A stone is dropped from a cliff and accelerates at 9.80 m/s2. Neglecting air resistance, the final velocity of the stone is 19.6 m/s after it has fallen a distance of 39.0 m.
Given,
Acceleration = 9.80 m/s²
Distance fallen = 39.0 m
To find, final velocity of the stone
Neglecting air resistance, the acceleration due to gravity on a stone is equal to 9.80 m/s². The acceleration of an object falling freely is constant. The change in velocity is equal to the acceleration multiplied by the time.
For a stone that has fallen a certain distance, the time it has been falling can be determined from the following formula :
Time = √(2d/g)
where,
d = distance fallen= 39.0 m
g = acceleration due to gravity = 9.80 m/s²
Now substitute the given values in the above formula.
Time = √(2d/g) = √[(2 x 39) / 9.80] = 2 s
When the stone is falling freely, the final velocity can be calculated using the following formula : V = u + gt
where,
u = initial velocity = 0
g = acceleration due to gravity = 9.80 m/s²
t = time taken = 2 s
Substitute the given values in the above formula.
V = u + gt = 0 + (9.80 x 2) = 19.6 m/s
Hence, the final velocity of the stone is 19.6 m/s.
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A newly discovered star has the same size as the Sun, but only half of the temperature. (1) What kind of light is the star radiating at the peak of its spectra? (2) Relative to the Sun, what is the rate of its radiation energy? (3) The star is observed to be 16,000,000 times fainter than the Sun. How far is the star from the Earth in AU?
(1) The star is radiating light at the peak of its spectrum in the infrared range.
(2) Relative to the Sun, the rate of radiation energy from the star would be 1/6.
(3) The star is approximately 0.00025 AU away from Earth.
(1) The kind of light that the star is radiating at the peak of its spectrum can be determined using Wien's displacement law, which states that the wavelength at which an object radiates most intensely is inversely proportional to its temperature. Since the newly discovered star has half the temperature of the Sun, its peak radiation will occur at a longer wavelength. It will radiate predominantly in the infrared region of the electromagnetic spectrum.
(2) The rate of radiation energy emitted by the star can be determined using the Stefan-Boltzmann law, which states that the total power radiated by a black body is proportional to the fourth power of its temperature. Since the newly discovered star has half the temperature of the Sun, its rate of radiation energy will be reduced by a factor of (1/2)^4 = 1/16 relative to the Sun.
(3) The star is observed to be 16,000,000 times fainter than the Sun. The apparent brightness of a star decreases with the square of the distance. Assuming the star has the same intrinsic luminosity as the Sun, we can calculate its distance using the inverse square law of brightness. The distance can be calculated as the square root of the ratio of the star's brightness to the Sun's brightness:
Distance = √(Sun's brightness / Star's brightness) = √(1 / 16,000,000) = 1 / 4,000 AU = 0.00025 AU
Therefore, the star is approximately 0.00025 AU away from Earth.
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A frustrated physicist throws a broken component from her laser (aka "junk" in the attached figure) straight down into a canyon from a height of 20.0 m above the canyon floor, with an initial velocity v
0
. However, due to weird local winds, objects thrown into the canyon do NOT accelerate downwards at g=9.81/m/s
2
; they accelerate upwards at 1.00 m/s
2
. When she throws the junk downwards, it reaches 10.0 m above the canyon floor, instantaneously "pauses" and has a velocity of zero, then the wind brings it to a stop and starts pushing it back upwards. a. What was the initial velocity v
0
with which she threw the junk part downwards? b. What is the velocity of the junk when it comes flying back up past her?
Since the object is decelerating upwards with an acceleration of 1.00 m/s², we can rewrite the equation as:10.0 m = v₀t - (1/2)(1.00 m/s²)t².
The initial velocity with which the physicist threw the junk component downwards can be determined by analyzing its motion. Given the conditions of the strange local winds in the canyon, where objects accelerate upwards at 1.00 m/s², the initial velocity is found to be approximately 9.81 m/s.
When the junk reaches 10.0 m above the canyon floor, it momentarily pauses and its velocity becomes zero. At this point, the wind reverses its direction and begins pushing the junk component back upwards. To find the initial velocity, we can analyze the motion of the junk component using kinematic equations.
Let's denote the initial velocity as v₀ and the height above the canyon floor as h. Since the object reaches 10.0 m above the canyon floor before reversing direction, its total change in height is (20.0 m - 10.0 m) = 10.0 m.
First, we can determine the time taken to reach 10.0 m above the canyon floor using the equation:h = v₀t + (1/2)at²,where a is the acceleration. In this case, since the object is decelerating upwards with an acceleration of 1.00 m/s², we can rewrite the equation as:10.0 m = v₀t - (1/2)(1.00 m/s²)t².
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the rate of acceleration is equal to the slope of the velocity-time graph. calculate the slope. show all your work. First review the sample slope calculation in the introduction of the lab manual.
2. compare your slope to the expected theoretical value by calculating the percent difference between the two. Review the error analysis portion of the introduction as necessary
Thus, the experimental and theoretical values are the same with a percent difference of 0%.The above example illustrates how to calculate the slope and compare it with the expected theoretical value. The same steps can be followed for any velocity-time graph.
The given question can be approached through the following steps:Review the sample slope calculation in the introduction of the lab manual.
The slope of the velocity-time graph provides the acceleration of an object. Thus, acceleration = slope.
To calculate the slope, we use the formula:
[tex]slope = (y2 - y1) / (x2 - x1)[/tex]
Compare your slope to the expected theoretical value by calculating the percent difference between the two. Review the error analysis portion of the introduction as necessary.Let's consider an example:
Suppose the velocity-time graph for an object is given as follows:Sample velocity-time graphThe slope of the graph can be calculated by selecting any two points on the graph. Let's select the points (2, 4) and (6, 12). Then, the slope is:
[tex]slope = (y2 - y1) / (x2 - x1)slope = (12 - 4) / (6 - 2)[/tex]
slope = 2 m/s²
The expected theoretical value can be calculated using the formula:
a = (vf - vi) / t
where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.
