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mash [69]
2 years ago
8

A water balloon is hovering directly above the line join points ANB which are 4.6 km apart if the angles of elevation to the bal

loon from point a to B or 28.8° and 52.2 respectively find the altitude of the balloon
Physics
1 answer:
Anna35 [415]2 years ago
7 0

Answer:

Drawing the triangle:

H / x = tan 52.2 = 1.29

H / (4.6 - x) = tan 28.8 = .550

H = 1.29 x

H = .55 * 4.6 - .55 x

1.84 x = 2.53        combining equations

x = 1.38

4.6 - 1.38 = 3.22

Total base of triangle = 1.38 + 3.22 = 4.6

H / x = tan 52,2 = 1.29

H = 1.29 * 1.38 = 1.78 height of triangle

Check:

1.78 / 3.22 = tan 28.9    

This agrees with the given value of 28.8

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An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo
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Answer:

a) The tangential component of acceleration at the edge of the motor at  t = 1.5\,s is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor (\alpha), measured in radians per square second, as a function of time (t), measured in seconds, is determined by the following formula:

\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right] (1)

The function for the angular velocity of the electric motor (\omega), measured in radians per second, is found by integration:

\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right] (2)

Where \omega_{o} is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration (a_{t}), measured in meters per square second, is defined by the following formula:

a_{t} = R\cdot \alpha (3)

Where R is the radius of the electric motor, measured in meters.

If we know that R = 7.165\times 10^{-2}\,m, \alpha = 10\cdot t and t = 1.5\,s, then the tangential component of the acceleration at the edge of the motor is:

a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)

a_{t} = -1.075\, \frac{m}{s^{2}}

The tangential component of acceleration at the edge of the motor at  t = 1.5\,s is -1.075 meters per square second.

b) If we know that \omega_{o} = 104.720\,\frac{rad}{s} and \omega = 26.180\,\frac{rad}{s}, then the time needed is:

26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}

5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}

t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}  }{5}

t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}  }{5} }

t \approx 3.963\,s

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

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Answer:

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