Answer:
[tex]971605.66\ \text{m/s}[/tex]
[tex]25.1\ \mu\text{m}[/tex]
Explanation:
m = Mass of electron = [tex]9.11\times 10^{-31}\ \text{kg}[/tex]
B = Magnetic field = 0.22 T
K = Kinetic energy of electron = [tex]4.3\times 10^{-19}\ \text{J}[/tex]
q = Charge = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
v = Velocity of electron
r = Radius of curved path
Kinetic energy is given by
[tex]K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4.3\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=971605.66\ \text{m/s}[/tex]
The speed of the electron is [tex]971605.66\ \text{m/s}[/tex]
The force balance of the system is given by
[tex]qvB=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{mv}{qB}\\\Rightarrow r=\dfrac{9.11\times 10^{-31}\times 971605.66}{1.6\times 10^{-19}\times 0.22}\\\Rightarrow r=0.0000251=25.1\ \mu\text{m}[/tex]
The radius of the curved path is [tex]25.1\ \mu\text{m}[/tex]
If a true bearing of a ship at sea is 227°, what is its direction angle?
A. 43°
B.313°
C. 223°
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switched off, it coasts to rest in 32 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.
Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:[tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]
Now, we need first to find the value of the angular acceleration, that we can get from the following expression:[tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]
Since the machine starts from rest, ω₀ = 0.We know the value of ωf₁ (the operating speed) in rev/min.Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:[tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]
Replacing by the givens in (2):[tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]
Solving for α:[tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]
Replacing (5) and Δt in (1), we get:[tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]
in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:[tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]
In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:[tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]
First of all, we need to find the value of the angular acceleration during the second period.We can use again (2) replacing by the givens:ωf =0 (the machine finally comes to an stop) ω₀ = ωf₁ = 57.5 rev/secΔt = 32 s[tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]
Solving for α in (9), we get:[tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]
Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:[tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]
In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:[tex]\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)[/tex]A string is wound around a uniform disc of radius 0.68 m and mass 3.7 kg. The disc is released from rest with the string vertical and its top end tied to a fixed support. calculate the speed of the center of mass when, after starting from rest, the center of mass has fallen 1.2 m. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Answer:
3.962
Explanation:
mass = 1.2
gravity g = 9.81
velocity = ?
mgh = 1/2mv² + 1/2(1/2)mv²
= 9.81 x 1.2 = 3/4v²
11.772 x 4 = 3/4v²
47.088 = 3v²
divide through both sides by 3
15.696 = v²
take the square root of both sides
√15.696 = √v²
= 3.6962 = v
the speed is therefore 3.692
Object A is negatively charged. Object A and Object B
attract. Object B and Object C repel. Object C and Object
D repel. What type of charge does Object B, Object C, and
Object D possess?
Answer:
Malrpr00qpq9owoowopwiaahaulaqkkkala9asoLHahababajjajalls
Explanation:
hhoootyiñlf7ogffyiklmhf
Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change
When sound waves travels from air to water
Answer:
Frequency of a sound wave remains independent of the medium through which it travels. Wavelength, velocity and phase change with the change in media.
Explanation:
Define the average acceleration of a particle
between two given instants.
*
Check all of the things that can affect power
Time
Work
Distance
Force
Answer:
fhddmvxmvydlyghclhchc
. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder
Answer:
As we can see, a string is attached with block A, and three string is folded with ply which is attached with B
x
B
=3x
A
Now differentiate with respect to x
V
B
=3V
A
Given,
V
A
=0.6m/s(totheright)
So,
V
B
=0.6×3
=1.8m/s(downward)
Explanation:
IF THE ANSWER IS RIGHT PLZ GIVE ME BRAINLIEST
THANK U
HAVE FUN AND BE SAFE
The circuit is working, and all three bulbs are lit. If a switch at B is opened, what will happen to the circuit? Photo included PLS ONLY ANSWER IF YOU KNOW 100%
Answer:
D. Bulb 2 will go out but bulbs 1 and 3 will remain lit.
Explanation:
If switch at B is opened then, Bulb 2 will go out but bulbs 1 and 3 will remain lit. This is because the positive and negative current flow still fully passes through the two closed switches at switch A and C which allows the bulbs to receive both currents and thus allowing it to turn on. When Switch B is opened it cuts off access for the negative charge to get to bulb 2 and without it, bulb 2 will not turn on.
Answer:all 3 bulbs will go out
Explanation:
Just did it and got it right
The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
Answer:
2π/[28 x (10^-3)]
Explanation:
Angular speed : ω=2π/T
T = 28ms = 28 x (10^-3) s
Angular speed = 2π/[28 x (10^-3)]
A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque black disk; the faces of the two disks are parallel. 40 cm beyond the black disk is a white viewing screen. The lighted disk illuminates the screen, but there's a shadow in the center due to the black disk. What is the diameter of the completely dark part of this shadow
Answer:
132
Explanation:
if you round it is correct
Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.2 T in 100 ms. Determine the magnitude of the induced emf within a circle of tissue of radius 1.3 mm and that is perpendicular to the direction of the field.
poste en français s’il vous plaît
What are the two main processes carried out by the excretory system?
Please help me!!! SHOW YOUR WORK to get BRAINLIEST
1. )Find the work done by a weightlifter lifting a 20kg barbell 2.2 m upward at a constant
speed.
2.) A 2700 kg car is moving across level ground at 3.5m/s when it begins an acceleration
that ends with the car moving at 12m/s. Is work done in this situation? How do you
know?
3.) An Alaskan huskie pulls a sled using a 700N force across an 8m wide street. The force of
friction on the 82kg sled is 160N. How much work is done by friction? How much work is
done by the huskie? How much by gravity?
4.) A 81kg man climbs a 4.2m tall flight of stairs. What work was done by the man against
the force of gravity?
5.) You are moving 6 books, each weighing 11 N, to a shelf 0.8 m higher. Do you
do more work by moving all 6 books at once, or by moving one book at a time?
6.) If you carry a U S. penny (weighing 0.023 N) to the top of the Empire State
Building (450 m), how much work have you done on the penny?
7.)Ramps for handicapped people to enter buildings are quite common and are usually
long, often having one or more turns. It seems that a shorter, straighter ramp would be
more convenient. Use the concepts of work to explain their design.
8.) The pyramid builders of ancient Egypt used a long ramp (18 m) to raise blocks
by one meter. Each block weighs 150,000 N. How much work is done to raise
one block to a height of one meter?
9.) When pumping up your bicycle tire you exert a force of 40. N to move the
handle down 0.18 m. If you do 200 Nm of work, how many times do you pump
the handle?
10.) How much work is done in holding a 800 N barbell motionless above your head for 40 s?
11.) A student takes a 200 N box from a shelf 1.5 m from the floor, and carries it 35 meters
and places it on another shelf 1.5 m from the floor. How much work against gravity has
she done?
1. Work = Force × Distance
Work = (20 kg) × (9.8 m/s^2) × (2.2 m) = 431.2 J
2. Yes, work is done as the car undergoes acceleration, indicated by a change in velocity.
3. Work done by friction = (force of friction) × (distance) = (160 N) × (8 m) = 1280 J
Work done by the huskie = (700 N) × (8 m) = 5600 J
Work done by gravity = (82 kg) × (9.8 m/s^2) × (8 m) = 6393.6 J
4. Work = Weight × Height = (81 kg) × (9.8 m/s^2) × (4.2 m) = 3290.44 J
5. The work done is the same whether moving all 6 books at once or one book at a time.
6. Work = Weight × Height = (0.023 N) × (450 m) = 10.35 J
7. Longer ramps with turns allow for a gradual increase in potential energy, minimizing the force required to ascend and providing convenience for individuals.
8. Work = Weight × Height = (150,000 N) × (1 m) = 150,000 J
9. Number of times = Work / (Force × Distance) = 200 Nm / (40 N × 0.18 m) ≈ 27.78 times
10. No work is done when holding a motionless object.
11. Work = Weight × Height = (200 N) × (1.5 m) = 300 J
1. To find the work done by the weightlifter, we can use the formula:
Work = Force × Distance
Given:
Mass of barbell = 20 kg
Distance lifted = 2.2 m
Acceleration due to gravity = 9.8 m/s^2 (assuming vertical motion)
Weight (force) = mass × acceleration due to gravity
Substituting the values, we get:
Work = (20 kg) × (9.8 m/s^2) × (2.2 m) = 431.2 J
2. Work is done when there is a displacement in the direction of the applied force. In this case, since the car is moving in the same direction as the force applied (acceleration), work is done. The work done can be calculated using the formula mentioned in question 1:
Work = Force × Distance
Since the mass of the car is given, we can use the equation:
Work = (1/2) × mass × (final velocity^2 - initial velocity^2)
Substituting the values, we get:
Work = (1/2) × (2700 kg) × [(12 m/s)^2 - (3.5 m/s)^2] = 219,660 J
3. Work done by friction can be calculated using the formula:
Work = Force of friction × Distance
Work done by friction = (force of friction on the sled) × (distance)
Work done by friction = (160 N) × (8 m) = 1280 J
Work done by the huskie = Force × Distance
Work done by the huskie = (700 N) × (8 m) = 5600 J
Work done by gravity = Weight × Height
Work done by gravity = (mass × acceleration due to gravity) × (distance)
Work done by gravity = (82 kg) × (9.8 m/s^2) × (8 m) = 6393.6 J
4. Work done against the force of gravity is given by:
Work = Weight × Height
Work = (mass × acceleration due to gravity) × (height)
Work = (81 kg) × (9.8 m/s^2) × (4.2 m) = 3290.44 J
5. The work done would be the same regardless of whether you move all 6 books at once or one book at a time. Work done is independent of the number of objects moved simultaneously.
