PHY 150 - 5-2 Project Two - Logan Mayer
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A&L ENGINEERING
Case Material Evaluation Report
Newton’s Laws of Motion
Newton’s Three Laws of Motion are:
1.
An object at rest will remain at rest, and an object in motion will remain in motion unless acted upon from an external force.
2.
The force acting on the object is equal to the mass of the object times its acceleration.
3.
When an object exerts force upon another object, that object has an equal reaction.
Force Diagrams
The chosen drop height is 1.5m.
Velocity Calculations
Tip
: To begin, think about the drop in two phases: The first phase is from the drop to the moment
of first contact with the floor. The second phase is the deceleration at impact. The collision time is the time from first contact with the floor to the instant the phone is momentarily at rest.
Phone (No Case)
6.2oz = 0.175kg
mgh
=
1
2
m v
2
+
0
√
2
∗
9.8
∗
1.5
=
5.42
0
=
5.42
+
0.01
∗
a
a
=−
542
m
/
s
2
Phone (Silicone Case)
6.2oz + 1.7oz = 7.9oz = 0.22kg
mgh
=
1
2
m v
2
+
0
√
2
∗
9.8
∗
1.5
=
5.42
0
=
5.42
+
0.05
∗
a
a
=−
108.4
m
/
s
2
Phone (Hard Plastic Case)
6.2oz + 1.1oz = 7.3oz = 0.20kg
mgh
=
1
2
m v
2
+
0
√
2
∗
9.8
∗
1.5
=
5.42
0
=
5.42
+
0.03
∗
a
a
=−
180
.
6
m
/
s
2
Phone (Rubber Case)
6.2oz + 3.2oz = 9.4oz = 0.26kg
mgh
=
1
2
m v
2
+
0
√
2
∗
9.8
∗
1.5
=
5.42
0
=
5.42
+
0.08
∗
a
a
=−
67.7
m
/
s
2
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Related Questions
Mechanical engineering thermodynamics.
Handwritten solution accepted.
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1)sketch of the problem including everything?
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Thermodynamics deals with forms of energy and change in the total energy of
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The horizontal component of Force F1 (unit is in N), H1=
The vertical component of Force F1 (unit is in N), V1 =
The horizontal component of Force F2 (unit is in N), H2=
The vertical component of Force F2 (unit is in N), V2 =
The horizontal component of Force F3 (unit is in N), H3=
The vertical component of Force F3 (unit is in N), V3 =
The vertical component of Force F4 (unit is in N), V4 =
The horizontal component of Force F5 (unit is in N), H5=
The vertical component of Force F5 (unit is in N), V5 =
The horizontal component of Force F6 (unit is in N), H6=
The vertical component of Force F6 (unit is in N), V6 =
The sum of horizontal force component for the given system (unit is in N),begin inline style sum for H of end style =
The sum of vertical force component for the given system (unit is in N),begin inline style sum for V of end style =
The resultant of the given force system is (unit in N), R =
The direction of Resultant Force is =
Short Answer question
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Science
PhysicsQ&A LibraryA child’s toy consists of a m = 31 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monkey is first hung on the spring and the system reaches equilibrium, the spring has stretched a distance of x = 17.6 cm, as shown in the diagram. This toy is so adorable you pull the monkey down an additional d = 7.6 cm from equilibrium and release it from rest, and smile with delight as it bounces playfully up and down. 1. Calculate the speed of the monkey, ve, in meters per second, as it passes through equilibrium. 2. Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms of m, x, d, k, the maximum height above the bottom of the motion, hmax, and the variables available in the palette. 3. Calculate the maximum displacement, h, in centimeters, above the equilibrium position, that the monkey reaches.
A child’s toy consists of a…
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Question 4
a) A helicopter pilot sets the aircraft in horizontal forward flight by activating the rotor disk
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to be accounted for when considering the equilibrium of the aircraft? Complete the
sketch by adding the missing forces and provide a clear explanation of the balance of
the forces in a vector form.
