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3 years ago

Why does a skydiver accelerate as she falls to the earth before opening her chute?

2 answers:

5 0

The earths gravity pulls down on her,causing her to accelerate. without her chute she would be dead within minitues.<span />

garik1379 [7]3 years ago

3 0

A skydiver accelerates because the force of the earth pulls on her. This force is the attraction between the earth and the skydiver, her weight due to gravity.

plz give the brainliest

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4 years ago

Read 2 more answers

A 3.00 kg mass is pushed against a spring and released. If the spring constant of the

Phoenix [80]

Answer:

Energy stored in the compressed spring: $37.5\; \rm J$ before the mass is released.

Maximum horizontal speed of the mass: $5.0\; \rm m \cdot s^{-1}$ .

(Assumption: the surface between the mass and the ground is frictionless until the mass separates from the spring.)

(Maximum horizontal acceleration of the mass: $250\;\rm m \cdot s^{-2}$ .)

Work that friction did on the mass: $13.5\; \rm J$ .

Maximum height of the mass: approximately $0.459\; \rm m$ (assuming that $g = 9.8\; \rm m \cdot s^{-2}$ .)

Explanation:

<h3>Energy stored in the spring</h3>

When an ideal spring of spring constant $k$ is compressed by a displacement of $x$ from the equilibrium position, the elastic potential energy stored in the spring is:

$\displaystyle \frac{1}{2}\, k \cdot x^{2}$ .

The question states that for this spring, $k = 7500\; \rm N \cdot m^{-1}$ (Newtons per meters) whereas $x = 10.0\; \rm cm$ (centimeters.) Convert the unit of $x$ to meters to match the unit of $k$ :

$\begin{aligned} x &= 10.0\; \rm cm\\ &= 10.0\; \rm cm \times \frac{1\; \rm m}{100\;\rm cm} = 0.100\; \rm m\end{aligned}$ .

Calculate the energy stored in this spring:

$\begin{aligned}& \text{Elastic Potential Energy} \\ &= \frac{1}{2}\, k \cdot x^{2} \\ &= \frac{1}{2} \times 7500\; \rm N \cdot m^{-1} \times 0.100\; \rm m \\ &=37.5\; \rm J \end{aligned}$ .

<h3>Maximum horizontal speed of the mass</h3>

Assume that the surface under the spring is frictionless. All that $37.5\; \rm J$ of elastic potential energy would be converted to kinetic energy by the time the mass separates from the spring.

The horizontal speed of the mass is largest at that moment. The reason is that immediately after this moment, friction (between the surface and the mass) would start to slow the mass down immediately.

Calculate the speed $v$ of the mass $m$ at that moment.

$\displaystyle \text{Kinetic Energy} = \frac{1}{2}\, m \cdot v^{2}$ .

Therefore:

$\displaystyle v = \sqrt{\frac{2\cdot (\text{Kinetic Energy})}{m}}$ .

The question states that $m =3.00\; \rm kg$ . At that moment, the kinetic energy of the mass is $37.5\; \rm J$ (or equivalently, $37.5\; \rm kg\cdot m^{2} \cdot s^{-2}$ in base units.)

Calculate the speed of the mass at that moment from the kinetic energy of the mass:

$\begin{aligned}v &= \sqrt{\frac{2\cdot (\text{Kinetic Energy})}{m}} \\ &= \sqrt{\frac{2 \times 37.5\; \rm kg \cdot m^{2}\cdot s^{2}}{3.00\; \rm kg}} = 5.00\; \rm m \cdot s^{-1}\end{aligned}$ .

<h3>Work friction did on the mass</h3>

Assume that the rough surface is level. Therefore, all the energy loss of the mass should be attributed to friction.

Kinetic energy of the mass before coming onto that rough surface: $37.5 \; \rm J$ .

Kinetic energy of the mass before leaving the rough surface:

$\begin{aligned} &\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 3.00\; \rm kg \times {\left(4.00\; \rm m \cdot s^{-1}\right)}^{2} \\ &= 24.0\; \rm J \end{aligned}$ .

If that rough surface is level, there would be no change to the gravitational potential energy of the mass. Calculate the change to the kinetic energy of the mass:

$37.5\;\rm J - 24.0\; \rm J = 13.5\; \rm J$ .

That should be equal to the size of the work that friction did on the mass.

<h3>Maximum height of the mass</h3>

Assume that the smooth ramp does not change the total energy of the block. Therefore, the total mechanical energy of the block would still be $24.0\; \rm J$ when the height of the block is maximized. However, all that energy would be in the form of gravitational potential energy.

Let $g$ denote the gravitational field strength ( $g \approx 9.8\; \rm m \cdot s^{2}$ near the surface of the earth.) The gravitational potential energy of an object of mass $m$ and height $h$ (relative to the surface of zero gravitational potential energy) would be:

$\begin{aligned}&\text{Gravitational Potential Energy} = m \cdot g \cdot h\end{aligned}$ .

Rewrite this equation to find an expression for $h$ :

$\displaystyle h = \frac{\text{Gravitational Potential Energy}}{m\cdot g}$ .

Assume that $g = 9.8\; \rm m \cdot s^{-2}$ . The question states that $m = 3.00\; \rm kg$ . $\text{Gravitational Potential Energy} = 24.0\; \rm J$ . Therefore:

$\begin{aligned} &h\\ &= \frac{\text{Gravitational Potential Energy}}{m\cdot g}\\ &= \frac{24.0\; \rm kg \cdot m^{2} \cdot s^{-2}}{3.00\; \rm kg \times 9.8\; \rm m \cdot s^{-2}} \approx 0.459\; \rm m\end{aligned}$ .

3 0

3 years ago