An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons
Answers
Answer:
50 protons 50 electrons and 69 neutrons...
Explanation:
the number of protons is equal to number of electrons. then mass number is the sum of protons and neutrons in a nucleus so for we to get the number of neutrons we take the mass number subtract the protons number.
Related Questions
Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.
Answers
Answer:
[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]
[tex]44.43^{\circ}[/tex], second order does not exist
Explanation:
n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]
[tex]\lambda[/tex] = Wavelength
m = Order
Distance between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]
[tex]\lambda=400\ \text{nm}[/tex]
m = 1
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]
The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].
[tex]\lambda=700\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]
Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.
The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].
Second order angle does not exist.
What is diffraction of light
Answers
Answer:
According to "http://ww2010.atmos.uiuc.edu" Diffraction is the slight bending of light as it passes around the edge of an object.
Some examples of Light Defraction would be..
-CD reflecting rainbow colours
-Sun appears red during sunset
-From the shadow of an object
What type of weather would a continental Polar air mass bring
Answers
Answer:
Continental polar ( cp):
Explanation:
Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.
PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium
Answers
Answer: B
Explanation:
Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answers
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
If you push with a power of 20 Watts
on a 150 Newton object, how long would
it take to push it over the 4.3 m?
Answers
Answer:
32.25 s
Explanation:
From the question,
P = W/t.............. Equation 1
Where P = Power, W = work done, t = time.
But
W = F×d................. Equation 2
Where F = force and d = distance
Substitute equation 2 into equation 1
P = F×d/t............... Equation 3
make t the subject of euqation 3
t = (F×d)/P............. Equation 4
Givn: F = 150 N, d = 4.3 m, P = 20 watts.
Substitute these values into equation 4
t = (150×4.3)/20
t = 32.25 s
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.
a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided
Answers
Answer:
a) p₂ = 1.88 kg*m/s
θ = 273.4 º
b) Kf = 37% of Ko
Explanation:
a)
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:[tex]p_{o1x} = 2.00 kg*m/s (1)[/tex]
[tex]p_{o1y} = 0 (2)[/tex]
We can do the same for the particle moving along the positive y-axis:[tex]p_{o2x} = 0 (3)[/tex]
[tex]p_{o2y} = 4.00 kg*m/s (4)[/tex]
Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:[tex]p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)[/tex]
[tex]p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)[/tex]
Now, the total initial momentum, along these directions, must be equal to the total final momentum.We can write the equation for the x- axis as follows:[tex]p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)[/tex]
We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:[tex]p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)[/tex]
Now, we can repeat exactly the same process for the y- axis, as follows:[tex]p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)[/tex]
We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:[tex]p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)[/tex]
Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:[tex]p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)[/tex]
We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)[/tex]
The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.b)
Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:[tex]\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)[/tex]
So, the final kinetic energy has lost a 37% of the initial one.An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.
Answers
Answer:
B. in the direction of the force
Explanation:
Sana nakatulong
HELP DUE TODAY BEST ANSWER GET BRAINLIE
One of the greatest dangers in a tornado is from flying objects. A 15 pound piece of lumber can turn into a flying missile that could severely damage walls and homes. A piece of steel with a mass of 200 pounds and travelling at the same velocity would cause even more damage. Select any evidence from the list below that you could use to explain why a 200 pound piece of steel would cause more damage than a 15 pound piece of wood travelling at the same velocity.
As the kinetic energy of an object increases, the force it can exert on another object decreases.
As the kinetic energy of an object increases, the force it can exert on another object increases.
Objects with more mass have less kinetic energy.
Objects with more mass have more kinetic energy.
As the velocity of an object increases, its kinetic energy decreases.
As the velocity of an object increases, its kinetic energy increases.
Answers
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
Cells use nutrients and oxygen to supply the body with the energy it needs. What three-body systems are working together in this situation?
A
nervous, digestive, and circulatory systems
B
digestive, circulatory, and excretory systems
C
circulatory, immune, and respiratory systems
D
digestive, respiratory, and circulatory systems
Answers
Answer:
The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.
