Question 6, Physics - equation sheet attached Physics 114 Equation SheetConstants and ConversionsKinematics Continuedg = 9.80 m/sFree-fall accelerationΔνInstantaneousainst. = limAt-o AtAcceleration1N = 1 kg m/s?NewtonUniform motion(v) = (v); = constantPosition in uniformX = x + (v)AtMathematics, Scaling and Vectorsb = a* + loga (b) = xmotionLogarithmConstant(v); = (v,); + azAtacceleration:1log(ab) = log(a) + log (b)x, = x, + (v,),At +a, (at)?log Ax" = n log x + log A(v,); = (v,)} + 2a,AxVolume of a sphereV =Surface area of aA = 4ar?ForcessphereNewton's second lawFnet = EF = mã%3DVolume of a cylinderV = arlNewton's second lawFnetx = EF = ma,%3DSurface area of aA = 2ar? + 2rlComponent formFnety = ER, = maycylinderMass densityp = m/VNewton's Third LawFA en =-ton AA, = A cos e (rel. to x-axis)Weightw = mgx -component of avector ÅApparent weightWapp = magnitude of supporting forcesy -component of aAy = A sin 8 (rel to x-axis)vector ÅKinetic frictionfk = HanMagnitude of vector ẢStatic frictionA = JA + A,Reynolds numberRe = pvl/nDirection of A relative8 = tan-(Ay/A,)1Drag (high Reynoldsnumber)=CopAv?to x-axisAddition of two vectors If = Å + B, thenC, = A, + B,D = 6nyrvDrag (low Reynoldsnumber)Cy = Ay + ByCircular MotionKinematicsCentripetal accelerationa =DisplacementAx = x - XAverage VelocityAxFrequency1VargT2arAtAxVinst. = limInstantaneous VelocityAt+0 AtAvAverage AccelerationdavgΔε A car drives from city A to city B and then back to city C at a constant speed of74.0 km/h. If the entire trip takes 2.10 h, and the distance between city A and city C is63.0 km, what is the distance between city A and city B? (The diagram is not drawn toscale).ABO 109 kmO The correct answer is not listed.O 248 kmO 46.2 kmO 126 km

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Description: Question 6, Physics - equation sheet attached Physics 114 Equation SheetConstants and ConversionsKinematics Continuedg = 9.80 m/sFree-fall accelerationΔνInstantaneousainst. = limAt-o AtAcceleration1N = 1 kg m/s?NewtonUniform motion(v) = (v); = consta

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A car drives from city A to city B and then back to city C at a constant speed of 74.0 km/h. If the entire trip takes 2.10 h, and the distance between city A and city C is 63.0 km, what is the distance between city A and city B? (The diagram is not drawn to scale). A B O 109 km O The correct answer is not listed. O 248 km O 46.2 km 126 km

A car drives from city A to city B and then back to city C at a constant speed of 74.0 km/h. If the entire trip takes 2.10 h, and the distance between city A and city C is 63.0 km, what is the distance between city A and city B? (The diagram is not drawn to scale). A B O 109 km O The correct answer is not listed. O 248 km O 46.2 km 126 km

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question 6, Physics - equation sheet attached

Physics 114 Equation Sheet
Constants and Conversions
Kinematics Continued
g = 9.80 m/s
Free-fall acceleration
Δν
Instantaneous
ainst. = lim
At-o At
Acceleration
1N = 1 kg m/s?
Newton
Uniform motion
(v) = (v); = constant
Position in uniform
X = x + (v)At
Mathematics, Scaling and Vectors
b = a* + loga (b) = x
motion
Logarithm
Constant
(v); = (v,); + azAt
acceleration:
1
log(ab) = log(a) + log (b)
x, = x, + (v,),At +a, (at)?
log Ax" = n log x + log A
(v,); = (v,)} + 2a,Ax
Volume of a sphere
V =
Surface area of a
A = 4ar?
Forces
sphere
Newton's second law
Fnet = EF = mã
%3D
Volume of a cylinder
V = arl
Newton's second law
Fnetx = EF = ma,
%3D
Surface area of a
A = 2ar? + 2rl
Component form
Fnety = ER, = may
cylinder
Mass density
p = m/V
Newton's Third Law
FA en =-
ton A
A, = A cos e (rel. to x-axis)
Weight
w = mg
x -component of a
vector Å
Apparent weight
Wapp = magnitude of supporting forces
y -component of a
Ay = A sin 8 (rel to x-axis)
vector Å
Kinetic friction
fk = Han
Magnitude of vector Ả
Static friction
A = JA + A,
Reynolds number
Re = pvl/n
Direction of A relative
8 = tan-(Ay/A,)
1
Drag (high Reynolds
number)
=CopAv?
to x-axis
Addition of two vectors If = Å + B, then
C, = A, + B,
D = 6nyrv
Drag (low Reynolds
number)
Cy = Ay + By
Circular Motion
Kinematics
Centripetal acceleration
a =
Displacement
Ax = x - X
Average Velocity
Ax
Frequency
1
Varg
T
2ar
At
Ax
Vinst. = lim
Instantaneous Velocity
At+0 At
Av
Average Acceleration
davg
Δε

A car drives from city A to city B and then back to city C at a constant speed of
74.0 km/h. If the entire trip takes 2.10 h, and the distance between city A and city C is
63.0 km, what is the distance between city A and city B? (The diagram is not drawn to
scale).
A
B
O 109 km
O The correct answer is not listed.
O 248 km
O 46.2 km
O 126 km

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