6. Work done on the penny can be calculated using the formula:
Work = Weight × Height
Work = (0.023 N) × (450 m) = 10.35 J
7. The design of long ramps with turns for handicapped people is based on minimizing the effort required to overcome the force of gravity while ascending. By providing a longer distance to cover, the incline of the ramp can be made shallower, reducing the force needed to ascend. This makes it more convenient for individuals using wheelchairs or other mobility aids to navigate the ramp.
8. The work done to raise one block to a height of one meter can be calculated as:
Work = Weight × Height
Work = (150,000 N) × (1 m) = 150,000 J
9. Work is calculated as the product of force and displacement. Given the work done as 200 Nm, we can rearrange the formula to solve for the force:
Work = Force × Distance
200 Nm = Force × 0.18 m
Force = 200 Nm / 0.18 m = 1111.11 N
The force required to move the handle down is 1111.11 N. As the force required to pump the handle is 40 N, we can calculate the number of times by dividing the force required:
Number of times = 1111.11 N / 40 N ≈ 27.78 times
10. When holding the barbell motionless above your head, there is no displacement, and hence no work is done against gravity.
11. The work done against gravity can be calculated using the formula:
Work = Weight × Height
Work = (mass × acceleration due to gravity) × (height)
Work = (200 N) × (1.5 m) = 300 J
Know more about Work done here:
https://brainly.com/question/21854305
#SPJ8
what makes a funnel appear black
Answer:
Air
Explanation:
The mixing of cooler air in the lower troposphere with air flowing in a different direction in the middle troposphere causes the rotation on a horizontal axis, which, when deflected and tightened vertically by convective updrafts, forms a vertical rotation that can cause condensation to form a funnel cloud.
In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find
an electron's acceleration in this field.
Answer:
a = 1.055 x 10¹⁷ m/s²
Explanation:
First, we will find the force on electron:
[tex]E = \frac{F}{q}\\\\F = Eq\\[/tex]
where,
F = Force = ?
E = Electric Field = 6 x 10⁵ N/C
q = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\[/tex]
F = 9.6 x 10⁻¹⁴ N
Now, we will calculate the acceleration using Newton's Second Law:
[tex]F = ma\\a = \frac{F}{m}\\[/tex]
where,
a = acceleration = ?
m = mass of electron = 9.1 x 10⁻³¹ kg
therefore,
[tex]a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\[/tex]
a = 1.055 x 10¹⁷ m/s²
PLEASEE HELPPP IM GONNA FAIL NEED THIS BEFORE 9:30 MIDDLE SCHOOL SCIENCE
Answer:
I think your soppose to multiply
Explanation:
What conversion takes place in a motor?
A. An electric current into a magnetic field
B. Mechanical energy into electric energy
C. Electric energy into mechanical energy
D. A lower voltage into a higher voltage
Thank you!! I will mark brainliest!!
PLEASE I NEED HELP CLICK ON THIS IMAGE
Which disciplines were developed based on the Greek questioning of the elements?
A. art and music
B. chemistry and biology
C. tragedy and comedy
effect of high pitch on humans
Answer:
High frequency sound causes two types of health effects: on the one hand objective health effects such as hearing loss (in case of protracted exposure) and on the other hand subjective effects which may already occur after a few minutes: headache, tinnitus, fatigue, dizziness and nausea.
A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is A charged particle is moving with speed perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2 perpendicular to the same magnetic field. If the frequency of revolution of the first particle is , the frequency of revolution of the second particle is:______.
a. f.
b. 4f.
c. f/4.
d. f/2.
e. 2f.
Answer:
A
Explanation:
Which of the following relationships is correct?
2 points
1 N = 1 kg
1 N = 1 kg·m
1 N = 1 kg·m/s
1 N = 1 kg·m/s2
A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it maintains a constant pressure of 1000 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500 o C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 280 o C, at which point the valve is closed. The pressure remains constant during the process. Determine:
Answer: Hello the missing piece of your question is attached
question : Determine mass of steam that has entered ( in kg )
answer : 0.206 kg
Explanation:
V1 = 0.1 m^3 ,
v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg
V2 = 0.2 m^3
using the steam tables
at ; P = 1000 kPa, v' = 0.167 m^3/kg
U1 = 2321 KJ/kg
at ; P = 1000 kPa , T2 = 280°C
v'2= 0.2481 m^3kg
U2 = 2760.6
at ; P = 5MPa , T = 500°C
h1 = 3434.7 KJ/Kg
calculate final mass ( m2 )
M2 = V2 / v'2
= 0.2 / 0.2481 = 0.806 kg
therefore the mass added = m2 - m1
= 0.806 - 0.6 = 0.206 kg
To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement s given by s=xf−xi.
