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Alika [10]
3 years ago
8

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

om the railing of the basket when it is 15 mm above the ground.
If the balloon continues to rise at the same speed, how high is the railing when the camera hits the ground?
Physics
1 answer:
fgiga [73]3 years ago
3 0

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}

The values are

\displaystyle y_o=15\ m

\displaystyle v_o=14\ m/s

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

\displaystyle y=0

\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

Multiplying by 2

\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

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Answer:

True.

Explanation:

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The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.

Hence, the given statement is true.

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A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its
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Answer:

The angular velocity is  w = 1.43\  rad/sec

Explanation:

From the question we are told that

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From the law of  conservation of angular momentum we express this question mathematically as

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The initial angular momentum of the Raven is m_r * u_r * \frac{L}{2}

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The initial angular momentum of the Gate is  zero

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  The final angular velocity  of the Raven is  m_r * v_r * \frac{L}{2}

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Substituting this formula

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  \frac{1}{3} m_g L^2 w   =    m_r * v_r * \frac{L}{2} -   m_r * u_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r *  \frac{L}{2} * [u_r - v_r]

Where w is the angular velocity

     Substituting value  

   \frac{1}{3} (4.5)(2)^2  w   =    1.2 *  \frac{2}{2} * [5 - 1.5]

     6w = 4.2

       w = \frac{6}{4.2}

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<u>Also the law of conservation of momentum is applicable in this case:</u>

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In this case the velocity of the lighter mass will get increases in the final condition.

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