Answer:
[tex]^{214} _{83} Bi[/tex] → [tex]^{210}_{81}Tl[/tex] + [tex]^4_2He[/tex]
Explanation:
[tex]^{214} _{83} Bi[/tex] → [tex]x[/tex] + [tex]^4_2He[/tex]
Subtract the [tex]^4_2He[/tex] from the [tex]^{214} _{83} Bi[/tex]
214 - 4 = 210
83 -2 = 81
Therefore, [tex]x[/tex] = [tex]^{210}_{81}Tl[/tex]
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the condenser. Suppose you have a discharging condenser and the instantaneous rate of change of the voltage is -0.01 of the voltage (in volts per second). How many seconds does it take for the voltage to decrease by 90 %?
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = [tex]e^{kt}[/tex]
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = [tex]e^{-0.01t}[/tex]
V = V₀[tex]e^{-0.01t}[/tex]
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀[tex]e^{-0.01t}[/tex]
0.1V₀ = V₀[tex]e^{-0.01t}[/tex]
0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]
0.1 = [tex]e^{-0.01t}[/tex]
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
URGENT
The component of solid-state physics that works with and/or heats metals and alloys to give them certain desired
shapes or properties is..
Answer:
the is metallurgy .....
An insulated tank contains 50 kg of water, which is stirred by a paddle wheel at 300 rpm while transmitting a torque of 0.1 kN-m. At the same time, an electric resistance heater inside the tank operates at 110 V, drawing a current of 2 A. Determine the rate of heat transfer after the system achieves steady state.
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW
The diagram shows the molecular structure of ethane. What is the chemical
formula for ethane?
Ethane
H H
H-C-C-H
| |
H H
Calculate the battery voltage to send 2.5 A of current through a light bulb with 3.6 ohms of resistance
Answer:
9 volts
Explanation:
[tex]v = i \times r \\ v = 2 .5 \times 3.6 \\ v = 9[/tex]
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 25 s interval
Answer:
The tub turns 37.520 revolutions during the 25-second interval.
Explanation:
The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:
[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)
Then, we expand the previous expression by kinematic equations for uniform accelerated motion:
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)
Where:
[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.
[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.
And each acceleration is determined by the following formulas:
Acceleration
[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)
Deceleration
[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)
Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.
If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:
[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]
[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]
[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]
[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]
[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]
[tex]\Delta n = 37.520\,rev[/tex]
The tub turns 37.520 revolutions during the 25-second interval.
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
A toy car of mass 600g moves through 6m in 2 seconds. The average kinetic energy of the toy car is
Answer:
12
Explanation:
I'm a beginner so am not sureeeeee
Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this lab that support your answer. Explain.
Answer:
Yes
Explanation:
Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,
Hence the same amount of charge flow is delivered to any circuit.
attached below are examples
A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).
Answer:
time is 32 s and speed is 304.3 m/s
Explanation:
Height, h = 146 m
speed, u = 14 m/s
Angle, A = 43 degree
Let it hits the ground after time t.
Use second equation of motion
[tex]h = u t +0.5 at^2\\\\- 146 =14 sin 43 t - 4.9 t^2\\\\4.9 t^2 - 9.5 t - 146 =0 \\\\t =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\\\\t=\frac{9.5\pm 54.3}{9.8}\\\\t = 32.05 s, - 22.4 s[/tex]
Time cannot be negative so the time is t = 32 s .
The vertical velocity at the time of strike is
v' = u sin A - g t
v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s
horizontal velocity
v'' = 14 cos 43 =10.3 m/s
The resultant velocity at the time of strike is
[tex]v=\sqrt{v'^2 + v''^2}\\\\v = \sqrt{304.1^2 +10.3^2 }\\\\v = 304.3 m/s[/tex]
Speeding up
Slowing down
Standing still
Holding at a constant non-zero velocity
Answer:
speeding up is the answer
Explanation:
from the graph it can be seen that as the time (horizontal axis) increases the speed of vehicle (vertical axis) increases
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m and we assume fully developed internal flow, find the pressure drop across this pipe length.
Answer:
[tex]\triangle P=1.95*10^{-4}[/tex]
Explanation:
Mass [tex]m=0.001[/tex]
Diameter [tex]d=1.2m[/tex]
Length [tex]l=10m[/tex]
Generally the equation for Volume flow rate is mathematically given by
[tex]Q=AV[/tex]
[tex]V=\frac{Q}{\pi/4D^2}[/tex]
[tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]
[tex]V=8.84*10^{-4}[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]F=\frac{64}{Re}[/tex]
Where Re
Re=Reynolds Number
[tex]Re=\frac{pVD}{\mu}[/tex]
[tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]
[tex]Re=1040[/tex]
Therefore
[tex]F=\frac{64}{Re}[/tex]
[tex]F=\frac{64}{1040}[/tex]
[tex]F=0.06[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]Head loss=\frac{fLv^2}{2dg}[/tex]
[tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]
[tex]H=19.9*10^{-9}[/tex]
Where
[tex]H=\frac{\triangle P}{\rho g}[/tex]
[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]
[tex]\triangle P=H*\rho g[/tex]
[tex]\triangle P=1.95*10^{-4}[/tex]
A car starts from rest and accelerates uniformly in a straight line in the positive x direction. After 25 seconds, its speed is 90 km/h.
a) Determine the acceleration of the object. [5]
b) How far does the object travel during the first 25 seconds? [3]
c) What is the average velocity of the object during the first 25 seconds?
Answer:
A. 1 m/s²
B. 312.5 m
C. 12.5 m/s
Explanation:
We'll begin by converting the velocity i.e 90 Km/h to m/s. This can be obtained as follow:
Velocity (Km/h) = 90 Km/h
Velocity (m/s) =?
Velocity (m/s) = Velocity (Km/h) × 1000 / 3600
Velocity (m/s) = 90 × 1000 / 3600
Velocity (m/s) = 90000 / 3600
Velocity (m/s) = 25 m/s
A. Determination of the acceleration.
Initial velocity (u) = 0 m/s
Final velocity (v) = 25 m/s
Time (t) = 25 s
Acceleration (a) =?
v = u + at
25 = 0 + (a × 25)
25 = 0 + 25a
25 = 25a
Divide both side by 25
a = 25/25
a = 1 m/s²
B. Determination of the distance travelled.
Initial velocity (u) = 0 m/s
Final velocity (v) = 25 m/s
Acceleration (a) = 1 m/s²
Distance travelled (s) =?
v² = u² + 2as
25² = 0 + (2 × 1 × s)
625 = 0 + 2s
625 = 2s
Divide both side by 2
s = 625 / 2
s = 312.5 m
C. Determination of the average velocity.
Total distance travelled = 312.5 m
Total time = 25 s
Average velocity =?
Average velocity = Total distance / total time
Average velocity = 312.5 / 25
Average velocity = 12.5 m/s
From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.
Answer:
time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.
Explanation:
Height, h = 15 m
Newton's second law
Force = mass x acceleration
The unit of gravitational force is Newton and the value is m x g.
where, m is the mas and g is the acceleration due to gravity.
Let the time of fall is t.
Use second equation of motion
[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]
Let the final speed is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]
which of the following is the correct description of momentum?
-the product of mass and acceleration -the product of mass and velocity
-velocity divided by mass
-acceleration divided by mass
Answer:
The product of mass and velocity is the correct answer
Explanation:
Momentum is defined as mass × velocity
p = mv
Answer:
The product of mass and velocity
Explanation:
I just did it and got it right with a 100%
Posted 1/3/23
Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object. Suppose a metal bar, initially at temperature 50 degrees Celsius, is placed in a room which is held at the constant temperature of 40 degrees Celsius. One minute later the bar has cooled to 40.18316 degrees . Write the differential equation that models the temperature in the bar (in degrees Celsius) as a function of time (in minutes). Hint: You will need to find the constant of proportionality. Start by calling the constant k and solving the initial value problem to obtain the temperature as a function of k and t . Then use the observed temperature after one minute to solve for k .
Answer:
Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),
Explanation:
Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says, "Wow that means everything cancels and nothing ever moves, it is all an illluussion! Wowwwwww" What statement best proves to him he is an idiot.
a) The equal and opposite forces act on different objects
b) If there is even a slight imbalance in the third law there will be a net force causing acceleration
C) he's right, all forces cancel, any motion I have ever seen is wrong
[tex] \huge \mathfrak{Answer.... }[/tex]
The Correct Answer is :
B. if there is even a slight imbalance in third law there will be a net force causing acceleration.
A slight difference in the forces can result in acceleration of an object.
[tex] \mathrm{✌TeeNForeveR✌}[/tex]
galileo was a contemporary of
Which shows the formula for converting from degrees Celsius to degrees Fahrenheit?
°F = (9/5 × °C) +32
°F = 5/9 × (°C – 32)
°F = °C – 273
°F = °C + 273
Answer:
the first answer
Explanation:
(32°F − 32) × 5/9 = 0°C
Answer:
Answer: A
Explanation:
A hockey puck is sliding across the ice with an initial velocity of 25 m/s. If the coefficient of friction between the hockey puck and the ice is 0.08, how much time (in seconds) will it take before the hockey puck slides to a stop
Answer: 31.89seconds
Explanation:
Based on the information given, we are meant to calculate deceleration which will be:
t = V/a
where, a = mg
Therefore, t = V/mg
t = 25/0.08 × 9.8
t = 25/0.784
t = 31.89seconds
Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height of 4.35 m above where it started. Using conservation of energy, find the height of the ball when it has a speed of 2.5 m/s.
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
At a rock concert, the sound intensity 1.0m in front of the bank of loudspeakers is 0.10 W/m^2. A fan is 30.0m from the loudspeakers. Her eardrums have a diameter of 8.4mm.How much energy is transferred to each eardrum in one second?
Answer: The amount of energy transferred to each eardrum in one second is [tex]6.2 \times 10^{-9} J[/tex].
Explanation:
Given: Intensity = 0.10 [tex]W/m^{2}[/tex]
Now, at 30.0 m the intensity will be calculated as follows.
[tex]Intensity = \frac{(30.0 m)^{2}}{(1 m)^{2}}\\= 900[/tex]
This means that the intensity is 900 times less assuming that the wave is spherical.
Hence, the new intensity is calculated as follows.
[tex]\frac{0.10 W/m^{2}}{900}\\= 1.11 \times 10^{-4} W/m^{2}[/tex]
The area of ear drum is expressed as follows.
[tex]Area = \pi \times r^{2}\\= 3.14 \times (4.2 mm)^{2}\\= 5.54 \times 10^{-5} m^{2}\\[/tex]
Formula used to calculate energy is as follows.
[tex]P = I \times a[/tex]
where,
P = power or energy
I = intensity
a = area
Substitute the values into above formula as follows.
[tex]P = I \times a\\= 1.11 \times 10^{-4} W/m^{2} \times 5.54 \times 10^{-5} m^{2}\\= 6.2 \times 10^{-9} W[/tex]
Thus, we can conclude that amount of energy transferred to each eardrum in one second is [tex]6.2 \times 10^{-9} J[/tex].
L Pretest Unit 7
Question 13 of 20
Andrew is trying to identify an unknown element. The element is shiny, but it
shatters when hammered and cannot be hammered into different shapes.
Where on the periodic table is this element most likely found?
A. On the left side
B. In one of the series below the main body of the table
C. On the right side
D. Along the metalloid stairstep line
SURMIT
Answer:
C
Explanation:
I think it would be there, it sounds like silicone and thats on the right side
The engine of a locomotive exerts a constant force of 6.8 105 N to accelerate a train to 80 km/h. Determine the time (in min) taken for the train of mass 1.1 107 kg to reach this speed from rest.
Answer:
t = 6 minutes
Explanation:
Given that,
Force,[tex]F=6.8\times 10^5\ N[/tex]
Initial speed of the train, u = 0
Final speed of the train, v = 80 km/h = 22.22 m/s
The mass of the train, [tex]m=1.1\times 10^7\ kg[/tex]
We need to find the time taken by the train to come to rest. We know that,
F = ma
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s[/tex]
or
t = 6 minutes (approx)
So, the required time is equal to 6 minutes.
what are the symptoms of hepatitis 'b'
What are the practical systems under continuos time system?
Answer:
REVIEWS
Wing-Kuen Ling, in Nonlinear Digital Filters, 2007
Controllability and observability
For a continuous time system, assume that x(0) = 0. ∀x1, if ∃t1 > 0 and u(t) such that x(t1) = x1, then the continuous time system is said to be reachable. Similarly, for a discrete time system, assume that x(0) = 0. ∀x1, if ∃n1 > 0 and u(n) such that x(n1) = x1, then the discrete time system is said to be reachable. For a continuous time system, if ∀x0, x1, ∃t1 > 0 and u(t) such that x(0) = x0 and x(t1) = x1, then the continuous time system is said to be controllable. Similarly, for a discrete time system, if ∀x0, x1, ∃n1 > 0 and u(n) such that x(0) = x0 and x(n1 = x1, then the discrete time system is said to be controllable. For LTI systems, the set of reachable state is R(|BAB… AnB|), where R(A) is defined as the range of A, that is R(A) = {y : y = Ax}. Also, the LTI systems are controllable if and only if R(A) = Rn Or in other words, rank(|BAB… AnB|) = n.
A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.
Answer:
Explanation:
Given that:
Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm
For objective lens of focal length f₁ = 20 mm
s₁' = 120 mm + 20 mm = 140 mm
∴
Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{140}{20}[/tex]
[tex]m_1 = 7 \ m[/tex]
For objective lens of focal length f₁ = 4 mm
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{124}{4}[/tex]
[tex]m_1 = 31 \ m[/tex]
For objective lens of focal length f₁ = 1.4 mm
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{121.4}{1.4}[/tex]
[tex]m_1 = 86.71 \ m[/tex]
The magnification of the eyepiece is given as:
[tex]m_e = 5X \ and \ m_e = 15X[/tex]
Thus, the largest angular magnification when [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]
[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]
= 86.71 × 15
= 1300.65
The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]
[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]
= 7 × 5
= 35
The largest magnification will be 1300.65 and the smallest magnification will be 35.
What is magnification?Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.
Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.
It is given in the question:
Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm
now we will calculate the magnification for all three focal lengths of the objective lens.
Also, each objective forms an image 120 mm beyond its second focal point.
(1) For an objective lens of focal length [tex]f_1=20 \ mm[/tex]
[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]
Magnification will be calculated as
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]
(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]
(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]
Now the magnification of the eyepiece is given as:
[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]
Thus, the largest angular magnification when
[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]
[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]
[tex]m_{large}=86.71\times 15=1300.65[/tex]
The smallest angular magnification derived when
[tex]m_1=7\ \ \ \ m_e=5[/tex]
[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]
[tex]m_{small}=7\times 5=35[/tex]
Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.
To know more about magnification follow
https://brainly.com/question/1599771
Ion how to do this at all
what is the application of a spherometer in the medical field?
Answer:
To correct the defects of vision by measuring the radius of curvature and thus the power of the lenses.
Explanation:
A spherometer is an instrument used to measure the curvature of objects such as lenses and curved mirrors.
Generally it consists of a fine screw which is moving in a nut carried on the center of a 3 small legged table or frame. The feet forms the vertices of an equilateral triangle. The lower end of the screw and those of the table legs are finely tapered and terminate in hemispheres.
If the screw has two turns of the thread to the milli meter the head is generally divided into 50 equal parts, so that differences of 0.01 millimeter may be measured without using a vernier scale.
The spherometer is used to measure the radius of curvature of the lenses so that the opthalmologist find the focal length of the lens and then give the power to the lens to correct the defects of vision.
who is corazon aquino?
Answer:
Maria Corazon Sumulong Cojuangco Aquino, popularly known as Cory Aquino, was a Filipino politician who served as the 11th President of the Philippines, the first woman to hold that office.
Answer:
Former President of the Philippines
Explanation: