We have that the magnification of each focal length is given respectively as
A) has [tex]u=3\frac{f}{2}[/tex]
B) has [tex]u=4\frac{f}{3}[/tex]
C) has [tex]u=5\frac{f}{4}[/tex]
From the question we are told that:
Focal Length F
Generally, the equation for Magnification is mathematically given by
[tex]M=\frac{-v}{u}[/tex]
Therefore
[tex]v=2u[/tex]
For A
[tex]M=-2[/tex]
Therefore
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}[/tex]
Therefore
[tex]u=3\frac{f}{2}[/tex]
For B
[tex]M=-3[/tex]
Therefore
[tex]v=3u[/tex]
Where
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}[/tex]
Therefore
[tex]u=4\frac{f}{3}[/tex]
For C
[tex]M=-4[/tex]
Therefore
[tex]v=4u[/tex]
Therefore
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}[/tex]
Therefore
[tex]u=5\frac{f}{4}[/tex]
Conclusion
From the calculations above we can rightly say that the magnifications of the values above are
A has [tex]u=3\frac{f}{2}[/tex]
B has [tex]u=4\frac{f}{3}[/tex]
C has [tex]u=5\frac{f}{4}[/tex]
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A train starts from rest (at position zero) and moves with constant acceleration. On the first observation, its velocity is 20m/s and 80seconds later the velocity became 60m/s. At 80s calculate the position, average velocity, and the constant acceleration over the interval.(7-points)
The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.
Acceleration:
We can use the fallowing kinematic equation to get the acceleration at 80 s.
[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]
Where:
v(i) is the initial velocity (20 m/s)v(f) is the final velocity (60 m/s)t is the interval (80 s)The, we have:
[tex]\vec{a}=\frac{60-20}{80}[/tex]
[tex]\vec{a}=0.5\: m/s^{2}[/tex]
Position:
Knowing the acceleration we can find the position using the falling equation.
[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]
[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]
[tex]\vec{x}=3200 m[/tex]
Average velocity:
The definition of the average velocity is:
[tex]\vec{v}=\frac{\Delta x}{t}[/tex]
[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]
[tex]\vec{v}=\frac{3200-0}{80}[/tex]
[tex]\vec{v}=40\: m/s[/tex]
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what Is accuracy ............
Answer:
Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem
《OAmalaOHopeO》
Answer:
In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.
Explanation:
_Hope it helps you_
The theory of the origin of the universe that is most popular among space scientists suggests that our universe originated how long ago
10-20 million years ago
10-20 trillion years ago,
10-20 billion years ago
10-20 thousand years ago
Answer:
10 - 20 million years ago
The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm
The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %
The known values are;
The increase in pressure per 10 meter increase in depth = 1.0 atm
The depth of the deepest ocean = 12 km = 12,000 m
The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm
The unknown
The percentage the density of water increased in the deepest ocean
Strategy;
Find the pressure at the deepest point of the deepest ocean and apply the compressibility
We have;
[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]
The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm
Therefore, we have for one cubic meter of water
[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]
Therefore;
[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³
The new volume = V - [tex]\mathbf{\partial}[/tex]V
∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³
The initial density = mass/(1 m³)
The new density = mass/(0.96 m³)
The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;
[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]
∴ [tex]\mathbf{\partial}[/tex]ρ% = 4.1[tex]\mathbf {\overline 6}[/tex] %
The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %
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Select the correct answer.
What is abstraction?
OA. the concept that software architecture can be separated into modules and that each module can be examined independently
OB. the process of containing information within a module, preventing any crossover or access to Irrelevant information
OC. the process of splitting a program both horizontally and vertically
OD. the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains
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Answer:
OD. The process of cutting down irrelevant information so only the information that is useful for particular purpose remains
Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.
What is abstraction?Abstraction is the practice of removing anything from a set of core features by eliminating or deleting attributes.
