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2 years ago

A tissue box has regular dimensions

Physics

1 answer:

vova2212 [387]2 years ago

5 0

Considering the volume of a rectangle, the volume of the tissue box is 3,239.1 cm³.

<h3>What is volume</h3>

Volume is a scalar-type metric quantity that is defined as the extension in three dimensions of a region of space. In other words, the volume corresponds to the space that the shape occupies.

<h3>Volume of a rectangle</h3>

To calculate the volume of a rectangle, it is necessary to multiply its 3 dimensions: length ×width×height. Volume is expressed in cubic units.

<h3>Volume of the tissue box</h3>

In this case, you know:

Length: 11.8 cm
Width: 12.2 cm
Height: 22.5 cm

Replacing in the definition of volume of a rectangle:

Volume of the tissue box= length ×width×height

Volume of the tissue box= 11.8 cm× 12.2 cm× 22.5 cm

Solving:

<u><em>Volume of the tissue box= 3,239.1 cm³</em></u>

Finally, the volume of the tissue box is 3,239.1 cm³.

Learn more about volume:

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3 years ago

A fireman is standing on top of a building 19 m high, holding a firehose 1 m above the top of the building. She finds that if sh

DENIUS [597]

Answer:

A 2 d vector model

The acceleration function is -9.8 m/s2 which is gravity

Initial velocity on the Y axis is 0, on the X axis is 12 m/s

Inital position is 20 mts above the ground.

It takes the water 1.01 seconds to reach the other building.

THe distace from one building to the other is 12.11 meters.

Explanation:

In order to solve this you just need to carefully read the problem and the data you are given, and use the formula for height in free fall:

$h=\frac{1}{2} g*t^{2}$

So first the data, we know that the water is coming out at a height of 20 meters since the building is 19 meters tall and the fireman is holding the firehose 1 meter above it, and the water is hitting the second building at a height of 15 meters, that means that the water is travelin -5meters.

Gravity as it doesn´t say otherwise would be 9.8m/s2 since that is gravity on earth, and water is leaving the firehose at 12m/s horizontally.

We can calculate the time by using the height formula fro free fall:

$h=\frac{1}{2} g*t^{2}\\5=\frac{1}{2} 9.81*t^{2}\\t^{2}= \frac{10}{9.8} \\t^{2}=1.0193\\t=1.009 seconds$

So it takes 1.009 seconds for the water to frop from 20 to 15 meters, as the horizontal velocity remains the same we just multiply it by the time and we get the horizontal distance between the two buildings and that would be:

12.11 meters.

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4 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant