Answer:
B
Explanation:
Because this oscillations occur when the restoring force is directly proportional to displacement, given as
F=-kx
Where k= force constant
X= displacement
The motors that drive airplane propellers are, in some cases, tuned by using beats. The whirring motor produces a sound wave having the same frequency as the propeller.
Part A
If one single-bladed propeller is turning at 574 rpm and you hear a 2.50 Hz beat when you run the second propeller, what are the two possible frequencies (in rpm) of the second propeller? f1,f2 =
Part B
Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.60 Hz . In part (A), which of the two answers was the correct one for the frequency of the second single-bladed propeller ?
Part C
How do you know the answer in part (B) to be correct? Explain.
Answer:
A) 2 possible frequencies of second propellar = 424 rpm or 724 rpm
B) Correct frequency is f2 = 724 rpm
C) Reason is stated in explanation
Explanation:
A) We are given;
Frequency of first propeller; f1 = 574 rpm
Beat frequency; f_beat = 2.5 Hz = 2.5 × 60 rpm = 150 rpm
Now, formula for the beat frequency is;
f_beat = |f1 - f2|
Now, |f1 - f2| means it is inside an absolute value.
Thus, it means,
f1 - f2 = 150 or f2 - f1 = 150
Thus;
574 - f2 = 150
Or f2 - 574 = 150
So,f2 = 574 - 150 = 424 rpm or f2 = 574 + 150 = 724 rpm
2 possible frequencies of second propellar = 424 rpm or 724 rpm
B) Now, if we increase the speed of the second propellar slightly, it means that f2 will increase as well.
Now, from the 2 possible values of f2 gotten, we can see that for f1 - f2 = 150, when we increase f2, the beat frequency will reduce while for f2 - f1 = 150, when we increase f2, the beat frequency will increase.
Thus, it's the frequency of f2 gotten in f2 - f1 = 150 that is the correct answer.
Thus, when we increase the speed of the second propellar slightly, it means that;
f2 = 724 rpm
C) Answer in part B is correct because as we increase the speed of the second propellar, the frequency will also increase and the value of f2 that corresponds with an increase in speed is 724 rpm.
A circular loop of wire 1.0 cm in radius carries a current of 40 A. The magnetic field at the center of the loop is
Answer:
The magnetic field at the center of the loop is 2.51 × 10⁻³ T
Explanation:
The magnetic field at the center of a circular loop is given by
B = μ₀I/2r
Where B is the magnetic field strength in Teslas (T)
μ₀ is the permeability of free space
μ₀ = 4π ×10⁻⁷ N/A²
I is the current in Amperes (A)
and r is the radius of the loop in meters (m)
From the question,
r = 1.0 cm
Convert this to meter (m)
1.0 cm = 1.0 × 10⁻² m = 0.01 m
∴ r = 0.01 m
I = 40 A
Hence, the magnetic field at the center of the loop is
B = μ₀I/2r
B = (4π ×10⁻⁷ × 40) / (2 × 0.01)
B = 5.0265 × 10⁻⁵ / 0.02
B = 2.51 × 10⁻³ T
Hence, the magnetic field at the center of the loop is 2.51 × 10⁻³ T
Which scientist demonstrated that light is the visible component of a larger spectrum of electromagnetic waves
Two point sources produce waves of the same wavelength that are in phase. At a point midway between the sources, what kind of interference would be observed?
Answer:
Constructive interference
Explanation:
Because the path difference between them is zero owing to the fact that they are in phase
Electric force on a dust particle having charge equal to 8X10-19 C when plates are separated by a distance of 2cm and have a potential difference of 5 kV is
Answer:
8×10⁻¹⁷ N
Explanation:
from the question, Electric force is given as
F = QV/r.............. Equation 1
Where F = Electric Force, Q = Charge, V = Electric potential, r = distance.
Given: Q = 8×10⁻¹⁹ C, V = 5 kV = 5000 V, r = 2 cm = 0.02 m.
Substitute into equation 1
F = 8×10⁻¹⁹(5000)(0.02)
F = 8×10⁻¹⁷ N
Hence the electric force on the dust particle is 8×10⁻¹⁷ N
Masses m and 2m approach each other at the same speed v and collide head-on. Find the final speed of mass 2m, if mass rebounds at speed 2v. g
Answer:
The rebound speed of the mass 2m is v/2
Explanation:
I will designate the two masses as body A and body B.
mass of body A = m
mass of body B = 2m
velocity of body A = v
velocity of body B = -v since they both move in opposite direction
final speed of mass A = 2v
final speed of body B = ?
