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1 year ago

During what time interval was the object traveling most quickly?

Physics

1 answer:

anygoal [31]1 year ago

3 0

ANSWER

0s to 3s

EXPLANATION

To find the time interval where the object was traveling more quickly, we have to find the velocity of the object during each interval.

From a position-time graph, we can obtain the velocity of the object from the slope of the graph in each interval. To find those slopes, we just have to divide the vertical difference by the time interval.

The interval from 0s to 3s:

$v=\frac{7-3}{0-3}=-\frac{4}{3}m/s$

The interval from 3s to 5s and the interval from 7s to 8s have a horizontal line, so the slope is zero and therefore the velocity is zero - meaning that the object was not moving during these periods.

The interval from 5s to 7s,

$v=\frac{5-3}{7-5}=\frac{2}{2}=1m/s$

And the interval from 8s to 12s,

$v=\frac{0-5}{12-8}=-\frac{5}{4}m/s$

Two of these three velocities are negative. Negative velocity indicates that the object is moving backward.

From these velocities, the greatest one, in absolute value, is the one between 0s to 3s. During this interval, the object is moving backward but at the greatest velocity.

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Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

2 years ago

A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

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3 years ago

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leonid [27]

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4 years ago

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Gekata [30.6K]

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Explanation:

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3 years ago

20 m

liubo4ka [24]

Answer:

1.9 km

Explanation:

The equation for vertical motion is ...

h(t) = -4.9t^2 +26t +20

This will have a zero near t = 5.988 seconds.*

The horizontal distance traveled in that time is ...

(310 m/s)(5.988 s) ≈ 1856 m ≈ 1.9 km

_____

* The root of a quadratic can be found numerous ways. We choose the quick and easy: let a graphing calculator show it to you.

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4 years ago