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stealth61 [152]
3 years ago
6

"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin

gs in a horizontal circle at a rate of 37.0 rev/min .
Physics
1 answer:
Bingel [31]3 years ago
3 0
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N. 

<span>Fx = [(233 + 840)/g]*v²/7.5 </span>

<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>

<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>

<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>

<span>233 + 840 = Ti*cos40º </span>

<span>solve for Ti. (This is the answer to the part b) </span>

<span>Horizontally </span>

<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>

<span>Solve for Th </span>

<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>

<span>using v and Ti computed above.</span>
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IgorLugansk [536]

Answer:

60/90

Explanation:

I think because the train's highest velocity is 60 n the time is 90

5 0
3 years ago
You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th
vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0
3 years ago
Read 2 more answers
A 2000 kg car slams on the brakes and slows down at a rate of -10 m/s2. How much force are the brakes applying?
Gemiola [76]

Answer:

-20,000N

Explanation:

Force (N) = mass (kg) x acceleration (m/s²)

So,

Force = 2000 x -10

= -20,000N (Newtons)

5 0
2 years ago
In the waves lab you change blank to see how it would affect wave speed​
GaryK [48]

where is the question in this ????

7 0
3 years ago
Two children are pulling on opposite sides of a blanket. The brother is pulling with a force of 3 N. The sister is pulling with
Citrus2011 [14]
Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)

Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N

-F

towards the sister (-F) (greater force applied)
7 0
2 years ago
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