A. Find the acceleration a of the particle.
B. Evaluate the integral W = integarvf,vi mudu.
Answer:
a) the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Explanation:
Given the data in the question;
force on particle F = ma
displacement s = x[tex]_f[/tex] - x[tex]_i[/tex]
work done on the particle W = Fs = mas
we know that; change in energy = work done { work energy theorem }
[tex]\frac{1}{2}[/tex]m( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = mas
[tex]\frac{1}{2}[/tex]( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = as
( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = 2as
a = ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
Therefore, the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) Evaluate the integral W = [tex]\int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W = \int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W =m[\frac{v^{2} }{2} ]^{vf}_{vi}[/tex]
W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Therefore, the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
how many atmospheres is at depth of 100 meters of ocean water?
Explanation:
At 100m above sea level, the air pressure is approximately 990mbar, so the air pressure has decreased by approximately 10mbar.
There would be 10 atmospheric pressure at a depth of 100 meters of ocean water because approximately 10 meters depth of water is equivalent to 1 atmospheric pressure.
What is pressure?The total applied force per unit of area is known as the pressure.
The pressure depends both on externally applied force as well the area on which it is applied.
As given in the problem we have to find out the atmospheric pressure equivalent at a depth of 100 meters of ocean water,
10 meters depth of water = 1 atmospheric pressure
100-meter depth of the ocean water = 10 atmospheric pressure
Thus, there would be 10 atmospheric pressure at a depth of 100 meters of ocean water.
To learn more about pressure, refer to the link given below ;
brainly.com/question/28012687
#SPJ2
3) A rather large fish is about to eat an unsuspecting small fish. The big fish has a mass of 5kg and is
swimming at 8 m/s, while the small fish has a mass of 1 kg and is swimming at -4 m/s. What is the
velocity of big fish after lunch?
Answer:
the velocity of the big fish after the launch is 6 m/s.
Explanation:
Given;
mass of the big fish, m₁ = 5 kg
velocity of the big fish, u₁ = 8 m/s
mass of the small fish, m₂ = 1 kg
velocity of the small fish, u₂ = -4 m/s
Let the final velocity of the big fish after launch = v
Apply the principle of conservation linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
5 x 8 + 1 x (-4) = v(5 + 1)
40 - 4 = 6v
36 = 6v
v = 36/6
v = 6 m/s.
Therefore, the velocity of the big fish after the launch is 6 m/s.
If
[tex] \binom{13}{7} [/tex]
N has a half-life of about 10.0 min, how long will it take for 20 g of the isotope to decay to 1.9 g? [2 marks] 2. Radon-22
Answer:
the time it will take the element to decay to 1.9 g is 34.8 mins.
Explanation:
Given;
half life of Nitrogen, t = 10 min
initial mass of the element, N₀ = 20 g
final mass of the element, N = 1.9 g
The time taken for the element to decay to final mass is calculated as follows;
time (min) mass remaining
0 ----------------------------------20 g
10 mins ------------------------- 10 g
20 mins ------------------------- 5 g
30 mins -------------------------- 2.5 g
40 mins --------------------------- 1.25 g
Interpolate between 2.5 g and 1.25 to obtain the time for 1.9 g
30 min ------------------------- 2.5 g
x ----------------------------------- 1.9 g
40 min -------------------------- 1.25 g
[tex]\frac{30 - x}{40- 30} = \frac{2.5 - 1.9}{1.25 - 2.5} \\\\-1.25(30-x) = 6\\\\-37.5 + 1.25x = 6\\\\1.25x = 6+37.5\\\\1.25x = 43.5\\\\x = \frac{43.5}{1.25} \\\\x = 34.8 \ mins[/tex]
Therefore, the time it will take the element to decay to 1.9 g is 34.8 mins.
3) if you have a convex (converging) lens (f#12 cm) that produced an image with a
magnification of -2.5, how far must the object be placed from the lens?
I
b) Draw a ray-tracing diagram of this situation below (label all points in cm)
Explanation:
step 1. since the magnification is negative this means the object is real and is farther away from the lens than the focal point
step 2. 1/f = 1/o + 1/i where f is the focal length, o is the object distance from the lens and i is the image distance from the lens
step 3. 1/12 = 1/o + 1/i, -i/o = -2.5 (these are our 2 equations or system
step 4. 1/12 = 1/o + 1/2.5o
step 5. o = 12 + 12/2.5 = 16.8cm
step 6. i = 2.5o = 2 5(16) = 42cm