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3 years ago

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You are waiting in the ticket line at a rock concert. Originally you were 50 meters from the ticket booth and now you are only 5

meters away. What is your change in position?

1 answer:

djverab [1.8K]3 years ago

7 0

here our distance is given with respect to ticket booth

initial distance is given 50 m from the booth

So our initial position is

$x_i = 50 m$

now after this finally we reach at final position

$x_f = 5 m$

now change in position is given as

$\Delta x = x_f - x_i$

$\Delta x = 5 - 50 = - 45 m$

<em>so change in position is 45 m in magnitude and since we are coming closer to the booth so its coming negative in sign</em>

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The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of

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Answer:

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(a)

Average velocity is given by;

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(2)

$y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s$

(3)

$y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s$

b. y = 235 - 16t²

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Answer:

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$m_{i} \cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})] + m_{w} \cdot c_{w}\cdot (T_{4}-T_{3}) = 0$ (1)

Where:

$m_{i}$ - Mass of ice, in grams.

$m_{w}$ - Mass of the juice, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of ice, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of the ice-juice system, in degrees Celsius.

$T_{4}$ - Initial temperature of the juice, in degrees Celsius.

If we know that $m_{i} = 100\,g$ , $c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334\,\frac{J}{g}$ , $T_{1} = -10\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ , $T_{3} = 10\,^{\circ}C$ and $T_{4} = 20\,^{\circ}C$ , then the mass of the juice is:

$m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}$

$m_{w} = \frac{(100\,g)\cdot \left[\left(2.090\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) - 334\,\frac{J}{g} +\left(4.18\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) \right]}{\left(4.180\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)}$

$m_{w} = 949.043\,g$

The mass of the juice responsible for melting the ice is 949.043 grams.

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3 years ago