One of the three core ideas of object-oriented programming is abstraction order to decrease complexity and maximize efficiency, a programmer uses abstraction to conceal all but the important facts about an object.
Abstraction is the process of cutting down irrelevant information so only the information that is useful for a particular purpose remains.
Hence option D is correct.
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What would happen if you changed the position of the screen, but kept the other factors the same?
Answer:
I wish I could help but iam srry
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike the ground
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
the treasure map gives the following directions to the buried treasure
Answer:
North
South
East
West
Explanation:
please mark me as brainliest
Which item will be shipped third?
—-
Answer:
I know it's groceries
Explanation:
electronics ship before clothing
electronics ship after groceries
urgent items are first so
order:
1.) A/Electronics
2.) Clothing/B
3.) Groceries(since groceries aren't urgent)
thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.
Describe the forces that act on a skydiver before
and after the parachute is opened.
I will give brainliest!!!!
Answer:
Before the parachute opens: Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.
Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.
I HOPE THIS WILL HELP YOU IF NOT THEN SORRYHAVE A GREAT DAY :)
A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
why do we go to hospital
Answer:
bcz we want to have fun there lolol
Answer:
for emergency, treatment, medicines,etc.....
Current is the rate at which charge is flowing.
a. True
b. Fals
Answer:
A. True
Explanation:
The angle of the resultant vector is equal to
the inverse tangent of the quotient of the x-component divided by the y-component of the resultant vector
the inverse cosine of the quotient of the y-component divided by the x-component of the resultant vector.
the inverse cosine of the quotient of the x-component divided by the y-component of the resultant vector.
the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
To find the angle of a resultant vector, the vector must be resolved into y-component and x-component.
The y-component of a vector is the product of the magnitude of the vector and the sine of the angle of the vector to the horizontal. The x-component of a vector is the product of the magnitude of the vector and the cosine of the angle of the vector to the horizontal.The angle of this resultant vector is also known as the direction of the vector.
Mathematically, the direction of a resultant vector is given as;
[tex]\theta = tan^{-1} (\frac{R_y}{R_x} )\\\\where;\\\\\theta \ is \ the \ direction \ of \ the \ resultant \ vetcor\\\\R_y \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ y - direction\\\\R_x \ is \ the \ magnitude \ of \ the\ vector \ resolved \ in \ x - direction[/tex]
Therefore, the angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector.
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What is tensor quantity?
Is Inertia a tensor? give reason
Answer:
A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts. Stress and strain are both tensor quantities. ... A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts.
Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.
Explanation:
Hope dis help
Length of spring is 8cm when mass is 4kg is hang on it its become 16cm what is its spring constant?
Answer:
constant is 500 N/m
Explanation:
Extension, e = (16 - 8) = 8 cm
Force, F = (4 × 10) = 40 N
from hooke's law:
[tex]F = kx \\ 40 = (k \times 0.08) \\ k = \frac{40}{0.08} \\ \\ k = 500 \: newton \: per \: metre[/tex]
14. If you are asked to analyze an issue for an essay assignment, what should you do? a. Nb. C. Divide the issue into its main parts and discuss each part. Express your opinion about the issue. Explain what you think the issue means and how you came to that interpretation Describe the main ideas or points about the issue. d.
Answer:
Option C
Explanation:
Divide the issue into its main parts and discuss each part.
A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
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SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
What is a benefit of joining the YMCA instead of a private health club?
The YMCA has nicer equipment and buildings.The YMCA has nicer equipment and buildings. , ,
The YMCA is likely to be in an upscale neighborhood.The YMCA is likely to be in an upscale neighborhood. , ,
The YMCA is much more affordable.The YMCA is much more affordable. , ,
The YMCA has restaurants and spa services.
Answer:
Explanation:
This depends on where you are.
The YMCA where I grew up was affordable. Very. Or my mother would not have let me go.
Earth’s Moon has a diameter of 3,474 km and orbits at an average distance of 384,000 km. At that distance it subtends and angle just slightly larger than half a degree in Earth’s sky. Pluto’s moon Charon has a diameter of 1,186 km and orbits at a distance of 19,600 km from the dwarf planet. Compare the appearance of Charon in Pluto’s skies with the Moon in Earth’s skies. Describe where in the sky Charon would appear as seen from various locations on Pluto.
The reason for arriving at the above solutions is as follows:
The given dimensions and distance from the Earth of the Moon are;
The diameter of the Moon, d = 3,474 km
The average distance of the Moon from the Earth, R = 384,000 km
Required:
The comparison between Charon's appearance in Pluto and the Moon's appearance on Earth Earth
Solution:
The distance of the Moon's travels in an orbit, C = 2·π·R
∴ C = 2 × π × 384,000 km
The angle subtended by the Moon, θ = d/C × 360°
∴ θ = 3,474/(2 × π × 384,000) × 360° ≈ 0.518°
Pluto's moon Charon, has the following parameters;
The diameter of the Charon, d₂ = 1,186 km
The average distance of the Charon from Pluto, R₂ = 19,600 km
Therefore, the distance of the Moon's travels in an orbit, C₂ = 2·π·R₂
∴ C₂ = 2 × π × 19,600 km
The angle subtended by the Moon, θ₂ = d₂/C₂ × 360°
∴ θ₂ = 1,186/(2 × π × 19,900) × 360° ≈ 3.415°
The angle subtended by Charon in Pluto's sky ≈ 3.415°
Charon therefore, appears 7 times larger in Pluto's skies than the Moon's appearance in Earth's skies
Required:
The appearance of Charon as seen from different locations on Pluto
Solution:
Charon is gravitationally locked to Pluto, therefore, the same side of Pluto is faced with the same side of Charon
Therefore;
Charon appears constantly overhead from the side of Pluto locked to CharonCharon appears constantly at the horizon from the poles on either side of the axis of rotation of Pluto and CharonLearn more about Pluto's moon Charon here:
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A plastic box with objects has a mass of 4 kg and is on a shelf at a height of 2.4 m. What will it's potential gravitational energy?
Answer:
potential energy=mgh
4×9.8×2.4
Explanation:
may be hope this will help you
calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is 35 degree Celsius and outside is -5 degree Celsius conductivity of Glass is 2.2 into 10 to the power 3 cal per s per cm per k .
Answer:
The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass
Explanation:
can you guys pls also solve for average speed.
Answer:
d_t = 3.05km
v_a = 4.3km/h
Explanation:
42mins*(2/3) = 28mins
42mins-28mins = 14mins
d = v*t
d_1 = (4km/h)*(1h/60mins)*(28mins)
d_1 = 1.87km
d_2 = (5km/h)*(1h/60mins)*(14mins)
d_2 = 1.17km
d_t = d_1+d_2
d_t = 1.87km+1.17km
d_t = 3.05km
v_a = (v_1+v_2)/2
v_a = [(2*4km/h)+5km/h)]/3
v_a = 4.3km/h
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?
Answer:
[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]
Explanation:
We are asked to calculate the work done to move a box.
Work is the product of force and distance or displacement.
[tex]W= F*d[/tex]
A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.
F= 10 N d= 10 mSubstitute the values into the formula.
[tex]W= 10 \ N * 10 \ m[/tex]
Multiply.
[tex]W= 100 \ N*m[/tex]
Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.
[tex]W= 100 \ J[/tex]
100 Joules of work was done to move the box.
Cho các máy cắt sử dụng trong công nghiệp có ký hiệu trên nhãn thiết bị: C350; B500. Hãy tính dòng điện bảo vệ ngắn mạch và dòng điện bảo vệ quá tải của từng thiết bị?
Answer:
ask in the English then I can help you
Explanation:
please mark me as brain list
Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave
Ez Physics question will mark brainliest.
Answer:
The answer is B. longitude wave
Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day
what do you mean about it
Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3
Answer:
When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.
When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.
Explanation:
