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Helen [10]
3 years ago
7

An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

nd of 2.41 m massless chains. When the system rotates, the chains make an angle of 14.5 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . θ l d What is the speed of each seat? Answer in units of m/s.
Physics
1 answer:
frozen [14]3 years ago
6 0

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

Tsin\theta= \frac{mv^2}{r}

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

Tcos\theta = mg

Matching both expressions with respect to the tension we will have to

T = \frac{\frac{mv^2}{r}}{sin\theta}

T = \frac{mg}{cos\theta}

Then we have that,

\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

\frac{mv^2}{r} = mg tan\theta

Rearranging to find the velocity we have that

v = \sqrt{grTan\theta}

The value of the angle is 14.5°, the acceleration  (g) is 9.8m/s^2 and the radius is

r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}

r = \frac{11.1}{2}+2.41

r = 7.96m

Replacing we have that

v = \sqrt{(9.8)(7.96)tan(14.5\°)}

v = 4.492m/s

Therefore the speed of each seat is 4.492m/s

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Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

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mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

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Given data;

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           force in Y direction (Fy) = 10-7 = 3 N

           Net force (Fnet) = Sq.root[(Fx)² + (Fy)²]

                                      = Sq root [ 4² + 3² ]

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        <em> Net force acting = 25 N</em>

                                   

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