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2 years ago

6

Which of the following is an example of a primary source?

1 answer:

nekit [7.7K]2 years ago

4 0

Answer:

An article written by a scientist .

Explanation:

Primary sources information are those that contain first hand information probably from the place of the activities or by the original contact of the information (source ) .

So primary information is normally contained in artifact , journals , letter , dissertations manuscripts , videos and audio recordings .

The fact that the article is written by the scientist means he is the one who has the original information hence he is the first person to access the information , meaning he is the original owner .

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Use the diagram below to answer the following question:

d1i1m1o1n [39]

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm$

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

5 0

2 years ago

2. Which of the following wavelength properties would require a stopwatch to measure?

katovenus [111]

Answer:

<u><em>A. wavelength</em></u>

Explanation:

The others are about sound and how high it is. That has nothing to do with time.

3 0

3 years ago

Read 2 more answers

Segmented mirrors sag under their own weight. their optical shape must be controlled by computer-driven thrusters under the mirr

madam [21]

Active Optics.

Hope that helps, Good luck! (:

6 0

3 years ago

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string is oscillated at 440 Hz, which is the frequency of t

Katarina [22]

Answer:

efsfefsfsdf

Explanation:

efsfesef

7 0

3 years ago

At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh

Anna35 [415]

We have the equation for electric field E = kQ/ $d^{2}$

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to $d^{2}$

So, $\frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$E_{1}$ = 485 N/C

$d_{1}$ = 0.208 cm

$d_{2}$ = 0.620 cm

$E_{2}$ = ?

$\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}$

$E_{2}$ = $\frac{485*0.208^{2}}{0.628^{2}}$

$E_{2}$ = 53.20 N/C

7 0

3 years ago

Read 2 more answers