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BaLLatris [955]
2 years ago
12

The base unit of length in the international system of units is the:

Physics
1 answer:
zavuch27 [327]2 years ago
5 0

The base unit of length in the international system of units is the: meter (m)

<h3>Meter</h3>

The meter, ( symbol: m ) is the SI unit of length. It is derived by taking the fixed numerical value of the speed of light in vacuum c to be 299 792 458 when expressed in the unit m s-1

<h3>International system of unit ( SI ) base units</h3>

The SI base units are the standard units of measurement defined which is defined by the International System of Units (SI) for the seven base quantities of what is now known as the International System of quantities:

Learn more about SI base units:

brainly.com/question/16393390

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An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

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A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
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\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

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In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a
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<span>In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a C. series of compressions and rarefactions.</span>
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