Answer:b
Explanation:
Help me with my physics, please
4. Paper is solid in packets labelled 80 g/m2. This means that a sheet of paper of area
10 000cm? has a mass of 80 g. The thickness of each sheet is 0.11mm. What is the
density of the paper?
A 0.073 g/cm?
B 0.088 g/cm
C 0.73 g/cm3
D 0.88 g/cm
B
с
Answer:
Option C. 0.73 g/cm³
Explanation:
From the question given above, the following data were obtained:
Mass = 80 g
Area (A) = 10000 cm²
Thickness = 0.11 mm
Density =?
Next, we shall convert 0.11 mm to cm. This can be obtained as follow:
10 mm = 1 cm
Therefore,
0.11 mm = 0.11 mm × 1 cm / 10 mm
0.11 mm = 0.011 cm
Thus, 0.11 mm is equivalent to 0.011 cm.
Next, we shall determine the volume of the paper. This can be obtained as follow:
Area (A) = 10000 cm²
Thickness = 0.011 cm
Volume =?
Volume = Area × Thickness
Volume = 10000 × 0.011
Volume = 110 cm³
Finally, we shall determine the density of the paper. This can be obtained as follow:
Mass = 80 g
Volume = 110 cm³
Density =?
Density = mass / volume
Density = 80 / 110
Density = 0.73 g/cm³
Therefore the density of the paper is 0.73 g/cm³
A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West
What would the separation between two identical objects, one carrying 4 C of positive charge and the other 4 C of negative charge, have to be if the electrical force on each was precisely 8 N
Answer:
7.46×10⁻⁶ m
Explanation:
Applying,
F = kqq'/r²............ Equation 1
make r the subject of the equation
r = √(F/kqq').......... Equation 2
From the question,
Given: F = 8 N, q' = q= 4 C
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 2
r = √[8/(4×4×8.98×10⁹)]
r = √(55.7×10⁻¹²)
r = 7.46×10⁻⁶ m
A boy is playing with a water hose, which has an exit area of
10 cm2 and has water flowing at a rate of 2 m/s. If he covers
the opening of the hose with his thumb so that it now has an
open area of 2 cm2, what will be the new exit velocity of the
water?
Answer:
The exit velocity of water is B. 15 m/s.
Explanation:
According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.
If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity is v₂, then,
Rewrite the expression for v₂.
Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.
The exit speed of water from the hose is 15 m/s.
A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate
(i) The speed of the train
(ii) The time it will taketo cover 240 m.
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]
A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops
Answer:
Angular displacement before it stops = 18 rev
Explanation:
Given:
Speed of fan w(i) = 6 rev/s
Speed of fan (Slow) ∝ = 1 rev/s
Final speed of fan w(f) = 0 rev/s
Find:
Angular displacement before it stops
Computation:
w(f)² = w(i) + 2∝θ
0² = 6² + 2(1)θ
0 = 36 + 2θ
2θ = -36
Angular displacement before it stops = -36 / 2
θ = -18
Angular displacement before it stops = 18 rev
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
Give reason why a man getting out of moving bus must run in the same direction for a certain distance.
Explanation:
Explanation: It's because when he stop down from a moving bus his feet come at rest while the upper portion of his body is still in motion and he falls in the forward direction.
Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s
Answer:
a
Explanation:
you take 32,000kg ÷2.0m
•. What is called the error due to the procedure and used apparatuses?
a. Random error
b. Index error
c. Systematic error
d. Parallax error.
Answer:
[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]
93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would result in an induced emf in the loop?
A. Moving the loop outside of the magnetic field region.
B. Change the diameter of the loop.
C. Change the magnitude of the magnetic field.
D. Spin the loop such that its axis does not consistently line up with the magnetic field direction.
Answer:
All the given options will result in an induced emf in the loop.
Explanation:
The induced emf in a conductor is directly proportional to the rate of change of flux.
[tex]emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta[/tex]
where;
A is the area of the loop
B is the strength of the magnetic field
θ is the angle between the loop and the magnetic field
Considering option A, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.
Considering option B, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.
Option C has a similar effect with option A, thus both will result in an induced emf.
Finally, considering option D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will change the angle between the loop and the magnetic field. This effect will also result in an induced emf.
Therefore, all the given options will result in an induced emf in the loop.
Your dog is running around the grass in your back yard. He undergoes successive displacements 3.20 m south, 8.16 m northeast, and 15.6 m west. What is the resultant displacement
Answer:
D1 = 3.50 m, south; D2 = 8.20 m, northeast; D3 = 15.0 m, west. Converting all these displacements from east where zero degrees is at east or + x-axis, the converted displacements are: D1 = 3.50 m 270°; D2 = 8.20 m 45° and D3 = 15.0 m 180°. We then tabulate these vectors including there x and y components. The x-components are solved by magnitudes * cos of direction angle while the y-components of the three vectors are solved by magnitudes * sin of direction angle.
The resultant is computed by summing the components algebraically. The direction in degrees is the arc tangent of the sum of all y divided by the sum of all x.
Explanation:
. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force
Answer:
elative magnitude of the two forces is the same and they are applied in a constant direction.
Explanation:
Newton's second law states that the sum of the forces is equal to the mass times the acceleration
∑ F = m a
in this case there are two forces on the x axis
F_applied - fr = 0
since they indicate that the velocity is constant, consequently
F_applied = fr
the relative magnitude of the two forces is the same and they are applied in a constant direction.
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
Physics help please
Answer:
i think the answer is 0.001m³
Two sinusoidal waves have the same frequency and wavelength. The wavelength is 20 cm. The two waves travel from their respective sources and reach the same point in space at the same time, resulting in interference. One wave travels a larger distance than the other. For each of the possible values of that extra distance listed below, identify whether the extra distance results in maximum constructive interference, maximum destructive interference, or something in-between.
a. 10 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
b. 15 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
c. 20 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
d. 30 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
e. 35 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
f. 40 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
Answer:
Explanation:
When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.
For constructive interference , path diff = n λ
For destructive interference path diff = ( 2n+ 1 ) λ /2
where λ is wave length of wave , n is an integer.
a )
path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.
b )
path diff = 15 cm which is neither half the wavelength nor full wavelength , so in between is the right option.
c )
path diff = 20 cm which is equal to the wavelength , so maximum constructive interference will occur.
d)
path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.
e)
path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.
f )
path diff = 40 cm which is 2 times the wavelength , so maximum constructive interference will occur
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field strength in volts per meter
Answer:
7.60× 10^6 V/m
Explanation:
electric field strength can be determined as ratio of potential drop and distance, I.e
E=V/d
Where E= electric field
V= potential drop= 74.0 mV= 0.07 V
d= distance= 9.20 nm = 9.2×10^-9 m
Substitute the values
E= 0.07/ 9.2×10^-9
= 7.60× 10^6 V/m
Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.
Answer:
Explanation:
English alphabets numbered fro 1 to 26
and the numbers 1 to10 so they are written in roman numbers as
1 - I
2 - II
3 - III
4 - IV
5 -V
6 - VI
7 -VII
8 - VIII
9 - IX
10 -X
11 - XI
12 - XII
13 - XIII
14 - XIV
15 - XV
16 - XVI
17 - XVII
18 - XVIII
19 - XIX
20- XX
21 - XXI
22 - XXII
23 - XXIII
24 - XXIV
25 - XXV
26 - XXVI
when blueshift occurs,the preceived frequency of the wave would be?
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
Define measurements.
Answer:
act or process of measuring
Explanation:
Explanation:
the comparison of an unknown quantity with a known quantity.
a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?
After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound
Answer:
64.20m
Explanation:
As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.
[tex]a^{2} +b^{2} = c^{2}[/tex]
where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x
[tex]51^{2} +39^{2} =x^{2}[/tex]
2,601 + 1,521 = [tex]x^{2}[/tex]
4,122 = [tex]x^{2}[/tex] ... square root both sides
64.20 = x
Finally, we see that the shortest distance is 64.20m
A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this
value express?
Answer:
the SI unit (meter per second square) indicates a linear type of motion.
Explanation:
Given;
acceleration of the car, a = 5.2 m/s² North
the SI unit of the car, = m/s²
The SI unit of the given value (acceleration), indicates a linear type of motion.
Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.
Therefore, the SI unit (meter per second square) indicates a linear type of motion.
Answer:
displacement
Explanation:
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, [tex]$\omega = \frac{1}{8}$[/tex] rev/sec
[tex]$=\frac{2 \pi \times 7.5}{8}$[/tex] rad/sec
= 5.89 rad/sec
Therefore, force required,
[tex]$F=m.\omega^2.r$[/tex]
[tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster?
Answer:
When an object is dropped, the "principal" force that acts on that object is the gravitational force.
Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:
g = 9.8m/s^2
So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)
Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.
But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)
And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)
Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.
A ball drops from a height, bounces three times, and then rolls to a stop when it reaches the ground the fourth time.
At what point is its potential energy greatest?
At what points does it have zero kinetic energy?
At what point did it have maximum kinetic energy?
Answer:
Greatest potential: moment before being dropped
Zero Kinetic: when it comes to rest
Greatest Kinetic: moment before first bounce
Explanation:
Which of the following categories of motion is mutually exclusive with each of the others? A. Translational motion B. Rectilinear motion C. Rotational motion D. Curvilinear motion
Answer:
C. Rotational motion
Explanation:
The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).
Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.
What is Rotational motion?"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."
Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.
What is Curvilinear motion?Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.
The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.
Learn more about motion here:
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