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
Lerok [7]
2 years ago
15

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

Physics
2 answers:
Andreas93 [3]2 years ago
8 0
<span>virtual, upright, and magnified</span>
Nastasia [14]2 years ago
4 0

Answer:

virtual, upright and magnified

Explanation:

As we know by lens formula

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

now we know that

d_o < f

now we have

\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}

now we have

d_i = -\frac{d_o f}{f - d_o}

now the magnification is given as

m = \frac{d_i}{d_o}

m = -\frac{f}{f - d_o}

Since here magnification is negative

so image is virtual and upright

also we can see from equation

m > 1

so here we have

virtual, upright and magnified

You might be interested in
What is the difference between a wedge and a screw?
Anvisha [2.4K]
<span>A wedge is a double sided screw basically. Therefore, the difference is the number of sides.</span>
7 0
3 years ago
A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket a light suddenly flashes. To an obs
Sedaia [141]

Answer:

B. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.

Explanation:

To an observer at rest in the rocket who can't see either sides of the rocket, the speed of the light is constant which means the distance to the front or the back is same and would appear to reach the rocket at the same time.

Although from the point of view of the person on the earth, the front of the rocket is travelling in opposite direction of the light while the back of the rocket is moving closer to the light. This means that the distance travelled by the light going forward will be longer going backwards. And since the speed of light is constant in both directions, the light will reach the back of the rocket before it reaches the front  for the observer on the earth.

5 0
3 years ago
Read 2 more answers
I'll mark brainliest
irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

                                  = 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

                                  = 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0
2 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
3 years ago
A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli
AleksandrR [38]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s

Time taken by the train to stop is 20 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m

∴ The engineer applied the brakes 500 m from the station

4 0
3 years ago
Other questions:
  • Describe the motion represented by a horizontal line on a distance time graph
    10·2 answers
  • Identify the techniques used to try to locate extra solar (exo) planets
    15·2 answers
  • Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south.
    13·1 answer
  • 10 points!HELP!!!! PLEASE, URGENT (PHYSICS)
    5·1 answer
  • Calculate the speed of the object from 5 seconds to 10 seconds. Show your work and don't forget to include the units. (HINT for
    14·1 answer
  • In a game of tug of war, Team A pulls with a force 850N, Team B pulls with force of 975N. Calculate the net force on the rope. B
    7·1 answer
  • Una gota de lluvia (m = 3.54 x 10-5 Kg.) cae verticalmente a velocidad constante bajo la influencia de la gravedad y la resisten
    11·1 answer
  • How many electron flow through a light bulb each second if the current flow through the light bulb 0.75A.The electric charge of
    12·1 answer
  • Which contributes most to our average dose of background radiation?
    6·1 answer
  • What is the answer?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!