Explanation:
so the answer is D
(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answers
Answer:
5.0/s
Explanation:
Answer:
b and a it is this that abewsr
Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?
Answers
Answer:
work done= force × displacement
=800×5
=4000J
Explanation:
The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.
A 1500-kg car travelling 90 km/h[N] collides with a 1200-kg minivan travelling 40 km/h[S]. After the collision, the two vehicles stick together.
a. Calculate the initial momentum of the car and the minivan.
b. Using the law of conservation of momentum, determine the total momentum of the two vehicles after the collision.
c. Calculate the final velocity of the two vehicles after the collision in metres per second.
Answers
Answer:
A) car - 37500 kg*m/s, minivan - 13332 kg*m/s
B) 50832 kg*m/s
C) 18.83 m/s
Explanation:
Realize that sticky collisions are modeled by: m1v1+m2v2=(m1+m2) vf
conevert to m/s....car going 25 m/s, minivan going 11.11 m/s
A) p=mv
p(car)=(1500)(25)
p(car)=37500 kg*m/s
p(minivan)=(1200)(11.11)
p(minivan)=13332 kg*m/s
B) 37500+13332=50832 kg*m/s
C) 37500+13332=(1500+1200) vf
50832=2700(vf)
18.83 m/s = vf
The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?
Answers
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. What is the speed of the tip of a blade 10 s after the fan is turned off
Answers
Answer:
the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Explanation:
Given;
diameter of the ceiling fan, d = 90 cm = 0.9 m
angular speed of the fan, ω = 64 rpm
time taken for the fan to stop, t = 28 s
The distance traveled by the ceiling fan when it comes to a stop is calculated as;
[tex]d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m[/tex]
The speed of the tip of a blade 10 s after the fan is turned off is calculated as;
[tex]v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s[/tex]
Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this change in momentum.
Answers
Answer:
Time = 0.58 seconds
Explanation:
Given the following data;
Initial momentum = 3 kgm/s
Final momentum = 10 kgm/s
Force = 12 N
To find the time required for the change in momentum;
First of all, we would determine the change in momentum.
[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]
[tex] Change \; in \; momentum = 10 - 3 [/tex]
Change in momentum = 7 kgm/s
Now, we can find the time required;
Note: the impulse of an object is equal to the change in momentum experienced by the object.
Mathematically, impulse (change in momentum) is given by the formula;
[tex] Impulse = force * time [/tex]
Making "time" the subject of formula, we have;
[tex] Time = \frac {impulse}{force} [/tex]
Substituting into the formula, we have;
[tex] Time = \frac {7}{12} [/tex]
Time = 0.58 seconds
Time of the day when the Sun does not shine (___time)
N____N
Answers
nighttime? So put night in the blank
A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.
Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?
Answers
Answer:
x = 0.056 m
ΔKE = 0.489 J
Explanation:
Given that
Angle, θ = 38°
Length, L = 1.7 m
Mass, m = 0.09 kg
Spring constant, K = 590 N/m
If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE
This is mathematically written as
1/2kx² = mgH
The height, H we can get by using the relation
H = L.Sinθ
H = 1.7 * Sin 38
H = 1.7 * 0.6157
H = 1.047 m
Next, we use the Work-Energy theorem
1/2kx² = mgH
1/2 * 590 * x² = 0.09 * 9.8 * 1.047
295 * x² = 0.9234
x² = 0.9235 / 295
x² = 0.00313
x = √0.00313
x = 0.056 m
If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8
Then we use the formula
ΔKE = mg(H - H1)
ΔKE = mg(xsinθ - x2.sinθ)
Where, x = 1.7 , x2 = 0.8
ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)
ΔKE = 0.882(1.047 - 0.493)
ΔKE = 0.882 * 0.554
ΔKE = 0.489 J
A 40 kg rock is rolling toward a town at 4 m/s after an earthquake. Calculate the KE.