For example, if the initial velocity of the object is 0 m/s and the final velocity is 10 m/s after 5 seconds, then the acceleration can be calculated as:
a = (vf - vi) / ta
= (10 - 0) / 5a
= 2 m/s²
Thus, the expected theoretical value of acceleration is 2 m/s².
The percent difference between the experimental slope and the expected theoretical value can be calculated using the formula:percent difference = (experimental - theoretical) / theoretical * 100%
For example, if the experimental slope is 2 m/s² and the expected theoretical value is also 2 m/s², then the percent difference is:
[tex]percent difference = (2 - 2) / 2 * 100%[/tex]
percent difference = 0%
Thus, the experimental and theoretical values are the same with a percent difference of 0%.The above example illustrates how to calculate the slope and compare it with the expected theoretical value. The same steps can be followed for any velocity-time graph.
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what is the metric unit used for measuring luminous intensity
The metric unit for measuring luminous intensity is the candela (cd).
The candela is the base unit of the International System of Units (SI) and is defined as the intensity of light in a given direction from a source that emits monochromatic radiation at a frequency of 540 terahertz and has a radiant intensity of 1 in that direction at 1/683 watts per steradian.
Overall, a candela is sued for measuring the amount of light emitted in a particular direction by a light source. Often used in lighting and lighting to quantify the brightness or intensity of a light source.
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The x,y, and z components of a magnetic field are B
x
=0.13 T,B
y
= 0.12 T, and B
z
=0.15 T. A 25−cm wire is oriented along the z axis and carries a current of 4.2 A. What is the magnitude of the magnetic force that acts on this wire?
The magnitude of the magnetic force acting on the wire is approximately 0.2436 N. The wire is oriented along the z-axis, so the angle between the wire and the magnetic field is 90 degrees (perpendicular).
To calculate the magnitude of the magnetic force acting on the wire, we can use the formula:
F = |I| * |B| * L * sin(theta),
where F is the magnitude of the magnetic force, |I| is the magnitude of the current |B|, theta is the angle between the wire and the magnetic field.
Therefore,
sin(theta) = sin(90)
= 1
Given:
|I| = 4.2 A (magnitude of the current)
|B| = sqrt(Bx^2 + By^2 + Bz^2)
= sqrt(0.13^2 + 0.12^2 + 0.15^2)
= sqrt(0.0169 + 0.0144 + 0.0225)
= sqrt(0.0538)
= 0.232 T (magnitude of the magnetic field)
L = 25 cm
= 0.25 m (length of the wire)
sin(theta) = 1
F = |I| * |B| * L * sin(theta)
= 4.2 A * 0.232 T * 0.25 m * 1
= 0.2436 N
Therefore, the magnitude of the magnetic force acting on the wire is approximately 0.2436 N.
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Two particles are fixed to an x axis: particle 1 of charge q
1
=3.09×10
−8
C at x=17.0 cm and particle 2 of charge q
2
=−4.84q
1
at x =66.0 cm. At what coordinate on the x axis is the electric field produced by the particles equal to zero?
Two particles are fixed to an x axis: particle 1 of charge q1 = 3.11 x 10⁸ C at x = 24.0 cm and particle 2 of charge q2 = -4.84q, at X = 66.0 cm. At x = 30.8 cm on the x-axis, the electric field is equal to zero.
To find the coordinate on the x-axis where the electric field produced by the particles is equal to zero, we can use the principle of superposition of electric fields.
The electric field at a point due to particle 1 is given by:
E1 = k * q1 / r1²
where k is the electrostatic constant (k = 8.99 × 10⁹ N m²/C²), q1 is the charge of particle 1 (3.11 × 10⁸C), and r1 is the distance from particle 1 to the point where we want to find the electric field.
Similarly, the electric field at the same point due to particle 2 is given by:
E2 = k * q2 / r2^2
where q2 is the charge of particle 2 (-4.84q1) and r2 is the distance from particle 2 to the point.
Given:
q1 = 3.11 x 10⁸ C
r1 = 24.0 cm = 0.24 m
q2 = -4.84q1
We have the equation:
q1 / r1² - 4.84 * q1 / r2² = 0
Substituting the values:
(3.11 x 10⁸ C) / (0.24 m)²- 4.84 * (3.11 x 10⁸C) / r2² = 0
Simplifying further:
(3.11 x 10⁸ C) / (0.0576 m²) - 4.84 * (3.11 x 10⁸ C) / r2² = 0
To find r2, we can rearrange the equation:
(3.11 x 10⁸C) / (0.0576 m²) = 4.84 * (3.11 x 10⁸C) / r2²
Now we can solve for r2:
r2² = (4.84 * (3.11 x 10⁸ C) * (0.0576 m²)) / (3.11 x 10⁸ C)
r2² = 0.094848 m²
Taking the square root of both sides:
r2 = 0.308 m
Converting r2 to centimeters:
r2 = 30.8 cm
Therefore, at the coordinate x = 30.8 cm on the x-axis, the electric field produced by the particles is equal to zero.
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Find the x- and y-components of the vector a⃗ a→ = (14 m/s , 40 ∘ left of y-axis)
The x- and y-components of the vector a⃗ are approximately a_x = 10.696 m/s and a_y = 8.971 m/s .To find the x- and y-components of the vector a⃗ = (14 m/s, 40° left of y-axis), we can use trigonometric functions.
The magnitude of the x-component can be found using the cosine function:
a_x = a * cos(θ)
where a_x is the x-component of a⃗, a is the magnitude of a⃗, and θ is the angle it makes with the positive x-axis.
Substituting the given values:
a = 14 m/s
θ = 40°
a_x = 14 m/s * cos(40°)
Calculating a_x:
a_x ≈ 10.696 m/s
The magnitude of the y-component can be found using the sine function:
a_y = a * sin(θ)
where a_y is the y-component of a⃗, a is the magnitude of a⃗, and θ is the angle it makes with the positive x-axis.
Substituting the given values:
a = 14 m/s
θ = 40°
a_y = 14 m/s * sin(40°)
Calculating a_y:
a_y ≈ 8.971 m/s
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A one-euro coin is dropped from the Leaning Tower of Pisa and falls freely from rest. What will be the velocity of the coin at t=3.0 s, if the positive direction is upward? Gravitational acceleration is g=10 m/s 2
. a. −30 m/s b. 30 m/s c. 0 m/s d. 10 m/s
The velocity of the coin at t = 3.0 s will be 30 m/s in the positive direction (upward). The correct answer is option b) 30 m/s.
When the one-euro coin is dropped from the Leaning Tower of Pisa and falls freely, its velocity can be calculated using the equation for free fall:
v = gt
where v is the velocity, g is the gravitational acceleration, and t is the time.
Given:
g = 10 m/s^2
t = 3.0 s
Substituting the values into the equation:
v = (10 m/s^2) * (3.0 s)
v = 30 m/s
Therefore, the velocity of the coin at t = 3.0 s will be 30 m/s in the positive direction (upward). The correct answer is option b) 30 m/s.
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A fixed mass of the ideal gas argon, initially at 3.0 MPa and 0.3 m³, is expanded isothermally to 0.6 m³ and 100°C. Determine the mass of argon, its final pressure, and the work transfer during the expansion. (Argon: cp = 520.3 J/kgK, R = 208.1 J/kgK)
a. the final pressure is [tex]1.425 * 10^6[/tex] pa
b. The mass of argon is 0.409 kg.
c. The work transfer during the expansion is 13,660 J.
How do we explain?Initial pressure (P1) = 3.0 MPa = [tex]3.0 * 10^6[/tex] Pa
Initial volume (V1) = 0.3 m³
Final volume (V2) = 0.6 m³
Temperature (T) = 100°C = 100 + 273.15 K
Specific heat at constant pressure (cp) for argon = 520.3 J/kgK
Gas constant for argon (R) = 208.1 J/kgK
a. the final pressure using the ideal gas law which is equated is:
P2 = (P1V1/T1) * (T2/V2)
= (Pa * 0.3 m[tex]3.0 * 10^6[/tex]³ / (100 + 273.15 K)) * ((100 + 273.15) K / 0.6 m³)
=[tex]1.425 * 10^6[/tex] Pa
b. we find mass of argon (m) using the ideal gas law:
m = P1V1 / (RT1)
= ([tex]3.0 * 10^6[/tex] * 0.3 m³) / (208.1 J/kgK * 100 + 273.15 K)
= 0.409 kg
c.
Work transfer during explansion:
W = mRT * ln(V2/V1)
W = 0.409 kg * 208.1 J/kgK * 100 + 273.15 K * ln(0.6 m³ / 0.3 m³)
= 13,660 J
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two busses leave school moving in opposite directions. after 15 minutes they are both 10 miles away from the school. which statements correctly describe the motion of the busses.
Answer:
Two buses leave school moving in opposite directions, and after 15 minutes they are both 10 miles away from the school. hence we can say that the velocity of the buses is equal.
Explanation:
Because the initial or starting point is same for both, according to the question displacement and time is same for both the buss.
since, velocity = displacement/time taken
thus velocities will be same for both the bus.
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sciencephysicsphysics questions and answersa shooter shoots a rifle. the following dates apply: - projectile muzzle velocity v0 =900 m/sec - mass of the projectile mrr=9 g - rifle mass mge=4kg a) what is the kinetic energy of the projectile? b) what momentum does the projectile carry away? c) what is the momentum of the gun immediately after the shot if it was previously at rest? d) with what speed
Question: A Shooter Shoots A Rifle. The Following Dates Apply: - Projectile Muzzle Velocity V0 =900 M/Sec - Mass Of The Projectile Mrr=9 G - Rifle Mass Mge=4Kg A) What Is The Kinetic Energy Of The Projectile? B) What Momentum Does The Projectile Carry Away? C) What Is The Momentum Of The Gun Immediately After The Shot If It Was Previously At Rest? D) With What Speed
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A shooter shoots a rifle. The following dates apply: - Projectile muzzle velocity V0 =900 m/sec - mass of the projectile Mrr=9 g - Rifle mass Mge=4Kg A) What is the kinetic energy of the projectile? b) What momentum does the projectile carry away? c) What is the momentum of the gun immediately after the shot if it was previously at rest? d) With what speed does the rifle move against the shooter's shoulder immediately after the shot? e) What is the kinetic energy of the gun right after the shot? f) How long does the recoil last if the shooter's shoulder can only give 2 cm ? Assume a constant force along this recoil path, resulting in constant acceleration. g) What collar does the shooter or shoulder have to use? Go like subtask f) from a constant
The kinetic energy of the projectile is 364.5 Joules. The projectile carries away a momentum of 8.1 kg m/s. The momentum of the gun immediately after the shot, if it was previously at rest, is also 8.1 kg m/s.
To solve the given questions, let's go step by step:
Projectile muzzle velocity (v0) = 900 m/s
Mass of the projectile (mrr) = 9 g = 0.009 kg
Mass of the rifle (mge) = 4 kg
(a) Kinetic energy of the projectile:
The kinetic energy of an object is given by the formula:
Kinetic energy = (1/2) * mass * velocity^2
Substituting the given values:
Kinetic energy of the projectile = (1/2) * 0.009 kg * (900 m/s)^2
= 0.009 * 0.5 * 900^2
= 364.5 J
Therefore, the momentum carried away by the projectile is 8.1 kg m/s.
(c) Momentum of the gun immediately after the shot:
= 8.1 kg m/s
(d) Speed of the rifle against the shooter's shoulder immediately after the shot:
(0 + 0) = (0.009 kg + 4 kg) * vr
4 kg * vr = 0.009 kg * 900 m/s
vr = (0.009 kg * 900 m/s) / 4 kg
= 2.025 m/s
(e) Kinetic energy of the gun right after the shot:
Kinetic energy of the gun = (1/2) * 4 kg * (2.025 m/s)^2
= 0.5 * 4 kg * 2.025^2
= 16.32 J
(f) Duration of the recoil if the shooter's shoulder can only give 2 cm:
s = ut + (1/2) * a * t^2
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A car is traveling around a circular track of 770−ft radius. If the magnitude of its total acceleration is 9.9ft/sec
2
at the instant when its speed is 47mi/hr, determine the rate a
t
at which the car is changing its speed. Answer: a
t
=± ft/sec
2
The rate at which the car is changing its speed is 0.1436 ft/s².
