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evablogger [386]
3 years ago
6

A hot air balloon is ascending at a rate of 7.5m/s when a passenger drops a camera. If the camera is 25m above the ground when i

t is dropped, how much time does it take for the camera to reach the ground, and what is the velocity just before it lands?
Physics
1 answer:
Gemiola [76]3 years ago
5 0
The time it takes for an object to reach a certain level given the distance and the initial velocity can be calculated through the equation,

    d = V₀t + 0.5gt²

Where d is the distance, V₀ is the initial velocity, g is the acceleration due to gravity and t is time. Substituting the known values from the given above,

   25 = (-7.5)(t) + 0.5(9.8 m/s²)(t²)

The value of t from the equation 3.15s. Hence, it will take the object 3.15 seconds to reach the ground. 
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Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber
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C. A rubber rod and a glass rod charged this way have opposite charges on them.

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
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With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

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\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

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4 years ago
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