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
DochEvi [55]
3 years ago
6

An object of height 2.2 cm is placed 5.1 cm in front of a diverging lens of focal length 19 cm and is observed through the lens

from the other side 50% Part (a) How many centimeters beyond the lens does the observer locate the object's image? Grade Summa Deductions Potential ry 0% 100% cosO asin0a Sin Submissions Attempts remaining:J5 (5% per attempt) detailed view cotana acosO acotan0sinh0 atan cosh0 tanh cotanhO Degrees Radians END DELI CLEAR Submit Hint I give up! Hints: deduction per hint. Hints remaining: Feedback: 296 deduction per feedback. là 50% Part (b) What is the height of the image, in centimeters?
Physics
1 answer:
monitta3 years ago
3 0

Answer:

a)   i = -4.02 cm , b)     h’= 1,576 cm

Explanation:

a) The constructor equation is

              1 / f = 1 / i + 1 / o

Where f is the focal length, i and o are the distance to the image and the object

Let's clear the distance to the image

             1 / i = 1 / f - 1 / o

             1 / i = 1 / -19 - 1 / 5.1

             1 / i = 0.2487

              i = -4.02 cm

b) let's use the expression of magnification

              m = h’/ h = - i / o

              h’= - h i / o

              h’= 2.2 4.02 /5.1

              h’= 1,576 cm

You might be interested in
A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e
Vesna [10]
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>

<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F  equal to:

F=qE

In our problem we have q=-2.5 nC=-2.5\cdot 10^{-9} C and F=18 nN = 18 \cdot 10^{-9} N, so we can find the magnitude of the electric field:

E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

e=1.6\cdot 10^{-19} C

Therefore, the magnitude of the force acting on the proton will be

F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

7 0
3 years ago
The caste system in India is illegal. True or False
anzhelika [568]
<span>Discrimination is illegal, but caste system is legal.
So answer: False</span>
4 0
3 years ago
A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =
Liula [17]
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam. 
1)What is the force the left support exerts on the beam? 
2)What is the force the right support exerts on the beam? 
3)How much extra mass could the gymnast hold before the beam begins to tip? 
Now the gymnast (not holding any additional mass) walks directly above the right support. 

4)What is the force the left support exerts on the beam? 
5)What is the force the right support exerts on the beam?</span>
6 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

3 0
3 years ago
Which feature of electromagnets makes them more useful than permanent
mezya [45]
I think it’s D................
4 0
3 years ago
Other questions:
  • Why is Earth’s North Pole a geographic north pole but a south seeking pole magnetically?
    6·1 answer
  • Which of the following weather conditions would result in the greatest rate of evaporation from the Earth's surface? A. hot and
    11·1 answer
  • A cook removes a one-gallon pot of hot soup from a stove and places it in an ice-water bath to cool. which is the best cooling p
    13·1 answer
  • An object moves 120 m in 15 seconds. Calculate<br> the object's speed.
    10·1 answer
  • A 210-kg woorden raft floats on a lake. When a 75-kg man stands on the raft, it sinks 5.0 cm deeper into the water. When he step
    5·1 answer
  • 1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film o
    9·1 answer
  • A box is sliding along a frictionless surface and gets to a ramp. Disregarding friction, how fast should the box be going on the
    12·1 answer
  • A 4kg object has a momentum of 12 kg*m/s, what is the objects velocity?
    8·1 answer
  • Bethany feels as though she cannot enjoy flowers properly, as she has lost her sense of smell. Bethany has _______
    11·1 answer
  • What is the angle of refraction when a ray of light passes frm ethanol to air
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!