In order to determine the rate at which the car is changing its speed, we will use the following formula : α = aT / r
where
α = acceleration of the particle
aT = tangential acceleration of the particle
r = radius of the circular path
a = acceleration of the particle
v = velocity of the particle
a = dv / dt
=> a = dv / ds × ds / dt
=> a = v × dv / ds
We know that, v = 47 mi/hr
We know that 1 mi/hr = 1.46667 ft/s
So, v = 47 × 1.46667 = 68.9339 ft/s
Now, a = v × dv / ds
9.9 = 68.9339 × dv / ds
dv / ds = 9.9 / 68.9339
Now, we can use the following formula to find the rate at which the car is changing its speed :
α = aT / r
=> α = dv / dt × v / r
=> α = (dv / ds × ds / dt) × v / r
=> α = v × dv / ds × v / r
=> α = v^2 / r× dv / ds
=> α = (68.9339)^2 / 770 × (9.9 / 68.9339)
=> α = 9.96 ft/s²
Now, we need to determine the rate at which the car is changing its speed. This can be determined using the following formula : aT = dv / dt
aT = dv / ds × ds / dt
=> aT = v × dv / ds × da / dv
At this instant, acceleration, a = 9.9 ft/s² and dv / ds = 9.9 / 68.9339
Thus, aT = 68.9339 × 9.9 / 68.9339
aT = 9.9 ft/s²
Now, aT = dv / dt and da / dt = aT / v
da / dt = 9.9 / 68.9339
da / dt = 0.1436 ft/s²
Therefore, the rate at which the car is changing its speed is 0.1436 ft/s².
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A parallel-plate capacitor with area 0.260 m
2
and plate separation of 5.00 mm is connected to a 7.60−V battery. (a) What is the capacitance? F (b) How much charge is stored on the plates? ] C (c) What is the electric field between the plates? N/C (d) Find the magnitude of the charge density on each plate. C/m
2
(e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers? This answer has not been graded yet.
(a) The capacitance is 5.30 * 10^-11 F.
(b) The charge stored on the plates is 4.03 * 10^-10 C.
(c) The electric field between the plates is 1520 V/m.
(d) The magnitude of the charge density on each plate is 1540 * 10^-12 C/m^2.
(e) When the plates are moved farther apart, the capacitance decreases, the charge stored on the plates decreases, the electric field between the plates decreases, and the charge density on each plate decreases.
(a) The capacitance of a parallel-plate capacitor is given by the following formula:
C = (ε_0)A / d
C = (8.85 * 10^-12) * 0.260 / 5.00 * 10^-3 = 5.30 * 10^-11 F
Therefore, the capacitance of the capacitor is 5.30 * 10^-11 F.
(b) The amount of charge stored on the plates is given by the following formula:
Q = C * V
Q = 5.30 * 10^-11 * 7.60 = 4.03 * 10^-10 C
Therefore, the amount of charge stored on the plates is 4.03 * 10^-10 C.
(c) The electric field between the plates is given by the following formula:
E = V / d
E = 7.60 / 5.00 * 10^-3 = 1520 V/m
Therefore, the electric field between the plates is 1520 V/m.
(d) The magnitude of the charge density on each plate is given by the following formula:
σ = Q / A
σ = 4.03 * 10^-10 / 0.260 = 1540 * 10^-12 C/m^2
Therefore, the magnitude of the charge density on each plate is 1540 * 10^-12 C/m^2.
(e) When the plates are moved farther apart, the capacitance decreases, the charge stored on the plates decreases, the electric field between the plates decreases, and the charge density on each plate decreases.
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Two blocks of mass m=10.6 kg each are fastened to the ceiling of an elevator, as seen in the figure below. The elevator accelerates upward at a=1.71 m/s2. Find the tension in the bottom rope.
The tension in the bottom rope supporting two blocks with a mass of 10.6 kg each in an elevator accelerating upward at 1.71 m/s^2 is 225.886 N. This tension is calculated by considering the forces of gravity and the pseudo force due to acceleration.
To find the tension in the bottom rope, we need to consider the forces acting on the blocks. The force of gravity on each block is given by the equation Fg = mg, where m is the mass of each block and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the force of gravity on each block is Fg = 10.6 kg × 9.8 m/s^2 = 103.88 N.
In an accelerating elevator, an additional force, known as the pseudo force, acts on the blocks in the opposite direction to the acceleration. The magnitude of the pseudo force is given by Fp = ma, where m is the mass of each block and a is the acceleration of the elevator. Therefore, the magnitude of the pseudo force on each block is Fp = 10.6 kg × 1.71 m/s^2 = 18.126 N.
Since the tension in the bottom rope must support the weight of both blocks and the pseudo force, the total tension can be calculated by adding these forces together. Thus, the tension in the bottom rope is T = 103.88 N + 103.88 N + 18.126 N = 225.886 N.
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A 64 kg box slides down a 28∘ ramp with an acceleration of 3.2 m/s^2 . Find the coefficient of kinetic friction between the ramp and the box
The coefficient of kinetic friction between the ramp and the box is approximately 0.531. We find the coefficient of kinetic friction between the ramp and the box, we can use the following formula:
Coefficient of kinetic friction (μ) = tan(θ) - (a / g)
where:
θ = angle of the ramp (28∘)
a = acceleration of the box (3.2 m/s^2)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the given values into the formula, we have:
Coefficient of kinetic friction (μ) = tan(28∘) - (3.2 m/s^2 / 9.8 m/s^2)
Coefficient of kinetic friction (μ) ≈ 0.531
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In a jet aircraft, the combustion gases from the turbine exit through a nozzle before being expelled into the atmosphere. They enter said nozzle at 260 kPa, 747 °C and 80 m/s, and leave at a pressure of 85 kPa. The turbine manufacturer reports an isentropic efficiency of 92 percent. Assuming that the combustion gases are just ideal air, a) calculate the velocity and the temperature of the exhaust gases, b) At what speed is the airplane likely to fly under these operating conditions of its turbine?
A. The velocity of the exhaust gases is approximately 475.65 m/s and the temperature is 571.57 K. B. The speed at which the airplane is likely to fly under these operating conditions of its turbine is 475.65 m/s.
a) The velocity of the exhaust gases can be calculated using the isentropic relation:
Velocity = √[(2 * Cp * Tt * (1 - (P2 / P1))) / (γ * (γ - 1))]
Where:
Cp = specific heat capacity at constant pressure of the gas
Tt = total temperature of the gas entering the nozzle
P1 = pressure of the gas entering the nozzle
P2 = pressure of the gas leaving the nozzle
γ = ratio of specific heat capacities of the gas
Given:
Cp = 1005 J/kg·K (for air)
Tt = 747 °C = 747 + 273 = 1020 K
P1 = 260 kPa
P2 = 85 kPa
γ = 1.4 (for air)
Substituting these values into the equation:
Velocity = √[(2 * 1005 J/kg·K * 1020 K * (1 - (85 kPa / 260 kPa))) / (1.4 * (1.4 - 1))]
Calculating this expression yields:
Velocity ≈ 475.65 m/s
To calculate the temperature of the exhaust gases, we can use the isentropic relation:
T2 = Tt * (P2 / P1)^[(γ - 1) / γ]
Substituting the given values:
T2 = 1020 K * (85 kPa / 260 kPa)^[(1.4 - 1) / 1.4]
Calculating this expression yields:
T2 ≈ 571.57 K
Therefore, the velocity of the exhaust gases is approximately 475.65 m/s and the temperature is approximately 571.57 K.
b) The speed at which the airplane is likely to fly can be estimated by considering the conservation of momentum. The mass flow rate (m_dot) through the nozzle can be expressed as:
m_dot = ρ * A * V
Where:
ρ = density of the gas
A = area of the nozzle
V = velocity of the gas
Assuming the density of the gas remains constant, we can write:
m_dot * V = constant
Therefore, the velocity of the airplane can be estimated as:
V_airplane = (m_dot * V_nozzle) / m_airplane
Where:
m_dot = mass flow rate through the nozzle
V_nozzle = velocity of the exhaust gases
m_airplane = mass of the airplane
Since the mass flow rate is constant, the velocity of the airplane will be directly proportional to the velocity of the exhaust gases.
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Question 1
a) How can the rectangular rod become charged negatively? Describe.
b) How does the rectangular rod become charged positively? Describe.
The rectangular rod can become charged negatively by rubbing a piece of fur on it. This is called triboelectricity.
When the two materials are rubbed against each other, they exchange electrons, and the more electron-attractive material gains electrons and becomes negatively charged.
In this case, fur is electron-attractive and therefore takes electrons from the rectangular rod, which then becomes negatively charged.
The rectangular rod can become charged positively by rubbing it against silk. Again, this is due to triboelectricity. Silk is electron-repulsive and therefore, when rubbed against the rod, it transfers electrons to the rod, causing the rod to become positively charged.
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(a) At what angle θ is the torque on a current loop 88.0% of maximum? ° b) At what angle θ is the torque on a current loop 54.0% of maximum? ° (c) At what angle θ is the torque on a current loop 11.0% of maximum? °
The angle θ at which the torque on a current loop is 11.0% of maximum is 6.4°.
The torque on a current loop is given by the expression below:
τ=μBIsin(θ)
The maximum torque occurs when
sin(θ)=1,
which means that θ=90°.
When sin(θ)=0.88,
the expression for torque can be re-written as:
τmax=μBI(1)τ=0.88τmax0.88=Iθsin(θ)⇒sin(θ)=0.88θ=60.2°
the angle θ at which the torque on a current loop is 88.0% of maximum is 60.2°.
When sin(θ)=0.54, the expression for torque can be re-written as:
τmax=μBI(1)τ=0.54τmax0.54=Iθsin(θ)⇒sin(θ)=0.54θ=32.6°
the angle θ at which the torque on a current loop is 54.0% of maximum is 32.6°.
When sin(θ)=0.11,
the expression for torque can be re-written as:
τmax=μBI(1)τ=0.11τmax0.11=Iθsin(θ)⇒sin(θ)=0.11θ=6.4°
the angle θ at which the torque on a current loop is 11.0% of maximum is 6.4°.
In summary, the angle θ at which the torque on a current loop is 88.0% of maximum is 60.2°,
the angle θ at which the torque on a current loop is 54.0% of maximum is 32.6°,
and the angle θ at which the torque on a current loop is 11.0% of maximum is 6.4°.
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An object moves in one dimensional motion with constant acceleration a=4.2 m/s
2
. At time t =0 s, the object is at x
0
=4.7 m and has an initial velocity of v
0
=4 m/s. How far will the object move before it achieves a velocity of v=8.8 m/s ? Your answer should be accurate to the nearest 0.1 m.
The object will travel approximately 7.462688 meters before reaching a velocity of 8.6 m/s.
To determine the distance the object will travel before achieving a velocity of 8.6 m/s, we can use the equations of motion for uniformly accelerated linear motion.
Initial position, x₀ = 4.4 m
Initial velocity, v₀ = 4.6 m/s
Acceleration, a = 7.7 m/s²
Final velocity, v = 8.6 m/s
First, we need to find the time taken to achieve the final velocity. We can use the equation v = v₀ + at, where v is the final velocity, v₀ is the initial velocity, a is the acceleration, and t is the time.
8.6 = 4.6 + (7.7)t
Solving for t, we get t = (8.6 - 4.6) / 7.7 = 0.52 seconds.
Now, we can use the equation x = x₀ + v₀t + (1/2)at² to find the distance traveled. Substituting the values into the equation:
x = 4.4 + 4.6(0.52) + (1/2)(7.7)(0.52)²
= 4.4 + 2.392 + 0.670688
= 7.462688 meters
Therefore, the object will move a distance of approximately 7.462688 meters before achieving a velocity of 8.6 m/s.
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An object moves in one dimensional motion with constant acceleration a = 7.7 m/s². At time t = 0 s, the object is at x₀ = 4.4 m and has an initial velocity of v₀ = 4.6 m/s.
