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3 years ago

How to use practical learning app

Physics

2 answers:

Llana [10]3 years ago

8 0

Download the 'Practically app' for FREE now.
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Start a FREE trial and enjoy 30 Free minutes to explore all that we've said.
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MissTica3 years ago

6 0

Answer:

Once done, upgrade as you like by choosing a suitable package.

Download the 'Practically app' for FREE now.

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Explanation:

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A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0

2 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

A train started its journey at 9:00. Its average speed between

Vera_Pavlovna [14]

Answer:

Explanation:

In the four hours before 13:00 the train traveled 4(80) = 320 km.

It needs to travel another 4(320) = 1280 km in the next 21:00 - 13:00 = 8 hrs.

1280 / 8 = 160 km/hr

5 0

3 years ago

What is the product of mass and velocity called?

Crank

The product of mass and velocity is momentum. P=mv. The 'P' in the equation stands for momentum.

3 0

3 years ago

According to the diagram of the electromagnetic spectrum shown, what would best represent the

iren2701 [21]

The best represent the size of visible light will be C. Protozoa

The electromagnetic spectrum, gives the overall distribution of electromagnetic radiation by the frequency or wavelength. All EM waves travel at the speed of light in a vacuum, but over a wide range of frequencies, wavelengths, and photon energies.

Visible light wavelengths cover the range of approximately 0.4 to 0.7 μm. electromagnetic spectrum that the human eye can see is the Visible light. Visible light is a form of electromagnetic (EM) radiation, along with radio waves, infrared, ultraviolet, X-rays, and microwaves. the wavelengths that are visible to most human eyes is generally known as Visible light

the best represent the size of visible light is Protozoa, According to the diagram of the electromagnetic spectrum shown,

Learn more about electromagnetic spectrum here brainly.com/question/25847009

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6 0

1 year ago

How to use practical learning app​

How to use practical learning app