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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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BARSIC [14]

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

7 0
3 years ago
A car speeds up from 12.0 m/s to 16.0 m/s in 8.0 s. the acceleration of the car is?
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6 0
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The diameter of Saturn is nearly ten times that of Earth. However, the density of Saturn is much less than that of Earth. What i
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Answer:

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Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is
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The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car  A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as  m_{1} \ and\ m_{2}

Let the velocity of cars A and B are denoted as v_{1} \ and\ v_{2}

The momentum before collision is-

                                                  p_{i} =m_{1} v_{1} +m_{2} v_{2}

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of  linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are v'_{1} \ and\ v'_{2}

Hence the final momentum of the system is-

                                                        p_{f} = m_{1} v'_{1} +m_{2} v'_{2}

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e      

                              p_{i} =p_{f}

                               m_{1} v_{1} +m_{2}v_{2} =m_{1} v'_{1} +m_{2} v'_{2}

Hence the option A  is right.

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3 years ago
Read 2 more answers
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mixas84 [53]

The force is still 1 pound.

The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)

8 0
3 years ago
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