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3 years ago

Please help!!!... Why does a parachute slow a person down?

1 answer:

7 0

A parachute is a device used to slow down an object that is falling towards the ground. without a parachute the gravity is more than air resistance. But, as the parachute opens, the air resistance increases.

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

Suppose you have thrown a rock nearly straight up at a coconut in a palm tree,and the rock misses on the way up but hits the coc

Margarita [4]

Most likely it would dislodge the coconut on the way down due to gravity because on the way up the gravity would slow down the rock but on they down the gravity pulls the rock

5 0

3 years ago

3. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the sl

Marina86 [1]

The slope of a speed-time graph is the acceleration represented by the graph.

All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.

6 0

3 years ago

Which of the following intermolecular forces explains why iodine (I2) is a solid at room temperature?

egoroff_w [7]

"Dispersion forces" is the one intermolecular force among the following choices given in the question that <span>explains why iodine (I2) is a solid at room temperature. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.</span>

3 0

3 years ago

1. ¿Cuál es el actor clave de del exceso de peso corporal?

Slav-nsk [51]

El factor mas importante para el exceso de peso es un exceso de energía creada por una alimentación excesiva

El peso de un cuerpo es definido por la relación entre la energía requerida para los procesos vitales del cuerpo, sus actividades físicas diarias y la energía suministra en forma de alimentos.

Cuando estos dos parámetros están en balance el peso es estable, pero cuando la cantidad de alimentos aumenta o el valor energético de los mismo aumenta se tiene un exceso de energía que el cuerpo almacena en forma de grasa corporal, este el el factor mas importante para el exceso de peso.

En conclusión el factor mas importante para el exceso de peso es un exceso de energía por una alimentación excesiva

aprende mas acerca del peso corporal aquí:

brainly.com/question/13032223

8 0

2 years ago