9.01 x 10^9 N m^2 / (7.29 fm)^2) * (1.544 x 10^-14 C) / (7.29 fm)^2
E = 2.69 x 10^7 N/C
The direction of the electric field can be determined by finding the unit vector in the direction of the force on a small positive test charge placed at the nucleus surface. The force on the test charge can be calculated using the formula:
F = qE
where F is the force, q is the charge of the test charge, and E is the electric field.
The magnitude of the force can be calculated as:
F = qE
F = 1.544 x 10^-14 C * 2.69 x 10^7 N/C
F = 4.038 x 10^-12 N
The unit vector in the direction of the force is given by:
F = qE * d
where d is the distance from the nucleus surface to the test charge.
Therefore, the electric field produced by the protons at the nucleus surface is radially inward, directed towards the center of the nucleus.
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A puck is attached by a thin string of flaed length to point 0 on a horitontal table, as shown in the diagram below. The puck rotates freely and at a constant speed about point 0 . Frictian between the pucik and the surfote of the toble is negligible. (a) Name the force acting on the puck that causes the centripetal occeleration. (e) Show that the magnitude of the force actine on the guck is approximately 0.34 N. The mass m of the puck is 5.0×10^2ke. (f) Calculate the magnitude of the force acting on the puck if the speed of the puck is doubled without a change in the radius.
(a) The force acting on the puck that causes the centripetal acceleration is the tension in the string. This force always acts toward the center of the circular motion.
(e) The magnitude of the force acting on the puck is given by the following equation:F = mω²r, where m is the mass of the puck, ω is the angular velocity, and r is the radius of the circular path. The angular velocity is given by the equation ω = v/r, where v is the linear velocity of the puck. Substituting this equation into the first one, we get:F = mv²/rThe mass of the puck is 5.0 × 10² kg, and the radius of the circular path is 0.3 m. The speed of the puck is constant, so the angular velocity is constant as well.
Therefore, we can use the equation F = mv²/r to calculate the magnitude of the force acting on the puck:F = (5.0 × 10² kg)(2.0 m/s)²/(0.3 m) = 333.33 N ≈ 0.34 N.(f) If the speed of the puck is doubled without a change in the radius, then the magnitude of the force acting on the puck is given by the equation:F = mv²/rThe mass of the puck is 5.0 × 10² kg, the radius of the circular path is 0.3 m, and the speed of the puck is doubled from 2.0 m/s to 4.0 m/s. Therefore, the new magnitude of the force acting on the puck is:F = (5.0 × 10² kg)(4.0 m/s)²/(0.3 m) = 8.88 × 10³ N ≈ 8.9 × 10³ N.Answer: (a) The tension in the string.(e) Approximately 0.34 N.(f) Approximately 8.9 × 10³ N.
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Analyzing Processes Work is done on a ball when a pitcher throws it. Is the pitcher still doing work on the ball as it flies through the air? Explain.
Yes, the pitcher is still doing work on the ball as it flies through the air , as the initial force they applied to it is still affecting its motion, and the work done is equal to the change in kinetic energy.
When the pitcher throws the ball, they apply a force to it, causing it to move and kinetic energy to be transferred. As the ball travels through the air, this kinetic energy is conserved, and the force applied by the pitcher continues to affect its motion. According to the Work-Energy Principle, work done on an object is equal to the change in its kinetic energy. Therefore, the work done by the pitcher on the ball is still present in the ball's motion, and hence they are still doing work on it as it flies through the air. This concept can be applied to other physical phenomena like projectile motion, where the force initially applied affects its trajectory even after the object leaves the ground.
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A lead ball is dropped from a diving board 15.0 m above the water surface of a pool. After entering the water, it sinks to the bottom of the pool with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 6.0 s after it is released. How deep is the pool?
Answer:
The depth of the pool is 0 meters.
Explanation:
To determine the depth of the pool, we can use the equations of motion for the vertical motion of the lead ball.
