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4 years ago

Juan Formababee runs at a speed of 17 mi/hr. How long will it take him to run 13.6 miles?

Physics

1 answer:

ryzh [129]4 years ago

6 0

The answer is 231.2

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Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$

The expression is now expanded:

$E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}$

Where $v_{x}$ is the horizontal speed of the arrow, measured in meters per second.

$v_{x} = v_{1}\cdot \cos \theta$

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the horizontal speed is:

$v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}$

$v_{x} \approx 74.911\,\frac{m}{s}$

If $m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{max} = 56.712\,m$ and $v_{x} \approx 74.911\,\frac{m}{s}$ , the total mechanical energy is:

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}$

$E = 201.720\,J$

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

$E = m\cdot g \cdot y_{max}$

$m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $y_{max} = 342.816\,m$

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)$

$E = 201.720\,J$

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0

4 years ago

On reducing the volume of gas at constant temperature the pressure of gas increases why

Illusion [34]

Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If volume increases, then pressure decreases and vice versa, when temperature is held constant.

8 0

3 years ago

Se realiza un experimento en el que se mide el alargamiento de un muelle debido a la acción de una pesa; se trata de comprobar l

mel-nik [20]

Answer:

m = 1,975 m / kg , b = 38.05 m

Explanation:

In this experiment, the elongation is plotted against the applied mass

getting a straight line

y = m x + b

where b would be the initial length of spring let's calculate the slope for which we use two well separated points

m = (56.3 -48.4) / (8 - 4)

m = 1,975 m / kg

the equation remains

y = 1,975 x + b

for x = 2 kg y = 42.0 m

we substitute in the equation

42 = 1,975 2 + b

b = 42 - 3.95

b = 38.05 m

4 0

3 years ago

Question 8, is it A or B?

Brrunno [24]

Answer:

Explanation:

protons are the postive charge in your nucleus in the cell

7 0

4 years ago

An object is placed near a plane mirror. What must be true about the image if the image distance is negative?

BartSMP [9]

Explanation:

The image is not on the same side of the reflecting surface as the outgoing light.

4 0

3 years ago