Physics, 27.10.2020 19:10 cxxcee

A train slows down as it rounds a sharp horizontal turn, going from 100 km/h to 60.0 km/h in the 14.0 s it takes to round the bend. The radius of the circle is 160 m. Compute the total acceleration and the angle at the moment the train speed reaches 60.0 km/h.

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Answer from: maddielr17

The magnitude of the total acceleration of the train is approximately 1.909 meters per square second.

The direction of total acceleration with respect to radial acceleration is approximately -24.578º.

Please note that negative sign indicates that train is decelerating.

Explanation:

From Physics, we know that magnitude of the acceleration experimented by the train (), measured in meters per square second, while rounding the bend is the Pythagorean identity involving radial and tangential accelerations ( $a_{r}$ , $a_{t}$ ), measured in meters per square second:

$a = \sqrt{a_{t}^{2}+a_{r}^{2}}$ (Eq. 1)

If we assuming that train decelerates at constant rate, then each component is determined by following expressions:

$a_{t} = \frac{v_{f}-v_{o}}{t}$ (Eq. 2)

$a_{r} = \frac{v_{f}^{2}}{R}$ (Eq. 3)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds, measured in meters per second.

- Deceleration time, measured in seconds.

- Bend radius, measured in meters.

If we know that $v_{o} = 27.778\,\frac{m}{s}$ , $v_{f} = 16.667\,\frac{m}{s}$ , $t = 14\,s$ and $R = 160\,m$ , then each acceleration component is calculated:

$a_{t} = \frac{16.667\,\frac{m}{s}-27.778\,\frac{m}{s} }{14\,s}$

$a_{t} = -0.794\,\frac{m}{s^{2}}$

$a_{r} = \frac{\left(16.667\,\frac{m}{s} \right)^{2}}{160\,m}$

$a_{r} = 1.736\,\frac{m}{s^{2}}$

Then, the magnitude of the total acceleration of the train is:

$a = \sqrt{\left(-0.794\,\frac{m}{s^{2}} \right)^{2}+\left(1.736\,\frac{m}{s^{2}} \right)^{2}}$

$a \approx 1.909\,\frac{m}{s^{2}}$

The magnitude of the total acceleration of the train is approximately 1.909 meters per square second.

After that, we obtain the angle of acceleration by using the following trigonometric expression:

$\theta = \tan^{-1} \frac{a_{t}}{a_{n}}$ (Eq. 4)

Where $\theta$ is the direction of total acceleration with respect to radial acceleration, measured in sexagesimal degrees.

If we get that $a_{t} = -0.794\,\frac{m}{s^{2}}$ and $a_{r} = 1.736\,\frac{m}{s^{2}}$ , the direction of total acceleration experimented by the train is:

$\theta = \tan^{-1}\left(\frac{-0.794\,\frac{m}{s^{2}} }{1.736\,\frac{m}{s^{2}} } \right)$

$\theta \approx -24.578^{\circ}$

Please note that negative sign indicates that train is decelerating.

The direction of total acceleration with respect to radial acceleration is approximately -24.578º.

A train slows down as it rounds a sharp horizontal turn, going from 100 km/h to 60.0 km/h in the 14.

Answer from: Quest

Iagree c would be the answer

Answer from: Quest

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j

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