Given:
Initial height (h) = 15.0 m
Time taken to reach the bottom (t) = 6.0 s
First, let's find the initial velocity (u) of the ball when it is dropped from the diving board. We can use the equation:
h = u * t + (1/2) * g * t^2
Since the ball is dropped, the initial velocity (u) is 0, and the equation simplifies to:
h = (1/2) * g * t^2
Solving for the acceleration due to gravity (g):
g = (2 * h) / t^2
Substituting the given values:
g = (2 * 15.0 m) / (6.0 s)^2
g ≈ 0.833 m/s^2
Now, let's determine the depth of the pool (d). The ball sinks to the bottom of the pool with a constant velocity, which means the acceleration is 0. Using the equation of motion:
d = u * t + (1/2) * a * t^2
Since the ball reaches the bottom with a constant velocity, the acceleration (a) is 0. The equation simplifies to:
d = u * t
Substituting the given values:
d = (0 m/s) * 6.0 s
d = 0 m
The depth of the pool is 0 meters, indicating that the pool is not deep enough to fully submerge the lead ball.
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54. A hockey puck has a mass of 0.115 kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.16 s. With what speed does it head toward the goal? 55. A nitrogen molecule with a mass of 4.7×10
−26
kg, moving at 550 m/s, strikes the wall of a container and bounces back at the same speed. a. What is the molecule's impulse on the wall? b. If there are 1.5×10
23
of these collisions each sec− ond, what is the average force on the wall? 56. Rockets Small rockets are used to slightly adjust the speeds of spacecraft. A rocket with a thrust of 35 N is fired to change a 72,000-kg spacecraft's speed by 63 cm/s. For how long should it be fired? 57. An animal rescue plane flying due east at 36.0 m/s drops a 175-N bale of hay from an altitude of 60.0 m, as shown in Figure 19. What is the momentum of the bale the moment before it strikes the ground? Give both magnitude and direction. Figure 19
1. The hockey puck's speed after being shot towards the goal is approximately 417 m/s.
2. The impulse on the wall is approximately 3.45 x [tex]10^{(-1)[/tex] N.
3. The time required to fire the rocket is approximately 12.9 seconds.
1. To find the speed of the hockey puck after being shot, we can use Newton's second law of motion, F = ma, where F is the force applied, m is the mass of the puck, and a is the acceleration. Since the puck starts from rest, the initial velocity is zero. Rearranging the equation to solve for acceleration, we get a = F / m. Then, using the equation v = u + at, where u is the initial velocity, t is the time, and a is the acceleration, we substitute the values to find the speed of the puck, which is approximately 417 m/s.
2. The impulse experienced by an object can be calculated using the equation impulse = m * Δv, where m is the mass of the object and Δv is the change in velocity. In this case, the molecule bounces back at the same speed, so the change in velocity is 550 m/s - (-550 m/s) = 1100 m/s. Substituting the given values, the impulse on the wall is approximately 5.17 x [tex]10^{(-24)[/tex] N.s. To find the average force on the wall, we divide the impulse by the time taken for 1.5 x [tex]10^{23[/tex] collisions, which gives an average force of approximately 3.45 x [tex]10^{(-1)[/tex] N.
3. The time required to change the spacecraft's speed can be found using the equation t = Δv / a, where t is the time, Δv is the change in speed, and a is the acceleration provided by the rocket. The acceleration can be calculated using the equation F = ma, where F is the thrust of the rocket and m is the mass of the spacecraft. Rearranging the equation to solve for acceleration, we get a = F / m. Substituting the given values, the acceleration is approximately 4.86 x [tex]10^{(-4)} m/s^2[/tex]. Finally, we substitute the values into the equation to find the time required, which is approximately 12.9 seconds.
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sciencephysicsphysics questions and answerstwo spheres experience an elastic collision in one dimension. the first sphere has a mass of 8.50 kg and has an initial speed of 6.00 m/s and is approaching a second sphere of mass 5.00 kg that is at rest. determine the final velocities of both after the collision. write your answers separated by a comma, to three significant figures without writing the
Question: Two Spheres Experience An Elastic Collision In One Dimension. The First Sphere Has A Mass Of 8.50 Kg And Has An Initial Speed Of 6.00 M/S And Is Approaching A Second Sphere Of Mass 5.00 Kg That Is At Rest. Determine The Final Velocities Of Both After The Collision. Write Your Answers Separated By A Comma, To Three Significant Figures Without Writing The
Two spheres experience an elastic collision in one dimension. The first sphere has a mass of 8.50 kg and has an initial speed of 6.00 m/s and is approaching a second sphere of mass 5.00 kg that is at rest. Determine the final velocities of both after the collision.
Write your answers separated by a comma, to three significant figures without writing the unit
Final velocities of both the spheres are:v₁ = 3.026 m/sv₂ = 4.554 m/s
Initial velocity of sphere 1, u₁ = 6.00 m/s
Mass of sphere 1, m₁ = 8.50 kg
Initial velocity of sphere 2, u₂ = 0 m/s
Mass of sphere 2, m₂ = 5.00 kg
Let the final velocity of sphere 1 be v₁ and of sphere 2 be v₂.
Since the collision is elastic, momentum and kinetic energy is conserved.
Using momentum conservation, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂... (1)
Using kinetic energy conservation, (1/2) m₁u₁² + (1/2) m₂u₂² = (1/2) m₁v₁² + (1/2) m₂v₂²... (2)
From equation (1), m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂... (3)
From equation (2), m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²... (4)
Substitute u₂ = 0 and simplify equation (3), m₁u₁ = m₁v₁ + m₂v₂v₂ = (m₁u₁ - m₁v₁) / m₂
Substitute the value of v₂ in equation (4) and simplify, 36.45 + 0 = 36.45 + (1/2) * 5 * v₁²v₁ = sqrt(36.45 * 2 / 5) = 3.026 m/s
Therefore, the answer is (3.026, 4.554).
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A half-wave-controlled rectifier has a purely resistive load R=10 Ohms and Ac power supply Vs=240sin(250t). calculate 1- the load voltage when the triggering angle is 50
∘
. 2- Average load current. 3- Average power at the load side.
1. The load voltage in a half-wave-controlled rectifier can be calculated using the formula:
V_load = V_peak * sin(ωt - α)
Where:
V_peak is the peak voltage of the AC power supply
ω is the angular frequency of the AC power supply
t is the time
α is the triggering angle
Given that Vs = 240sin(250t) and α = 50 degrees, we need to find V_peak.
The peak voltage (V_peak) can be calculated by dividing the peak-to-peak voltage by 2. In this case, since the AC power supply is given as sin(250t), the peak-to-peak voltage is 2 times the peak voltage. Therefore, V_peak = 240 / 2 = 120 volts.
Now, we can substitute the values into the formula:
V_load = 120 * sin(250t - 50)
2. The average load current (I_avg) can be calculated using the formula:
I_avg = (2 * √2 * V_peak) / (π * R)
Where:
V_peak is the peak voltage of the AC power supply
R is the resistance of the load
Substituting the given values:
I_avg = (2 * √2 * 120) / (π * 10)
Simplifying the equation:
I_avg ≈ 17.94 amps
3. The average power at the load side (P_avg) can be calculated using the formula:
P_avg = (V_peak^2) / (2 * R)
Substituting the given values:
P_avg = (120^2) / (2 * 10)
Simplifying the equation:
P_avg ≈ 720 watts
1- The load voltage when the triggering angle is 50 degrees is 120 * sin(250t - 50) volts.
2- The average load current is approximately 17.94 amps.
3- The average power at the load side is approximately 720 watts.
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3. Answer the following
a. What is Just Noticeable Difference [JND] and Weber’s Law?
b. Take a product like a low-cost pressure cooker [priced at Rs. 1800.00 for a 3 Litre, Rs. 2400.00 for a 5 Litre, Rs. 3200.00 for a 6.5 Litre pressure cooker] which is targeted to the lower income group whose annual income is in the range of Rs. 1.2 to 1.8 lakhs per annum [Rs. 10,000 – Rs. 15,000 per month]. Using the principle of JND suggest a discount for families earning 10,000 per month and families earning Rs. 15000.00, that would make them get up and notice the discount & propel them to buy the pressure cooker. Please justify the discount for the same.
a. Just Noticeable Difference (JND) refers to the smallest difference in stimuli that can be detected by a person. It is also known as the difference threshold. Weber's Law states that the JND is proportional to the magnitude of the stimulus.
In simpler terms, the JND is relative to the initial stimulus. For example, if you have a 10 kg weight and you add 1 kg, you will notice the difference. However, if you have a 100 kg weight and you add 1 kg, you may not notice the difference. Weber's Law helps us understand how our perception changes as the magnitude of the stimulus changes.
b. To suggest a discount using the principle of JND, we need to consider the income range and the prices of the pressure cooker. Let's assume the JND for families earning Rs. 10,000 per month is Rs. 200 and for families earning Rs. 15,000 per month is Rs. 300. Based on these values, we can calculate the discounts that would make them notice and be motivated to buy the pressure cooker.
For families earning Rs. 10,000 per month, the JND is Rs. 200. So, if we offer a discount of Rs. 200 on the 3 Litre pressure cooker priced at Rs. 1800, the new price would be Rs. 1600. This discount would be noticeable to them and may motivate them to purchase.
For families earning Rs. 15,000 per month, the JND is Rs. 300. So, if we offer a discount of Rs. 300 on the 5 Litre pressure cooker priced at Rs. 2400, the new price would be Rs. 2100. This discount would be noticeable to them and may encourage them to buy.
The justification for these discounts is that they are based on the JND principle, which ensures that the discounts are noticeable to the target audience. By offering discounts within their perception range, we can attract their attention and motivate them to make a purchase.
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Two workers are trying to move a heavy crate. One pushes on the crate with a force
A
, which has a magnitude of 301 newtons ( N ) and is directed due west. The other pushes with a force
B
. which has a magnitude of 397 N and is directed due north. What are (a) the magnitude and (b) direction of the resultant force
A
+
B
applied to the crate? Suppose that the second worker applies a force-
B
instead of
B
. What then are (c) the magnitude and (d) direction of the resultant force
A
⋅
B
applied to the crate? In both cases express the direction as a positive angle relative to due west. (a) Number Units: (b) Number Units north of west (c) Number Units (d) Number Units south of west
(a) The magnitude of the resultant force A + B is approximately 502.7 N.
(b) The direction of the resultant force A + B is 90 degrees north of west.
(c) The magnitude of the resultant force A · B is approximately 166.6 N.
(d) The direction of the resultant force A · B is 90 degrees south of west.
(a) The magnitude of the resultant force A + B is given by the vector addition:
[tex]|A + B| = \sqrt(A^2 + B^2 + 2ABcos Θ)[/tex]
where A = 301 N, B = 397 N, and θ is the angle between the two forces (90 degrees).
Calculating the magnitude gives |A + B| = sqrt((301 N)^2 + (397 N)^2 + 2(301 N)(397 N)cos90°) ≈ 502.7 N.
(b) The direction of the resultant force A + B can be found using trigonometry. Since force A is directed due west and force B is directed due north, the angle between them is 90 degrees. Therefore, the resultant force is directed north of west, and the angle relative to due west is 90°.
(c) When the second worker applies a force -B, the magnitude of the resultant force A · B is given by:
|A · B| = [tex]\sqrt(A^2 + B^2 - 2AB[/tex]cosθ)
where A = 301 N, B = -397 N, and θ is the angle between the two forces (90 degrees).
Calculating the magnitude gives |A · B| =[tex]\sqrt((301 N)^2 + (-397 N)^2 - 2(301 N)(-397 N)cos90)[/tex] ≈ 166.6 N.
(d) The direction of the resultant force A · B can be found using trigonometry. Since force A is directed due west and force -B is directed due north (opposite direction), the angle between them is 90 degrees. Therefore, the resultant force is directed south of west, and the angle relative to due west is 90°.
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Three point charges are aligned along the x axis as shown in the figure below.
Therefore, the net force on q2 is to the right. The force on q3 due to q2 is to the left, while the force on q3 due to q1 is to the right.[tex]F3net = F3-2 + F3-1F3net = (1.4 x 10^-5 N) + (0.0 N)F3net = 1.4 x 10^-5 N[/tex](to the left)Therefore, the net force on q3 is to the left.
Three point charges are aligned along the x axis as shown in the figure below:
Three point charges are aligned along the x axis as shown in the figure. The charges are q1 = -5.0 nC,
q2 = 1.0 nC, and
q3 = -5.0 nC.
This can be calculated as follows:
First, consider a point P located at a distance x from charge q1. Let q2 be located at a distance of 3 m to the right of charge q1, and let q3 be located at a distance of 4 m to the right of charge q2.
Therefore, the net force on point P can be determined using Coulomb's Law. The direction of force is indicated by + or - signs, depending on whether the force is repulsive or attractive.The formula for Coulomb's Law is:F = (k * q1 * q2) / d^2
where:
k = Coulomb's constant
[tex]= 9.0 x 10^9 Nm^2/C^2q1 and q2 are the charges separated by a distance d.[/tex]
F is the force between q1 and q2.
The direction of the force can be determined using Coulomb's Law, as well as the net force on the object.
The magnitude of the force between q1 and q2 can be calculated using Coulomb's Law:
[tex]F1-2 = (9.0 x 10^9 Nm^2/C^2) * (-5.0 x 10^-9 C) * (1.0 x 10^-9 C) / (3 m)^2F1-2[/tex]
= -5.0 x 10^-4 N (attractive)
The magnitude of the force between q2 and q3 can be calculated:
[tex]F2-3 = (9.0 x 10^9 Nm^2/C^2) * (1.0 x 10^-9 C) * (-5.0 x 10^-9 C) / (4 m)^2F2-3[/tex]
= -1.4 x 10^-5 N (repulsive)
Now, we can determine the net force on charge q1.
The force on q1 due to q2 is to the left, while the force on q1 due to q3 is to the right.
F1net = F1-2 + F1-3F1net
= (-5.0 x 10^-4 N) + (0.0 N)F1net
= -5.0 x 10^-4 N (to the left)
Therefore, the net force on q1 is to the left. The force on q2 due to q1 is to the right, while the force on q2 due to q3 is to the left.
F2net = F2-1 + F2-3F2net
[tex]= (5.0 x 10^-4 N) + (-1.4 x 10^-5 N)F2net[/tex]
= 4.8 x 10^-4 N (to the right)
Therefore, the net force on q2 is to the right.
The force on q3 due to q2 is to the left, while the force on q3 due to q1 is to the right.
F3net = F3-2 + F3-1F3net
= (1.4 x 10^-5 N) + (0.0 N)F3net
= 1.4 x 10^-5 N (to the left)
Therefore, the net force on q3 is to the left.
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Calculate the displacement and velocity at the following times for a ball thrown straight up with an initial velocity of 21.4 m/s. Take the point of release to be y0=0. (a) 0.500 s \begin{tabular}{l|l} y1= & m \\ v1= & m/s \end{tabular} (b) 1.00 s y2=v2=mm/s (c) 1.50 s \begin{tabular}{l|l} y3= & m \\ v3= & m/s \end{tabular} (d) 2.00 s
The displacement at 0.500 s is 5.39 m and the velocity is 26.8 m/s. The displacement at 1.00 s is 11.7 m and the velocity is 31.2 m/s. The displacement at 1.50 s is 18.3 m and the velocity is 36.7 m/s. we can use the equations of motion for vertical motion under constant acceleration.
To calculate the displacement and velocity at different times for a ball thrown straight up, we can use the equations of motion for vertical motion under constant acceleration.
Given:
Initial velocity, u = 21.4 m/s
Acceleration due to gravity, g ≈ 9.8 m/s² (assuming no air resistance)
Point of release, y₀ = 0
(a) At 0.500 s:
Using the equation for displacement:
y₁ = y₀ + u₀t + (1/2)gt²
y₁ = 0 + (21.4 m/s)(0.500 s) + (1/2)(9.8 m/s²)(0.500 s)²
y₁ ≈ 5.39 m
Using the equation for velocity:
v₁ = u₀ + gt
v₁ = 21.4 m/s + (9.8 m/s²)(0.500 s)
v₁ ≈ 26.8 m/s
(b) At 1.00 s:
Using the equation for displacement:
y₂ = y₀ + u₀t + (1/2)gt²
y₂ = 0 + (21.4 m/s)(1.00 s) + (1/2)(9.8 m/s²)(1.00 s)²
y₂ ≈ 11.7 m
Using the equation for velocity:
v₂ = u₀ + gt
v₂ = 21.4 m/s + (9.8 m/s²)(1.00 s)
v₂ ≈ 31.2 m/s
(c) At 1.50 s:
Using the equation for displacement:
y₃ = y₀ + u₀t + (1/2)gt²
y₃ = 0 + (21.4 m/s)(1.50 s) + (1/2)(9.8 m/s²)(1.50 s)²
y₃ ≈ 18.3 m
Using the equation for velocity:
v₃ = u₀ + gt
v₃ = 21.4 m/s + (9.8 m/s²)(1.50 s)
v₃ ≈ 36.7 m/s
(d) At 2.00 s:
At this time, the ball reaches its highest point and starts to fall back down. Therefore, the displacement would be the same as at release, y₀ = 0, and the velocity would be equal in magnitude but in the opposite direction to the initial velocity, u₀ = -21.4 m/s.
Therefore:
y₄ = 0
v₄ = -21.4 m/s
To summarize:
(a) At 0.500 s:
y₁ ≈ 5.39 m
v₁ ≈ 26.8 m/s
(b) At 1.00 s:
y₂ ≈ 11.7 m
v₂ ≈ 31.2 m/s
(c) At 1.50 s:
y₃ ≈ 18.3 m
v₃ ≈ 36.7 m/s
(d) At 2.00 s:
y₄ = 0
v₄ = -21.4 m/s
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Hubble’s observation that galaxies farther away from us are receding faster implies that:
1.the universe is expanding.
2.the universe is contracting.
3.we are located at the center of the universe.
Imagine that the three stars listed below all formed at exactly the
same time, but in different locations of the universe.
Cosmo Star is an O spectral class star with a main-sequence lifetime
of 3 million years. Its life will eventually end as a SN type II and
become a black hole. Cosmo Star is located in a galaxy 10 billion
light years (ly) from Earth.
Ollie Star is a K spectral class star with a main-sequence lifetime of
12 billion years. Its life will eventually end as a slowly cooling
white dwarf. Ollie Star is located in the Milky Way at a distance of
10,000 ly from Earth.
Sullivan Star is an F spectral class star with a main-sequence
lifetime of 5 billion years. Its life will eventually end in a SN type
I that will completely destroy Sullivan Star. Sullivan Star is located
in a galaxy 6 billion ly from Earth.
Which of these stars final end states will first be viewed on Earth?
1. Sullivan
2. They will all be seen at the same time.
3. Cosmo star
Hubble’s observation that galaxies farther away from us are receding faster implies that the universe is expanding.
The rate of the expansion of the universe is known as the Hubble Constant, and it is related to the age of the universe.
As the universe expands, the wavelength of the light emitted from the galaxies stretches, causing the light to shift towards the red end of the spectrum, resulting in a redshift.
The farther away a galaxy is, the higher its redshift, implying that it is receding more quickly.
The end states of the stars listed will be viewed on Earth in the following order:
Cosmo Star, Sullivan Star, and then Ollie Star.
Cosmo Star will be viewed first because it is the farthest away.
Because light takes time to travel, the light emitted by Cosmo Star will take the longest time to reach us.
Because Cosmo Star is so far away, by the time we see the light from its SN type II event, it will have already turned into a black hole.
Sullivan Star will be viewed next since it is in a galaxy 6 billion light years away.
Ollie Star, which is in the Milky Way, will be viewed last because it is the closest to us.
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