1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
iris [78.8K]
2 years ago
14

If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced

?
The image produced is virtual and enlarged.
The image produced is virtual and smaller than the object.
The image produced is real and enlarged.
The image produced is real and smaller than the object.
The image produced is virtual and of the same size as the object.

Physics
2 answers:
lys-0071 [83]2 years ago
6 0

Answer:

The image produced is real and enlarged

Explanation:

When we draw the image of an object through ray diagram, we use 3 rules, for the path of ray passing through convex lens, which are as follows:

1. The ray which is coming parallel to the axis, is converged on the focus after passing through convex lens.

2. The ray which comes passing the focus, becomes parallel to axis after passing the lens.

3. A ray that passes through the center of lens, goes straight without the change of direction.

These, rules are displayed in the diagram attached.

Drawing the ray diagram of an object between f (focus) and 2f (center of curvature) of lens, we find the following characteristics of the image formes:

<u>The image produced is real and enlarged</u>

The ray diagram is attached.

valentinak56 [21]2 years ago
3 0
<span>The image produced is real and enlarged.

Check for various positions of objects and Images for convex lens.

Note at position of 2F, the image is same as the object, and once it is between 2F and F, the image becomes bigger than the object. </span>
You might be interested in
True or false. Students with a Learners License may not receive a motorcycle endorsement.
viktelen [127]
This is a true statement 
3 0
3 years ago
You repeated a measurement 6 times and recorded the time using a stop watch: 5.8s, 4.6s, 4.8s, 5.1s, 4.3s, 4.6s. What average va
iVinArrow [24]

Answer and Explanation

Arranging the measured values in increasing order;

4.3s, 4.6s, 4.6s, 4.8s, 5.1s, 5.8s

The two outliers are obviously 4.3s and 5.8s; An outlier is a value in a statistical sample which does not fit a pattern that describes most other data point. Outliers make the average value complicated. So, it is usually better for data to be precise with data points spreading out around a small area.

So, the mean is the average of the four remaining data points after removing the outliers.

Mean = (4.6 + 4.6 + 4.8 + 5.1)/4

Mean = 4.775s

So, the value recorded should be 4.775s, 4.78s or 4.8s depending on the number of decimal places allowed.

QED!

8 0
3 years ago
I need help with this question how to solve it for Brass and Cooper
Ksenya-84 [330]

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

3 0
1 year ago
How are metals identified i the periodic table??
zaharov [31]

Answer:

okay here is a thing I learned when I was younger in my middle school:

Explanation:

my teacher would tell me that metals are considered a weak metals are on the left side and the good metals are located on the right side because the only way I remembered was the right means it is really strong and the left is weak and not that supportive. but I think that's how I still think it is or other people may have their own opinions. but hope this helped out with your question!

8 0
3 years ago
Read 2 more answers
Why is figure 5 an unhelpful visualization tool for this data set? <br><br> Please help!
Paraphin [41]

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

8 0
2 years ago
Other questions:
  • How is a habitat different from an ecosystem?
    14·2 answers
  • Which heavenly bodies are involved in causing the earths tides?
    9·1 answer
  • What is the “lag of seasons”?
    5·1 answer
  • Why is it important we uncover and study fossils?
    15·2 answers
  • 4.
    15·2 answers
  • How would the atmosphere of earth differ if the earth was much larger
    9·2 answers
  • Which of the following is a force or motion that
    6·1 answer
  • Write out the preamble that's all so ill give crown if you do
    6·1 answer
  • PLS I NEED TO TURN THIS IN SOON
    8·1 answer
  • An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!