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PHYS2091
Unit 4 AS: Homework 4
Description:
The purpose of this assignment is to develop problem-solving skills using real-life examples. The
students will engage in problems ranging from everyday activities to the applications of physics in biology.
Instructions:
This assignment has 15 questions of various complexities. Upload your answers as a single PDF file on Blackboard.
35 points CSLO 1 (a – e), 2, 3
1.
(3 points)
(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. [(1500)(9.8)+100] x 40= 5.92x10^2J
(b) What is the work done on the lift by the gravitational force in this process? (1500)(9.8)(40)= 5.88x10^5J
(c) What is the total work done on the lift? 0J
2.
(5 points)
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against
a 35.0 N frictional force. He pushes in a direction 35.0° below the horizontal. (a) What is the work done on the cart by friction? (35)(20)cos(180)= -700J
(b) What is the work done on the cart by the gravitational force? (20)cos(90)= 0J
(c) What is the work done on the cart by the shopper? (-700)-0 = 700J
(d) Find the force the shopper exerts, using energy considerations. 700/(20)cos(35)= 38.7N
(e) What is the total work done on the cart? 0J
3.
(2 points)
(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). –(950)[(90)(1000/1)
(1/3600)]^2/2(120)= -2.474x10^3N
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a). -(950)
[(90.0) (1000/1) (1/3600)]/ 2(2.0) = -1.484 x 10^5N
-1.484 x 10^5/ -2.474 x10^3= 60 N
4.
(1 point)
A car’s bumper is designed to withstand a 4.0-km/h (1.12-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by
absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an initial speed of 1.12 m/s. 900/2(0.200)((0)-1.1)^2)= 2722.5N
5.
(2 points)
Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (75x10^-3)(9.8)(2.5)= 1.837J
(b) How much work did it do to raise its own center of mass to the branch? (0.350)(9.8)(2.5)= 8.575J
6.
(1 point)
A 5
×
10
5
kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant k
of the spring? (5.00x10^5)(0.5)^2/(0.400)= 7.812x10^5N/m
7.
(2 points)
A car is traveling at 10 m/s. (a) How fast would the car need to go to double its kinetic energy? 14.14m/s
(b) By what factor does the car’s kinetic energy increase if its speed is doubled to 20 m/s? Factor=
K
1/
K
2 = 200/50
= 4
8.
(2 points)
(a) What is the kinetic energy of a 1,500 kg car traveling at a speed of 30 m/s? ½ (1500)(30)^2= 67.5x10^5J
(b) From what height should the car be dropped to have this same amount of kinetic energy just before impact? (30)^2/2(9.8)= 45.92m
9.
(1 point)
Melissa drags a 23 kg duffel bag 15 m across the gym floor. If the coefficient of kinetic friction between the floor and the bag is 0.18, how much thermal energy does Melissa create? W
= 40.6134 ×15=609.201 J
10.
(4 points)
A 2.3 kg box, starting from rest, is pushed up a ramp by a a10 N force parallel to the ramp. The ramp is 2.0 m long and tilted at 17°. The speed of the box at the top is 0.80 m/s. (a) How much work is done by the force on the system (box + ramp + Earth)? 20J
(b) What is the change in kinetic energy of the system? ½(2.3)(0.8)^2= 0.736J
(c) What is the change in gravitational potential energy of the system? 2.3(9.8)(2.0sin17)= 13.194J
(d) What is the change in thermal energy of the system? 6.07J
11.
(3 points)
A 50 kg sprinter, starting from rest, runs 50 m in 7.0 s at constant acceleration. (a) What is the magnitude of the horizontal force acting on the sprinter? 2(50)/(7.0)^2= 2.04m/s^2
(b) What is the sprinter’s average power output during the first 2.0 s of his run? (102)(4.08)/2= 208.1W
(c) What is the sprinter’s average power output during the final 2.0 s? (102)(24.5)/2= 1249.5W
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Pls solve this question correctly instantly in 5 min i will give u 3 like for sure
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dTglass = -0.75Tolass + 0.75TpvtT
dt
(1)
dTpyT
= -
dt
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dTwax
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dt
(3)
Where, Tatass, Tpyr, and Twax, are temperatures illustrated in Figure 1(b). At time t=0 the initial
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Call seppty
Ia tree water
Tank
Glass
PVT
Expu
Nane PCMPVT Callector
Tub
Tepe
Sterg Tak
e Ma
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Hest Eschanger
Container
Teket
vae
Pump
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(b)
Figure 1. (a) Schematic diagram of the nano-PCM and nanofluid PVT system, (b) mathematical model in context of the
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Part A
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a- Give the solution and calculate V, E.
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HYS1800 A - Engineering Mechanics (Fall 2021-2022)
My courses / SC/PHYS1800 A - Engineering Mechanics (Fall 2021-2022) / Work and Energy / Assignment 6
A car travels at a steady 30.3 m/s around a horizontal curve of radius 111 m. What is the minimum coefficient of static
friction between the road and the car's tires that will allow the car to travel at this speed without sliding?
Answer:
Next page
MacBook Air
DD
08
F3
64
F5
F6
F4
F2
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Problem: (The answer to this question is in the given picture)
A person with a black belt in karate has a fist that has a mass of 0.70 kg. Starting from rest, this fist attains a velocity of 8.0 m/s in 0.15 s. What is the magnitude of the average net force applied to the fist to achieve this level of performance?
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Problem: (The answer to this question is in the given picture)
1. A car travels to the east at a constant 50 km/h for 100 km. It then speeds up to 100 km/h and is driven another 100 km.
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Problem: (The answer to this question is in the given picture)
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b.
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d.
F.
(V-V)
F = m
e.
(1,-4)
where
F-force (N)
mmass (kg)
a acceleration (m's)
V velocity (m's)
R-radius (m)
lime (s)
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Force on q, by q, = 79.889N
91
O cm
3
4
5
7
8
10
Charge 1
Charge 2
Force Values
-4 μC
2 μ
Scientific Notation
-10
10
-10
10
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