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3 years ago

10

cheAt an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without fric

tion, and falls through a vertical height of 2.77 m, what is her speed at the bottom of the slide

1 answer:

KiRa [710]3 years ago

3 0

Answer:

$v \approx 7.371\,\frac{m}{s}$

Explanation:

The swimmer can be modelled by the Principle of Energy Conservation. The speed is derived herein:

$m_{swimmer}\cdot g \cdot h = \frac{1}{2}\cdot m_{swimmer}\cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.77\,m)}$

$v \approx 7.371\,\frac{m}{s}$

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Work done(as a measure of energy)=force x distance. Use this equation to show that the SI base units of energy are kg m^2 s^-2

Arturiano [62]

Force [kgms^-2] = mass [kg] x acceleration [ms^-2]
Work = force x distance
Work = [kgms^-2] x [m]
Work = [kgm^2s^-2]

4 0

3 years ago

You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0

3 years ago

A thermodynamic system undergoes a process in which its internal energy decreases by 1,477 J. If at the same time 678 J of therm

Andrews [41]

Answer:2155 J

Explanation:

Given

Change in Internal energy $\Delta U=-1477 J$ i.e. decrease in Internal Energy

Heat added to system $Q=678 J$

According First law for a system

$dQ=dU+dW$

$678=-1477+dW$

$dW=2155 J$

Thus 2155 J of work is done by system

5 0

3 years ago

natulia [17]

Answer:

$\% Error = 2.6\%$

Explanation:

Given

$x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45$

Required

Determine the percentage error

First, we calculate the mean

$\bar x = \frac{\sum x}{n}$

This gives:

$\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}$

$\bar x = \frac{9.06}{6}$

$\bar x = 1.51$

Next, calculate the mean absolute error (E)

$|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}$

This gives:

$|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}$

$|E| = \sqrt{\frac{1}{6}*0.0132}$

$|E| = \sqrt{0.0022}$

$|E| = 0.04$

Next, calculate the relative error (R)

$R = \frac{|E|}{\bar x}$

$R = \frac{0.04}{1.51}$

$R = 0.026$

Lastly, the percentage error is calculated as:

$\% Error = R * 100\%$

$\% Error = 0.026 * 100\%$

$\% Error = 2.6\%$

4 0

3 years ago

David lifts a book 1.2 meters from the floor to the top of his desk. If the book weighed 0.50 kg, what is the gravitational pote

Anika [276]

Answer:

5.886 J

Explanation:

Given:

The mass of the book is, $m=0.5$ kg

Height of lift is, $\Delta h=1.2$ m

Acceleration due to gravity is, $g=9.81$ m/s²

Now, gain in gravitational potential energy is a function of change in position and is given as:

$\Delta PE=mg\Delta h$

Here, $\Delta PE$ is the change in gravitational potential energy.

Plug in 0.5 kg for $m$ , 9.81 for $g$ and 1.2 for $\Delta h$ . Solve for $\Delta PE$

$\Delta PE=0.50\times 9.81\times 1.2=5.886\ J$

Therefore, the gain in gravitational energy of the book is 5.886 J.

7 0

3 years ago