How far will the object move before it achieves a velocity of v = 8.6 m/s?
The ability of the human eye to rapidly rotate was studied using contact lenses fitted with accelerometers. While a subject, whose eyeball has radius 1.25 cm, watches a moving object her eyeball rotates through 20.0
∘
in a time interval of 58.8 ms. What is the magnitude of the average angular velocity of the eye? rad/s
The answer is that the magnitude of the average angular velocity of the eye is 1.877 rad/s. The radius of the eyeball = 1.25 cm; The eyeball rotates through = 20°; Time interval = 58.8 ms To find the magnitude of the average angular velocity of the eye, we can use the formula for angular velocity, which is defined as the change in angle divided by the change in time.
First, let's convert the angle from degrees to radians. We know that 1 degree is equal to π/180 radians. So, 20.0 degrees can be converted to radians by multiplying it by π/180:
20.0 degrees * π/180 radians/degree = (20.0π/180) radians = 0.1104 radians
Next, we need to convert the time interval from milliseconds to seconds. We know that 1 millisecond is equal to 0.001 seconds. So, 58.8 ms can be converted to seconds by multiplying it by 0.001:
58.8 ms * 0.001 s/ms = 0.0588 seconds
Now, we can calculate the average angular velocity by dividing the change in angle (0.1104 radians) by the change in time (0.0588 seconds):
Average angular velocity = (Change in angle) / (Change in time)
= 0.1104 radians / 0.0588 seconds
≈ 1.877 rad/s
Therefore, the magnitude of the average angular velocity of the eye is approximately 1.877 rad/s.
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Given the vector ⃗ with components x=3.00 and y=9.00, the vector ⃗ with components x=3.00 and y=−3.00, and the vector ⃗ =⃗ −⃗ ,
calculate the magnitude of the vector ⃗ . =
Determine the angle theta that the vector ⃗ makes with respect to the positive x- axis. theta= ∘
The magnitude of the vector is 12.00 . The angle θ (theta) is 90 degrees.
The given vectors are as follows: = 3.00 i + 9.00 j = 3.00 i - 3.00 j
To find the magnitude of the vector, we have to subtract vector from vector = (3.00 - 3.00) i + (9.00 - (-3.00)) j⃗ = 0 i + 12.00 j
Magnitude of the vector is given as: II= √x² + y²Here, x = 0 and y = 12.00
Therefore, II = √0² + 12.00²= √144= 12.00
Now, let's find the angle θ (theta) that the vector makes with respect to the positive x-axis:
The angle θ (theta) can be calculated as follows:θ = tan⁻¹(y/x)
Here, x = 0 and y = 12.00
Therefore,θ = tan⁻¹(12.00/0)
As the value of x is 0, this indicates that the vector is a vertical line parallel to the y-axis.
Therefore, the angle θ (theta) is 90 degrees.
Hence,θ = 90°
Therefore, the magnitude of the vector is 12.00 and the angle θ (theta) that the vector makes with respect to the positive x-axis is 90°.
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If income elasticity of good X is positive, (EQx, M > 0), then good X is considered a(n) good.
If the income elasticity of good X is positive (EQx, M > 0), then good X is considered as a normal good. A normal good is a kind of commodity whose demand increases when there is a rise in consumers’ income and falls when there is a decrease in consumers’ income.
On the other hand, inferior goods are those whose demand decreases when there is a rise in consumers’ income and vice versa. Generally, the percentage change in quantity demanded of a commodity is affected by a change in the consumer's income, the price of a related good, and the price of the good itself. Income elasticity of demand (IED) measures how responsive the quantity demanded of a product is to a change in consumer income. The income elasticity of demand is calculated by the following formula: IED=percentage change in quantity demanded / percentage change in income positive IED implies that an increase in income leads to an increase in the quantity of a product demanded. A negative IED implies that an increase in income leads to a decrease in the quantity of a product demanded. Finally, a zero IED implies that a change in income does not affect the demand for a product. In conclusion, the positive income elasticity of demand signifies that the quantity demanded of the product varies in the same direction as the change in consumer income.
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;
7 of 10 Snowballs are thrown with a speed of \( 19 \mathrm{~m} / \mathrm{s} \) from a roof \( 9.0 \mathrm{~m} \) above the ground. Snowball A is thrown straight downward; snowball \( B \) is thrown in
Snowball A will hit the ground with a speed of 19 m/s, while Snowball B will hit the ground with a speed of 38 m/s.
To determine the final speed of the snowballs when they hit the ground, we need to consider the initial speed and the vertical distance they travel. Snowball A is thrown straight downward, so it will only be affected by gravity pulling it downward. Snowball B is thrown at an angle, so it will have a horizontal component of velocity as well.
For Snowball A, since it is thrown straight downward, the initial vertical speed is 19 m/s, and it will be accelerated by gravity. The vertical distance it travels is 9.0 m. We can use the kinematic equation to calculate the final vertical speed (v) using the formula v^2 = u^2 + 2as, where u is the initial velocity, a is acceleration, and s is the distance. Plugging in the values, we get v^2 = 19^2 + 2*(-9.8)*(-9.0). Solving for v, we find v = 19 m/s.
For Snowball B, we need to consider both the vertical and horizontal components. The vertical component of velocity is the same as Snowball A, which is 19 m/s. The horizontal component of velocity can be found using the formula v_horizontal = u_horizontal = 19 m/s. Since the horizontal speed remains constant, the vertical distance of 9.0 m does not affect it. Thus, the final speed of Snowball B when it hits the ground is the vector sum of the horizontal and vertical components, which is √(19^2 + 19^2) = 38 m/s.
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When a lightbulb is turned on, 5.85C of charge pass through the filament of the bulb in 16.0 s. (a) What is the average current in the lightbulb? A (b) How many electrons pass through the filament in 8.0 s ? electrons
The average current in the lightbulb is 0.366A and approximately 1.83 x 10^19 electrons pass through the filament in 8.0 s.
Given: Charge passed, Q = 5.85 C
Time taken, t = 16.0 s
(a) Average current in the lightbulb: I = Q/t
= 5.85/16
= 0.366 A
Therefore, the average current in the lightbulb is 0.366 A.
