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professor190 [17]
1 year ago
12

what image that is produced when the object is placed between the focal point and the lens. your statement must contain three ch

aracteristics (i.e., type, orientation, size, etc.) about the image to receive full credit. (10 points)
Physics
1 answer:
hodyreva [135]1 year ago
5 0

When the object is placed between the focal point and a convex lens, the image it produces contains these characteristics:

  • Magnified
  • Virtual
  • Upright

A convex lens (converging lens) is a type of lens that is thicker at the center and thinner at the edges. It bends light from distant objects inward toward a point (focal point). A convex lens is generally used to correct farsightedness.

There's three rules for the ray path passing through convex lens

  1. The ray coming parallel to the axis converged on the focus after passing the lens.
  2. The ray coming passing the focus becomes parallel to the axis after passing the lens.
  3. The ray that passes through the center of the lens goes straight without any direction change.

When those rules is applied to know how the image is produced when an object is placed between the focal point and the lens, you'd find out that the image produced is virtual (located on the same side as the object), upright, and magnified.

Learn more about convex lenses at brainly.com/question/12323990

#SPJ4

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In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

8 0
4 years ago
A 2kg blob of putty moving at 4 m/s slams into a 6kg blob of putty at rest what is the speed of the to stick together blobs imme
andrew11 [14]
<h2>Answer</h2>

1m/s

<h2>Explanation</h2>

Given that:

<em>Mass of first blob = 2kg = m1</em>

<em>Velocity of blob = 4m/s = v1</em>

<em>Mass of second blob = 6kg = m2</em>

<em>Velocity of blob = 0m/s = v2</em>

<em />

To find:

<em>Final velocity = Vf</em>

<em />

<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>

<em />

<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>

(2*4) + (6*0) = Vf(2+6)

8 + 0 = Vf(8)

8 = Vf(8)

Vf = 1 m/s

So the speed of two blobs immediately after colliding = 1 m/s

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3 years ago
The speed of a car is increased uniformly from 11 meters per second to 19 meters per second. The average speed of the car during
dsp73
15 m/s average is (11+19)/2
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3 years ago
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otez555 [7]
Sliding and Static.

Would be the right one here.
3 0
4 years ago
An object with a mass of 2 kg is moving with a velocity of 15 m/s.
7nadin3 [17]
KE= 1/2MV^2  - equation
KE= 1/2 (2 kg)(15 m/s)^2 - plug it to the equation
KE= (1 kg)(225 m/s) - multiply 
KE= 225 J - Answer (letter D)
Hope this helps 
8 0
4 years ago
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