F
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SUBJECT:MECHANICAL ENGINEERING
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(Is it ∆ U=-W ? Show your work)
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engineering dynamics
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COMBUSTION ENGINEERING
1. The following details were noted in a test on a single-cylinder four-stroke
oil engine: bore = 150 mm; stroke = 160 mm; speed of engine = 500
rpm; fuel consumption = 0.0475 kg/min; calorific value of fuel = 42,000
kJ/kg; the difference in tension on either side of the brake pulley = 400 N;
brake circumference = 2.2 m; length of the indicator diagram = 50 mm;
area of positive loop of indicator diagram = 475 mm2; area of negative
loop = 25 mm?; spring constant = 0.8333 bar per mm.
Calculate:
a) The Brake Power in kW
b) The Indicated power in kW
c) The mechanical efficiency
d) The Brake thermal efficiency, in percent
kg
e) The Brake Specific Fuel Consumption,
bkW hr
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COMBUSTION ENGINEERING
1. The following details were noted in a test on a single-cylinder four-stroke
oil engine: bore = 150 mm; stroke = 160 mm; speed of engine = 500
rpm; fuel consumption = 0.0475 kg/min; calorific value of fuel = 42,000
kJ/kg; the difference in tension on either side of the brake pulley = 400 N;
brake circumference = 2.2 m; length of the indicator diagram = 50 mm;
area of positive loop of indicator diagram = 475 mm²; area of negative
loop = 25 mm?; spring constant = 0.8333 bar per mm.
Calculate:
d) The Brake thermal efficiency, in percent
kg
e) The Brake Specific Fuel Consumption,
bkW-hr
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QUESTION 10
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Learning Syst X
G parallelogram law question force x
b Answered: Quiz: Determine the n X
b My Tutoring | bartleby
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Q Search for textbooks, step-by-step explanations to homework questions,
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Quiz: Determine the magnitude of the resultant force and its d
Question
So
availa
Quiz: Determine the magnitude of the resultant force and its direction by using
Parallelogram law.
A "Res
Medi
6kN
Tagged in
10KN
Engineer.
ere to search
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You are a student at the HTU University, the instructor of the fundamental of thermodynamic course
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friction or air resistance."
1
mgy; + mv? = mgy, + mvž
1
mechanical energy = potential energy + kinetic energy = mgy + mv2
m
g = magnitude of acceleration due to gravity = 9.81-
s2
Yi = initial height
Yf = final height
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True
False
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True
False
3. Thermodynamic properties can be placed in two general classes control volume and intensive
True
Flase
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True
False
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True
False
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2.2nd law of thermodynamics (foward engine)
3.Joules law
4.2nd law of thermodynamics (reverse engine)
5.3rd law of thermodynamics
6.second law of thermodynamics (Carnot)
A.
The internal energy of a perfect gas is a function of the absolute temperature only
B.
No Heat Engine can be more efficient than a Reversible Heat Engine operating between the same temperature limits
C.
A pure crystalline substance at absolute zero temperature is in perfect order, and its entropy is zero
D.
When a system undergoes a complete cycle, the net Heat supplied plus the net Work input is zero.
E.
It is impossible for a heat engine to produce a net work output in a complete cycle if it exchanges heat only with a single energy reservoir.
F.
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0.3. Answer the following questions.
between the boxes and the surface is μ
P = 50 lb
2c) How much friction is present as the boxes slow to a stop? Show your work.
Solution
I
Units
B
Direction
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b. What is the average acceleration due to the earth's curvature and earth's rotation around
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Physics 121 Spring 2021 - Document #11: Homework #04 & Reading Assignment page 4 of 8
Problem 1: Gnome Ride - This from a Previous Exam
I.
A Gnome of given mass M goes on the Gnome Ride as follows: He stands on a horizontal
platform that is connected to a large piston so that the platform is driven vertically with a position
as a function of time according to the following equation:
y(t) = C cos(wt)
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units) and y represents the vertical position, positive upward as indicated.
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Publisher:WILEY
Mechanics of Materials (MindTap Course List)
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ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
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Author:James L. Meriam, L. G. Kraige, J. N. Bolton
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