The equation of conservation of momentum for this system is
mv - 2mv = -2mv + x
where x is the final momentum of the mass B
x = mv - 2mv + 2mv
x = mv
to get the speed, we divide the momentum by the mass of mass B
x/2m = v = mv/2m
speed of mass B = v/2
An old-fashioned incandescent lamp is basically just a resistor that gets hot enough to glow when a current passes through it. A modern strand of Christmas lights consists of a series small incandescent lamps wired together. Should these be connected in series or in parallel
Answer:
They should be connected in parallel
Explanation:
Because In Parallel circuit, voltage across all lamps are same and current will flow independent each other such that a malfunction of one will nor affect the rest unlike in series connection
If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s , determine the constant deceleration of the car.
Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
TRICARE is health insurance for what part of the population?
Answer:
TRICARE for Life (TFL), a program for Medicare-eligible military retirees and their dependents, acts as a supplement to Medicare.
Explanation:
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated impedance.
Answer:
16∠45° Ω
Explanation:
Applying,
Z = V/I................... Equation 1
Where Z = Impedance, V = Voltage output, I = current input.
Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°
Substitute these values into equation 1
Z = 120cos(10t+75°)/7.5cos(10t+30)
Z = 120∠75°/ 7.5∠30°
Z = 16∠(75°-30)
Z = 16∠45° Ω
Hence the impedance of the linear network is 16∠45° Ω
A fishing pole is an example of a compound machine. What simple machines are used to make up this compound machine?
Answer:
Corkscrew, bicycle, scissors
Explanation:
they are made from many simple machine's
Answer:
compound meachiun include bicycles, cars, scissors, and fising rods with reels.
has current 150 A and a height of 6.0 m above the ground, what magnetic field does the line produce at ground level
Complete question:
Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 6.0 m above the ground, what magnetic field does the line produce at ground level?
Answer:
The magnetic field the line produces at ground level is 5 x 10⁻⁶ T.
Explanation:
Given;
current in the dc transmission line, I = 150 A
height above ground level, R = 6 m
The transmission line will be treated as current in a long straight wire.
Thus, the magnetic field produced at the ground level is given as magnetic field in a long straight conductor.
[tex]B = \frac{\mu_o I}{2\pi R}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A
[tex]B = \frac{4\pi*10^{-7} * 150}{2\pi * 6}\\\\B = 5*10^{-6} \ T[/tex]
Therefore, the magnetic field the line produces at ground level is 5 x 10⁻⁶ T.
g What are the magnitude and direction of the electric field at a distance 43.5 mmmm from the center of the shells
Complete Question
Two spherical shells have a common center. A -1.6 x 10-6 C charge is spread uniformly over the inner shell, which has a radius of 0.030 m. A +5.1 x 10-6 C charge is spread uniformly over the outer shell, which has a radius of 0.15 m.What are the magnitude and direction of the electric field at a distance 43.5 mm from the center of the shells.
Answer:
The magnitude is [tex]E = 7.6021*10^{6} \N/C[/tex]
The direction is radially inward toward the center
Explanation:
From the question we are told that
The charge on the inner shell is [tex]q_i = -1.6*10^{-6} \ C[/tex]
The radius of the inner shell is [tex]c_1 = 0.030 \ m[/tex]
The charge on the outer shell is [tex]q_o = 5.1*10^{-6}\ C[/tex]
The radius of the outer shell is [tex]c_2 = 0.15\ m[/tex]
The distance considered is [tex]r = 43.5 \ mm = 0.0435 \ m[/tex]
Generally the electric field at the position considered is mathematically represented as
[tex]E = \frac{Q_c }{4\pi r^2 \epsilon_o }[/tex]
Here [tex]Q_c[/tex] is the charge which is enclosed by the distance considered which in this case is the charge on the inner shell
So [tex]Q_c =q_i = -1.6*10^{-6} \ C[/tex]
Hence
[tex]E = \frac{-1.6 *10^{-6} }{4* 3.142 *0.0435^2* 8.85*10^{-12} }[/tex]
=> [tex]E = -7.6021*10^{6} \N/C[/tex]
The negative sign is show that the direction of the field is radially inward toward the center
Determine the nuclear radius (in fm) for each of the following nuclei.
a) 126C
b) 147N
c) 6027Co
d) 20882Pb
Answer:
(a) 2.75 fm
(b) 2.89 fm
(c) 4.70 fm
(d) 7.12 fm
Explanation:
For a given element, the radius r of its nuclei is given by;
r = r₀[tex]A^{(1/3)}[/tex]
Where;
A = Atomic mass of the element
r₀ = 1.2 x 10⁻¹⁵m = 1.2fm
Now let's solve for the given elements
(a) ¹²₆C
Carbon element => This has an atomic mass number of 12
Therefore its radius is given by;
r = 1.2 x [tex]12^{1/3}[/tex]
r = 1.2 x 2.29
r = 2.75 fm
(b) ¹⁴₇N
Nitrogen element => This has an atomic mass number of 14
Therefore its radius is given by;
r = 1.2 x [tex]14^{1/3}[/tex]
r = 1.2 x 2.41
r = 2.89 fm
(c) ⁶⁰₂₇Co
Cobalt element => This has an atomic mass number of 60
Therefore its radius is given by;
r = 1.2 x [tex]60^{1/3}[/tex]
r = 1.2 x 3.92
r = 4.70 fm
(d) ²⁰⁸₈₂Pb
Lead element => This has an atomic mass number of 208
Therefore its radius is given by;
r = 1.2 x [tex]208^{1/3}[/tex]
r = 1.2 x 5.93
r = 7.12 fm
Convert an acceleration of 1km/h^2 into cm/8^2?
Answer:
1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]
Explanation:
1km = 10^3 m = 10^5 cm
1h = 60x60 = 3600s
--> 1km/h^2 = 10^5 cm / (3600^2) s^2
<=> 1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]
Add the vectors for: Two soccer players kick a ball simultaneously from opposite sides. Red #3 kicks
with 50 N of force while Blue #5 kicks with 63 N of force. What is the net force on
the ball?
Answer:
force applied by red = 50 N
force applied by blue = - 63 N (since it is in the opposite direction)
net force = force by red + force by blue
net force = 50 + (- 63)
net force = - 13 N
(b) A Blu-ray laser has a power of 5 milliwatts (1 watt = 1 J s−1). How many photons of light are produced by the laser in 1 hour? Energy=(Power)x(time) 5x10-3 J s-1x 1 x 3600s=18.0 J
Answer:
The number of photons of light produced by the laser in 1 hour is
1 Photon / hour
Explanation:
Number of photons of light produced is given by
[tex]Number of photons = \frac{Power}{Energy}[/tex]
From the question,
Power = 5 mW (milliwatts) = 5 × 10⁻³ W
Since 1 Watt = 1 Js⁻¹
Then, 5 × 10⁻³ W = 5 × 10⁻³ Js⁻¹
For the Energy,
As given from the question
Energy=(Power)x(time)
Time = 1 hour = (1 × 60 × 60) s = 3600 s
∴ Energy = 5 × 10⁻³ Js⁻¹ x 1 x 3600s
Energy =18.0 J
Now for the Number of photons produced,
[tex]Number of photons = \frac{Power}{Energy}[/tex]
Power = 5 × 10⁻³ Js⁻¹ = 0.005 Js⁻¹
[tex]Number of photons = \frac{0.005}{18}[/tex]
Number of photons = 2.78 × 10⁻⁴ Photons / sec
This is the number of photons produced in 1 second.
For the number of photons produced in 1 hour, we will multiply the result by 3600
(NOTE: 1 sec = [tex]\frac{1}{3600}[/tex] hour)
Number of photons = 2.78 × 10⁻⁴ Photons / sec
= 2.78 × 10⁻⁴ × 3600 Photons / hour
= 1.0008 Photons / hour
≅ 1 Photon / hour
Hence, the number of photons of light produced by the laser in 1 hour is
1 Photon / hour
Have you donated blood ? If so describe your experience. If not, explain whether you would or would not donate blood and why.
What accepted idea was replaced by copernicus's new approach to the solar system?
Answer:
The geocentric model of Ptolemy in which the sun and other planets were believed to move round the Earth.
Explanation:
The Copernicus heliocentric model of the solar system was put forward by Nicolaus Copernicus and was published in 1543. The model proposed by Copernicus puts the Sun near the center of the Universe, motionless, with Earth and the other planets orbiting around the sun in circular paths, modified by epicycles, and at uniform speed. This is contrary to Ptolemy's geocentric solar system that puts the Earth at the center of the solar system, in which the sun and other planets were believed to move round the Earth.
We all know that punching a solid wall can badly hurt your hands. Using Newton's Third Law, how can you explain this happening? a. According to Newton's Third Law, the force you put into the wall with your punch causes the wall to put that same force back onto your hand. b. According to Newton's Second Law, the harder you hit the wall, the more you need to exert yourself and your hand to do so, which damages your hand c. According to Newton's First Law, you cannot stop your hand once your cause it to start moving, thus your hand will be damaged by hitting the wall because you can't stop it in time d. Newton's Laws cannot explain this
The answer is B
Explanation:
Every action has an equal reaction. If you punch a wall, the wall is going to exert the wall with exert the same amount of force back towards your hand, causing your hand to hurt.
Answer:
a. According to Newton's Third Law, the force you put into the wall with your punch causes the wall to put the same force back onto your hand.
Explanation:
I got it correct.
A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 5.12 s later. If the speed of sound is 340 m/s, how high is the cliff
Answer:
height of cliff (h) = 112.38m
Explanation:
The time 5.12 s is the total time it takes for the rock to fall, and for the soundwave to travel back to the top of the cliff before it is heard.
[tex]5.12 = t_f\ +\ t_s - - - - -(1)\\where:\\t_f = time\ of\ fall\ of\ the\ piece\ of\ rock\\t_s = time\ travelled\ by\ the\ return\ sound[/tex]
Let h be the height of the cliff in meters, the time taken for the rock to fall is given by:
[tex]t_f=\sqrt{\frac{2h}{g} } \\where:\\t_f = time\ of\ fall\\h = height\ of\ cliff\\g= acceleration\ due\ to\ gravity= 9.8 m \slash s^2[/tex]
[tex]\therefore t_f = \sqrt{\frac{2h}{9.8}} \\squaring\ both\ sides\\(t_f)^2 = \frac{2h}{9.8}\\ 2h = 9.8 \timess\ (t_f)^2\\h = \frac{9.8 \timess\ (t_f)^2}{2} \\h= 4.9(t_f)^2 - - - - - (2)[/tex]
Next, let us calculate the time taken fot the sound to return
[tex]t_s = \frac{h}{v} \\where:\\t_s = time\ for\ sound\ to\ travel\ up\ the\ cliff\\h= distance\ tavelled\ = height\ of\ cliff\\v= speed\ = 340m \slash s\ (speed\ of\ sound)\\\therefore t_s = \frac{h}{340} - - - - - (3)\\[/tex]
now putting the values of h from equation 2 into equation (3)
[tex]t_s = \frac{4.9(t_f)^2}{340}[/tex]
Putting the value of [tex]t_s[/tex] into equation (1)
[tex]5.12 = t_f +\frac{4.9(t_f)^2}{340} \\[/tex]
multiplying through by 340
[tex]1740.8 = 340(t_f) + 4.9(t_f)^2\\4.9(t_f)^2 + 340 (t_f) - 1740.8 = 0[/tex]
now let us solve the quadratic equationsss;
[tex]Let\ (t_f) = x[/tex]
[tex]4.9x^2 + 340x - 1740.8 = 0\\using\ quadratic\ formula\\x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} \\x = \frac{-340 \pm\sqrt{(340)^2 -\ 4 \times4.9 \times(-1740.8)} }{2\times4.9}\\x = \frac{-340\ \pm\ 386.936}{9.8} \\x =\frac{386.938 - 340}{9.8} \\x = \frac{46.936}{9.8}\\ x = 4.789\\x = t_f\\t_f=4.789s[/tex]
note, time cannot be negative, so we ignored the negative answer
putting the value of [tex]t_f[/tex] into equation (2) to find height of cliff (h)
[tex]h= 4.9(t_f)^2\\h = 4.9 \times(4.789)^2\\h = 112.38m[/tex]
Therefore, height of cliff (h) = 112.38m
What are the laws of aerodynamics?
Answer:
Lift, weight, thrust and drag.
Answer:
According to Newton's first law of motion (inertia), an object at rest will remain at rest, or an object in motion will continue in motion at the same speed and in the same direction, until an outside force acts on it. For an aircraft to taxi or fly, a force must be applied to it. It would remain at rest without an outside force. Once the aircraft is moving, another force must act on it to bring it to a stop. It would continue in motion without an outside force. This willingness of an object to remain at rest or to continue in motion is referred to as inertia.
Newton's Second Law of Motion
The second law of motion (force) states that if a object moving with uniform speed is acted upon by an external force, the change of motion (acceleration) will be directly proportional to the amount of force and inversely proportional to the mass of the object being moved. The motion will take place in the direction in which the force acts. Simply stated, this means that an object being pushed by 10 pounds of force will travel faster than it would if it were pushed by 5 pounds of force. A heavier object will accelerate more slowly than a lighter object when an equal force is applied.
Newton's Third Law of Motion
The third law of motion (action and reaction) states that for every action (force) there is an equal and opposite reaction (force). This law can be demonstrated with a balloon. If you inflate a balloon with air and release it without securing the neck, as the air is expelled the balloon moves in the opposite direction of the air rushing out of it.
g a small metal sphere, carrying a net charge is held stationary. what is the speed are 0.4 m apart
Answer:
The final speed of small metal sphere is 12.6 m/s.
Explanation:
Given that,
Distance = 0.4 m
Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.
We need to calculate the final speed of small metal sphere
Using conservation of energy
[tex]\dfrac{1}{2}mv_{1}^2+\dfrac{kq_{1}q_{2}}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{kq_{1}q_{2}}{r_{2}}[/tex]
[tex]\dfrac{1}{2}m(v_{1}^2-v_{2}^2)=kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{2}^2)=9\times10^{9}\times(-2)\times10^{-6}\times(-8)\times10^{-6}(\dfrac{1}{0.4}-\dfrac{1}{0.8})[/tex]
[tex]400-v_{2}^2=240[/tex]
[tex]-v_{2}^2=240-400[/tex]
[tex]v_{2}=\sqrt{160}\ m/s[/tex]
[tex]v_{2}=12.6\ m/s[/tex]
Hence, The final speed of small metal sphere is 12.6 m/s.
If someone told you that they traveled 50 km east, are they describing the distance they traveled or their displacement? How do you know? *
Answer:
they are describing their displacement since displacement is nothing but distance along with the direction of motion
If this circuit were connected to a standard 120 V ac outlet, what would the rms current in the circuit be
Answer:
Irms =226A
Explanation:
The current is high because the total impedance is relatively low. Actually, plugging such a circuit into a 120-V outlet would most likely burn out the circuit elements
To which layer can geologists apply the principle of faunal succession to determine the age of the layer?
A. Layer C
B. Layer A
C. Layer D
D. Layer B
A bullet is fired horizontally with initial velocity of 800 m/s as a target located 200 mfrom the rifle.(a).How much time is required for the bullet to reach the target
Answer:
t = 0.25 seconds
Explanation:
Given that,
Initial speed of a bullet, v = 800 m/s
Distance from the target is 200 m
We need to find the time required for the bullet to reach the target. Time is simply calculated by the definition of velocity i.e.
[tex]t=\dfrac{d}{v}\\\\t=\dfrac{200\ m}{800\ m/s}\\\\t=0.25\ s[/tex]
So, it will take 0.25 seconds to reach the target.
If your speedometer has an uncertainty of 2.5 km/h at a speed of 92 km/h, what is the percent uncertainty
Answer:
2.7%
Explanation:
Given:
Uncertainty of the speedometer (u)= 2.5km/h
Speed measured at that uncertainty (v) = 92km/h
Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e
p = [tex]\frac{u}{v} * 100[/tex]%
p = [tex]\frac{2.5}{92} * 100[/tex]%
p = 2.7%
Therefore, the percent uncertainty is 2.7%
A sound source producing 1.00 kHz waves moves toward a stationary listener at one-half the speed of sound.
a) What frequency will the listener hear?
(b) Suppose instead that the source is stationary and the listener moves toward the source at one half the speed of sound. What frequency does the listener hear? How does your answer compare with that in par a? Did you expect to get the same answer in both cases? Explain on physical grounds why the two answers differ
Answer:
2000Hz and 1500Hz
Explanation:
Using
a) f = f0((c+vr)/(c+vs))
=>>> f0((c)/(c-0.5c))
=>>>1000/0.5 = 2000Hz
b) f = f0((c+vr)/(c+vs))
=>>>f0((c+0.5c)/(c))
=>>>>1000 x 1.5 = 1500Hz
Here we have a problem referring to the Doppler effect, the solutions are:
a) f = 2.0 kHzb) f = 1.5 kHzThe Doppler effect:
The Doppler effect is an effect that explains how the perception of waves changes as the source moves or as the listener moves.
The formula, for sound, is:
[tex]f = \frac{v + v_0}{v - v_1}*f_0[/tex]
where:
v = velocity of the wave = 340 mf = perceived frequencyf₀ = actual frequency = 1.00kHzv₀ = velocity of the observerv₁ = velocity of the source.a) First we have that the source moves towards a stationary listener, then we have:
[tex]f = \frac{v }{v - v/2}*1.0 kHz\\\\f = \frac{v }{v/2}*1.0 kHz = 2.0 kHz[/tex]
b) in this case, the listener moves towards the source, so we have:
[tex]f = \frac{v + v/2 }{v }*1.0 kHz\\\\f = \frac{ (3/2)*v }{v}*1.0 kHz = 1.5 kHz[/tex]
So in this case the perceived frequency is smaller than in the point a.
This is because the waves will move at a fixed rate in air, in one case, the successive waves are emitted from different points in space (each time closer to the listener) while in the other case the waves are emitted from a fixed point, and the listener moves towards them, thus feels that the waves move faster.
If you want to learn more about the Doppler effect, you can read:
https://brainly.com/question/3826119
The tires of a car make 90 revolutions as the car reduces its speed uniformly from 88.0 km/hkm/h to 56.0 km/hkm/h. The tires have a diameter of 0.84 mm.
1. What was the angular acceleration of the tires?
2. If the car continues to decelerate at this rate, how much more time is required for it to stop?
3. If the car continues to decelerate at this rate, how far does it go? Find the total distance.
Answer:
1) α = 2.2 rad / s² , 2) t = 7.068 s , 3) in this interval s = 23.096 m
total distance s = 57.4 m
Explanation:
For this exercise we use the angular kinematic relations, before starting the problem we reduce all the magnitude to the SI system
v₀ = 88 km / h (1000 m / 1km) 1h / 3600s) = 24.44 m / s
v = 56.0 km / h = 15.55 m / s
θ = 90 rev (2pi rad / 1 rev) = 565,487 rad
d = 0.84 m
r = d / 2
r = 0.84 / 2
r = 0.42 m
1) ask for angular acceleration
w² = w₀² - 2 α Δθ
α = (w₀² -w²) / 2 Δθ
To find the angular velocities we use the acceleration between the linear and angular velocity
v = w r
w = v / r
w₀ = 24.44 / 0.42
w₀ = 58.20 rad / s
w = 15.55 / 0.42
w = 30.037 rad / s
we calculate
α = (58.20² - 30.037²) / (2 565.487)
α = 2.2 rad / s²
2) how much longer does it take to stop
w₂ = 15.55 rad / s
w = 0
w = w₂ - α t
t = (w₂ -0) / α
t = 15.55 / 2.2
t = 7.068 s
3) the distance that the car travels from the beginning of the movement, we can find it by looking for the number of revolutions until it stops and then using the relationship between the angular and linear variable
w² = w₀² - 2 α θ
at the end of the movement speed is zero
0 = w₀² - 2 α θ
θ = w₀² / 2 α
θ = 24.44² / (2 2.2)
θ = 135.80 rad
If the angles are measured in radians, we can apply the relation
θ = s / R
s = R ttea
s = 0.42 135
s = 57.4 m
this is the distance from when the movement starts
the distance for the final part of the movement is
w = 15.55 rad / s
θ = w² / 2 α
θ = 15.55 2 / (2 2.2)
θ = 54.99 rad
the distance in this interval is
s = 0.42 54.99
s = 23.096 m