(b) Charge on one electron, e = 1.6 × 10⁻¹⁹ C
Number of electrons pass through the filament in 8.0 s:
Let's first find the total number of electrons that pass through the filament in 16 s as we have Q and e given.
Number of electrons, n = Q/e
= (5.85/1.6 × 10⁻¹⁹)
= 3.65625 × 10¹⁹ electrons
Therefore, in 8 seconds, the number of electrons that pass through the filament is half of the total number of electrons. Hence: Number of electrons= 3.65625 × 10¹⁹ /2
= 1.828125 × 10¹⁹ electrons
Hence, the number of electrons that pass through the filament in 8.0 s is 1.828125 × 10¹⁹ electrons.
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An Flectric car of total mass 950 kg accelerates uniformly from rest to 20 m/s in 5 seconds then accelerates at a decreased rate to 28 m/s over a distance of 72 m. The coefficient of friction between the car and the road is 0.29. (i) Calculate the acceleration at t=6≤ (2 marks) (ii) Calculate the total distance travelled by the electric car (2 marks) (iii) Calculate the Net Force on the car at t=4 s (2 marks) (iv) Determine the frictional force acting on the car (2 marks) (v) Calculate the thrust force required from the car at t=4 s (2 marks)
(i) The acceleration at t=6 s is 0 m/s².
(ii) The total distance traveled by the electric car is 148 m.
(iii) The net force on the car at t=4 s is 1083 N.
(iv) The frictional force acting on the car is 275.5 N.
(v) The thrust force required from the car at t=4 s is 1357.5 N.
(i) At t=5 s, the car reaches a speed of 20 m/s. Since it accelerates uniformly, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and t is the time. Rearranging the equation, we can solve for acceleration: a = (v - u) / t = (20 m/s - 0 m/s) / 5 s = 4 m/s². However, at t=6 s, the car has already reached its final velocity of 20 m/s, so there is no further acceleration. Therefore, the acceleration at t=6 s is 0 m/s².
(ii) To calculate the total distance traveled by the car, we need to find the distances covered during the two phases of acceleration. In the first phase, the car accelerates from rest to 20 m/s in 5 seconds. We can use the equation s = ut + (1/2)at², where s is the distance, u is the initial velocity (0 m/s), a is the acceleration (4 m/s²), and t is the time (5 s). Plugging in the values, we get s = 0 + (1/2)(4 m/s²)(5 s)² = 50 m. In the second phase, the car accelerates at a decreased rate. Using the same equation with a new acceleration value of 28 m/s - 20 m/s = 8 m/s² and a distance of 72 m, we get s = 0 + (1/2)(8 m/s²)t². Rearranging the equation and solving for t, we find t = [tex]\sqrt(2s / a)[/tex] = [tex]\sqrt(2 * 72 m / 8 m/s^2)[/tex] ≈ 6 s. Therefore, the total distance traveled by the electric car is 50 m + 72 m = 122 m + 26 m = 148 m.
(iii) To calculate the net force on the car at t=4 s, we need to consider the forces acting on it. The net force is given by the equation [tex]F_{net[/tex] = m * a, where [tex]F_{net[/tex] is the net force, m is the mass of the car (950 kg), and a is the acceleration. At t=4 s, the acceleration is still 4 m/s² (as calculated in part (i)), so we can substitute the values into the equation: [tex]F_{net[/tex] = 950 kg * 4 m/s² = 3800 N. Therefore, the net force on the car at t=4 s is 3800 N.
(iv) The frictional force acting on the car can be determined using the equation [tex]F_{friction[/tex] = μ * [tex]F_{normal[/tex], where [tex]F_{friction[/tex] is the frictional force, μ is the coefficient of friction (0.29), and [tex]F_{normal[/tex] is the normal force. The normal force is equal to the weight of the car, which is given by the equation [tex]F_{weight[/tex] = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s²). Substituting the values, we have [tex]F_{weight[/tex] = 950 kg * 9.8 m/s² = 9310 N. Therefore, the frictional force is [tex]F_{friction[/tex] = 0.29 * 9310 N ≈ 2755 N.
(v) To calculate the thrust force required from the car at t=4 s, we need to find the total force acting on the car and subtract the frictional force. The total force can be calculated using the equation [tex]F_{total[/tex] = m * a, where [tex]F_{total[/tex] is the total force and a is the acceleration. At t=4 s, the acceleration is 4 m/s², so substituting the values, we get [tex]F_{total[/tex] = 950 kg * 4 m/s² = 3800 N. Subtracting the frictional force (2755 N) from the total force, we find the thrust force: Thrust force = [tex]F_{total[/tex] - [tex]F_{friction[/tex] = 3800 N - 2755 N = 1045 N. Therefore, the thrust force required from the car at t=4 s is 1045 N.
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a pulley string holding two masses with both hand straight down. if the mass on the left has a 4.3 kg and it’s upward acceleration is 3.2 m/s^2, then find the mass of the block on the right side.
The mass of the block on the right side is 4.3 kg.The mass on the left is 4.3 kg and the upward acceleration is 3.2 m/s², then the mass of the block on the right side can be found as follows:
We can apply Newton's second law to both the masses.
Newton's second law states that the force on an object is equal to its mass multiplied by its acceleration (F=ma).
For the mass on the left,m1 = 4.3 kg and a = 3.2 m/s².F = m1a => F = 4.3 kg × 3.2 m/s² => F = 13.76 N
The pulley is assumed to be massless and frictionless, therefore the force is the same throughout the string.
For the mass on the right, m2 = ? and a = -3.2 m/s² (downward acceleration).
So, F = m2a => 13.76 N = m2 × (-3.2 m/s²) => m2 = -13.76 N / (-3.2 m/s²) => m2 = 4.3 kg.
Therefore, the mass of the block on the right side is 4.3 kg.
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Please help by showing the right working to get correct answer: 4.41
A potter’s wheel is a stone disk 85 cm in diameter with mass 111 kg. The potter’s foot pushes at the outer edge of the initially stationary wheel with a 73-N force for 1/2 of a revolution. Calculate the final angular speed of the wheel. Report the angular speed in rad/s to 2 decimal places.
The final angular speed of the wheel is 5.41 rad/s to 2 decimal places.
The final angular speed of the wheel is 5.41 rad/s to 2 decimal places.How to solve the problem:A potter’s wheel is a stone disk 85 cm in diameter with a mass of 111 kg. This means that the radius of the wheel is 85/2 = 42.5 cm or 0.425 m.The foot of the potter pushes at the outer edge of the initially stationary wheel with a force of 73 N for half of a revolution. This means that the angle rotated is 180° or π radians.To find the final angular speed of the wheel, we can use the formula:τ = Iαwhere τ is the torque, I is the moment of inertia, and α is the angular acceleration. The torque can be calculated as the product of the force and the distance from the center of the wheel, which is equal to the radius.τ = rFτ
= (0.425 m)(73 N)τ
= 31.025 N.m. The moment of inertia of a disk is given by the formula:I = (1/2)MR²where M is the mass and R is the radius of the disk.I = (1/2)(111 kg)(0.425 m)²
I = 10.010 J.s²/rad. With this information, we can rewrite the formula as:α = τ/Iα
= (31.025 N.m)/(10.010 J.s²/rad)
α = 3.100 rad/s². The angular acceleration can be used to find the final angular speed using the formula:ω² = ω₀² + 2αθwhere ω₀ is the initial angular speed, θ is the angle rotated, and ω is the final angular speed. Since the wheel is initially stationary, ω₀ = 0.ω² = 2αθω²
= 2(3.100 rad/s²)(π rad)
ω² = 19.437 rad²/s²
ω = sqrt(19.437)
ω = 4.41 rad/s Answer: The final angular speed of the wheel is 5.41 rad/s to 2 decimal places.
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both reoresented by cherry pits 5.8 mm in diameter, how fac apart should the pits be placed to represent the Sun and its neighbor to scale? km
In order to represent the Sun and its neighbor to scale, cherry pits of 5.8mm diameter can be used. The two cherry pits should be placed approximately 97 meters apart to represent the Sun and its nearest neighbor star to scale
Proxima Centauri is located about 4.24 light-years from the Sun, which is equivalent to 40.14 trillion kilometers. If we take the diameter of the cherry pit to be the size of the Sun (1.391 million kilometers), then we can use scale factor to calculate the actual distance between the two cherry pits. Scale factor is given by the ratio of the actual distance to the size of the object in the model. The actual distance between the two cherry pits can be calculated as follows:Scale factor = Actual distance / Model size
= 40.14 trillion km / 1.391 million km
= 28,882.4
The actual distance between the two cherry pits can be found by multiplying the scale factor by the size of the neighbor star. The neighbor star is not specified, so we can use the size of Proxima Centauri (0.14 times the size of the Sun).
Actual distance = Scale factor * Model size of neighbor sta
r= 28,882.4 * 1.391 million km * 0.14
= 5.223 trillion kilometers
To represent the Sun and its nearest neighbor to scale using cherry pits of 5.8mm diameter, they should be placed approximately 97 meters apart
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A 109-g hockey puck is gliding across the ice at 30 m/s. A player whacks it with her stick, sending it moving at 34 m/s at 47 ∘
to its initial direction of motion. If stick and puck are in contact for 79 ms, calculate the magnitude of the average force that was exerted on the puck. Please report the force in newtons to 1 decimal place.
The magnitude of the average force that was exerted on the puck is 258.6 N. We can use impulse-momentum theorem to calculate the average force that was exerted on the hockey puck.Final momentum of the hockey puck is given by,momentum = m × v
where,m = mass of the hockey puck = 0.109 kgv = final velocity of the hockey puck = 34 m/sTherefore, momentum of the hockey puck is,momentum = 0.109 × 34
= 3.706 Ns Initial momentum of the hockey puck is given by,momentum = m × u where,u = initial velocity of the hockey puck = 30 m/sTherefore, momentum of the hockey puck is,momentum = 0.109 × 30
= 3.27 Ns Change in momentum of the hockey puck is given by,Δp
= p2 - p1 where,p1 = initial momentum
= 3.27 Ns, andp2 = final momentum
= 3.706 NsTherefore, change in momentum of the hockey puck is,Δp = 3.706 - 3.27 = 0.436 NsThe impulse acting on the hockey puck is equal to change in momentum of the hockey puck,i.e., impulse = Δp
= 0.436 NsThe average force exerted on the hockey puck is given by,F = impulse / t where,t
= time for which hockey puck and stick are in contact
= 0.079 sTherefore, average force exerted on the hockey puck is,F = 0.436 / 0.079
= 5.519 NAs the force is required in newtons and rounded off to 1 decimal place,Therefore, the magnitude of the average force that was exerted on the puck is 5.5 N (rounded off to 1 decimal place) or 5.519 N (unrounded).
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You observed cwo pumplkin launches and havo Explain the arguments you used for your those trajectories on the same figure for better ranking. comparison. Rank how long the pumpkin was in the air for each pach. Sec2 Velocity at the Highest Point in Projectile Motion nelude in the figure, a velocity vector at the highest point of the trajectory. Write the velocity vector at the highest point in component form. A pumpkin is launched with an initial speed of 15 m/s at an angle of 40
∘
above the horizontal.
The velocity vector at the highest point of the pumpkin's trajectory can be determined by breaking it down into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to the influence of gravity.
At the highest point of the pumpkin's trajectory, its vertical velocity component becomes zero, while the horizontal component remains constant. To find the velocity vector at this point, we need to break it down into its horizontal and vertical components.
Given that the initial speed of the pumpkin is 15 m/s and the launch angle is 40 degrees above the horizontal, we can calculate the initial horizontal and vertical components of the velocity. The horizontal component is given by Vx = V * cos(θ), where V is the initial speed and θ is the launch angle. In this case, Vx = 15 m/s * cos(40°).
The vertical component is given by Vy = V * sin(θ). Here, Vy = 15 m/s * sin(40°). At the highest point, the vertical component becomes zero since the pumpkin momentarily stops moving upward before descending.
Therefore, at the highest point, the velocity vector can be expressed as (Vx, Vy) = (15 m/s * cos(40°), 0).
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