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3 years ago

11

A plane has a takeoff speed of 88.3 m/s and can accelerate at a rate of 3m/s to reach that speed. How long does the plane take t

o take-off.

1 answer:

Mrrafil [7]3 years ago

3 0

v = $\frac{d}{t}$

and

a = $\frac{v}{t}$

We have acceleration and velocity so:

3 = $\frac{v}{t}$

88.3 = $\frac{d}{t}$

In the acceleration equation we can isolate for v and then plug it back into the other equation to solve...

So...

$3t = v$

$88.3 = 3t$

Divide by three and

t = 29.4 s

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2 years ago

A current of 1.8 A delivers 2.5 C of charge. How much time was required? 0.70 s 0.72 s 1.4 s 4.5 s

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Answer:

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Read 2 more answers

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with

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Answer:

Part a)

$I = 1879.7 kg m^2$

Part b)

$\alpha = 0.70 rad/s^2$

Part c)

$I = 153.8 kg m^2$

Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

$I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)$

So we have

$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

$I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)$

$I = (\frac{16}{3} + 127)m_r R^2$

$I = (\frac{16}{3} + 127)(6.85)(1.44)^2$

$I = 1879.7 kg m^2$

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

$\tau = F\frac{L}{2}$

$\tau = 460(2R)$

$\tau = 460 \times 2 \times 1.44$

$\tau = 1324.8 Nm$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1324.8}{1879.7}$

$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

$I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})$

$I = 6.85(1.44)^2\times 10.83$

$I = 153.8 kg m^2$

Part 4)

If force is applied parallel to the length of rod

then we have

$\tau = \vec r \times \vec F$

$\tau = 0$

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)$

$I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)$

$I = 6.85(1.44)^2\times 24.33$

$I = 345.6 kg m^2$

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