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3 years ago

True or False: Dimensional analysis can give you the numerical

1 answer:

4 0

Dimensional analysis is a method of checking the dimensions of each value in an equation. I will give you an example.

$E=mc^2$

Is a know equation which equates energy and mass. So our question is, is this true or not? The method is following:

knowing that c has the dimension of [m/s] and m has [kg], what is the dimension of E? So here the dimensional analysis begins.

$E = mc^2 \Rightarrow [E] = \text{kg} \cdot \left( \frac{\text{m}} {\text{s}}\right)^2 = \text{kgms}^{-2}$

This can be used also to solve "equations" to prove certain dimensions of unknown constants.

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Explain how columns can be used and set up to increase the effectiveness of business documents.

Usimov [2.4K]

You can put the name of the product and the price and add another column and add all of your expenses<span />

3 0

3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

Which of these best describes how an appropriate star chart is selected to locate objects in the sky?

Sergio [31]

The layout of the stars in the sky is determined by the date, time of night, and your location (mainly latitude). So to pick the best star chart, you should go with the one that's closest to the present date and your location, then make allowance for what time it is. Everything in the sky moves about a degree every 4 minutes.

6 0

3 years ago

Read 2 more answers

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Savatey [412]

We are given the equation:

<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s ) = 56.65 m/s

</span>For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24<span> meters) / (2.40 s - 2.20 s ) = 50.6 m/s</span>

4 0

3 years ago

An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00

Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq