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3 years ago

15

To demonstrate an elastic collision, a teacher places a tennis ball on the floor. She wants to hit the ball so that the followin

g conditions are satisfied.
1. After collision, the tennis ball should roll with the velocity of the ball that hit it
2. At the instant of collision, the ball that hit the tennis ball should stop.
What ball should she use?
Golf ball, tennis ball, basketball, or bowling ball?

1 answer:

kodGreya [7K]3 years ago

0 0

A tennis ball

The two balls need to have the same mass to meet this condition.

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A 100-lb load hangs from three cables of equal length that are anchored at the points (minus4,0,0), (2,2 StartRoot 3 EndRoo

Anarel [89]

Answer:

$\vec{F}_{cable_1} = (-19.245 \ lbf ,0, 33.333 \ lbf)$
$\vec{F}_{cable_2} = (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)$
$\vec{F}_{cable_3} = (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)$

Explanation:

The mass of the load is

$m_{load} = 100 \ lb$

As the mass hangs, the cables must be tight, so, we can obtain the vector parallel to the cable as:

$\vec{r}_{cable} = \vec{r}_{anchored} - \vec{r}_{load}$

where $\vec{r}_{load}$ is the position of the load and $\vec{r}_{anchored}$ is the point where the cable is anchored.

So, for our cables

$\vec{r}_{cable_1} = (-4,0,0) - (0,0,-4\sqrt{3})=(-4,0,4\sqrt{3})$

$\vec{r}_{cable_2} = (2,2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,2\sqrt{2},4\sqrt{3})$

$\vec{r}_{cable_3} = (2,-2\sqrt{3},0) - (0,0,-4\sqrt{3})=(2,-2\sqrt{2},4\sqrt{3})$

We know that the forces must be in this directions, so we can write

$\vec{F}_i=k_i \vec{r}_{cable_i}$

We also know, as the system is in equilibrium, the sum of the forces must be zero:

$\vec{F}_{cable_1}+\vec{F}_{cable_2}+\vec{F}_{cable_3}+\vec{W}=0$

where $\vec{W}$ is the weight,

$\vec{W} = (0,0,-100 \ lbf)$

So, we get:

$k_1 (-4,0,4\sqrt{3}) + k_2 (2,2\sqrt{2},4\sqrt{3}) + k_3 (2,-2\sqrt{2},4\sqrt{3}) + (0,0,-100 \ lbf) = (0,0,0)$

This gives us the following equations:

$-4 \ k_1 + 2 \ k_2 + 2 \ k_3 = 0$
$2\sqrt{2} \ k_2 -2\sqrt{2} \ k_3 = 0$
$4\sqrt{3} \ k_1 + 4\sqrt{3} \ k_2 + 4\sqrt{3} \ k_3 -100 \ lbf = 0$

From equation [2] is clear that $k_2 = k_3$ , we can see that

$2\sqrt{2} \ k_2 = 2\sqrt{2} \ k_3$

$\frac{2\sqrt{2} \ k_2}{2\sqrt{2} } = \frac{2\sqrt{2} \ k_3}{2\sqrt{2} }$

$k_2 = k_3$

Now, putting this in equation [1]

$-4 \ k_1 + 2 \ k_2 + 2 \ k_3 = -4 \ k_1 + 2 \ k_3 + 2 \ k_3 = -4 \ k_1 + 4 \ k_3 = 0$

$4 \ k_1 = 4 \ k_3$

$\ k_1 = \ k_3$

Taking this result to the equation [3]

$4\sqrt{3} \ k_1 + 4\sqrt{3} \ k_2 + 4\sqrt{3} \ k_3 -100 \ lbf = 0$

$4\sqrt{3} \ k_3 + 4\sqrt{3} \ k_3 + 4\sqrt{3} \ k_3 = 100 \ lbf$

$3 * (4\sqrt{3} \ k_3) = 100 \ lbf$

$k_3 = \frac{100 \ lbf}{ 12 \sqrt{3}}$

$k_1 = k_2 = k_3 = \frac{100 \ lbf}{ 12 \sqrt{3}}$

So, the forces are:

$\vec{F}_{cable_1} = k_1 (-4,0,4\sqrt{3})$

$\vec{F}_{cable_1} = \frac{100 \ lbf}{ 12 \sqrt{3}} (-4,0,4\sqrt{3})$

$\vec{F}_{cable_1} = (-\frac{100 \ lbf}{ 3 \sqrt{3}},0,\frac{100 \ lbf}{3})$

$\vec{F}_{cable_1} = (-19.245 \ lbf ,0, 33.333 \ lbf)$

$\vec{F}_{cable_2} = k_2 (2,2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_2} = \frac{100 \ lbf}{ 12 \sqrt{3}} (2,2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_2} = (9.622 \ lbf , 13.608 \ lbf ,33.333 \ lbf)$

$\vec{F}_{cable_3} = k_3 (2,-2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_3} = \frac{100 \ lbf}{ 12 \sqrt{3}} (2,-2\sqrt{2},4\sqrt{3})$

$\vec{F}_{cable_3} = (9.622 \ lbf , -13.608 \ lbf ,33.333 \ lbf)$

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Select the appropriate unit for each type of measurement. Amplitude: Frequency: Wavelength: Wave speed: Loudness:.

blondinia [14]

The amplitude and wavelength are measured in meters while frequency is measured in decibels.

<h2>Units of measurement:</h2>

Amplitude: It is the maximum displacement of the points of the wave from its mean position. Hence it is measured in meters.

Frequency: It is measured in Hertz. A hertz is a number of waves passed from a single point per second.

Wavelength: It is the distance between two consecutive crests or troughs. hence it is measured in meters.

Wave speed: It is measured in m/s.

Loudness: The loudness of the sound is measured in decibels.

Therefore, the amplitude and wavelength are measured in meters while frequency is measured in decibels.

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The kinetic energy when it returns to its original height is 100 J

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

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Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

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Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

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