Be sure to show your work and include units!
Use the formula KE = 1/2mv2 will brainlist
Answers
Answer:
320 J
Explanation:
From the question,
KE = 1/2mv².................. Equation 1
Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock
Given: m = 40 kg, v = 4m/s
Substitute these values into equation 1
KE = 1/2(40)(4²)
KE = 20×16
KE = 320 J
Hence the kinetic energy of the rock is 320 J
What is the acceleration of a 4,000 kg car pushed with a
force of 12,000 N?
Answers
Answer:
3 m/s
Explanation:
A= F/m
12,000/ 4000 = 3
Answer:
3 m/s^2
Explanation:
The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.
Our known values are:
F ( Force ) = 12,000 N
m ( mass ) = 4,000 kg
a ( acceleration ) = ?
Now we plug in the known values into the equation and solve
F=ma
12,000=4,000a
We have to divide 4,000 by both sides to isolate the a value
12,000/4,000=4,000/4,000a
The 4,000s on the right of the equation cancel.
And 12,000 divided by 4,000 equals 3
The acceleration (a) is 3 meters per second squared (m/s^2)
Next, check to make sure 3 does work by plugging it back into the equation.
12,000=4,000*3
12,000=12,000 ✔
As you can see, the acceleration will be 3 m/s^2
5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136
J/kg °C
Answers
Answer:
Q = 636.48 J
Explanation:
Given that,
The mass of gold, m = 0.052 kg
The temperature increase from 30°C to 120°C.
The specific heat of gold is 136 J/kg °C.
We need to find the heat needed to warm the gold. The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]
So, 636.48 J of heat is needed to warm gold.
A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed
Answers
Answer:
The angular speed is 23.24 rad/s.
Explanation:
Given;
mass of the disk, m = 7 kg
radius of the disk, r = 0.2 m
applied force, F = 42 N
distance moved by disk, d = 0.9 m
The torque experienced by the disk is calculated as follows;
τ = F x d = I x α
where;
I is the moment of inertia of the disk = ¹/₂mr²
α is the angular acceleration
F x r = ¹/₂mr² x α
The angular acceleration is calculated as;
[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]
The angular speed is determined by applying the following kinematic equation;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]
initial angular speed, ωi = 0
angular distance, θ = d/r = 0.9/0.2 = 4.5 rad
[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]
Therefore, the angular speed is 23.24 rad/s.
When6-2 He He-6 undergoes beta decay, the daughter is?
Answers
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength ?
Answers
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;
[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]
Therefore, the electric field strength at the mid-point between the two rings is zero.
If matter cannot be created or destroyed, then how do you end up with
rust? Below is the equation for rust.
4Fe + 302 → 2Fe203
oxygen from the air
water in the atmosphere
oxygen from in the metal
there shouldn't be any oxygen
Answers
Your question is a "non sequitur", which means "it doesn't follow".
Your "then" doesn't contradict your "If", so no mystery is implied.
Maybe you're trying to say that matter is somehow not conserved in the equation . . . 4Fe + 302 → 2Fe203 . But it is. There are 4 Irons and 6 Oxygens on each side, so conservation is not violated here.
I looked up "rust" on Floogle, and got slapped with pages and pages of chemistry that I don't completely understand. But what it's saying is that rusting is a very complex chemical process, AND it doesn't happen unless there's some water involved.
So the bottom line is that there's a lot more going on than simply
4Fe + 302 → 2Fe203 ,
there's water going in and out of the process at every stage, and when it's all over, you have rusty iron, and mass has been conserved.
Which of these is the BEST answer for why science is important?
Science can take us to other planets, even if it’s only through a telescope.
Science is part of human nature; it helps answer questions about how the world works.
Science helps us learn to think more critically and weigh evidence better.
Science gives us better tablet computers and games.
QQQUUUUCCCKKKK!!!!!!
Answers
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×106 N , one at an angle 14.0 ∘ west of north, and the other at an angle 14.0 ∘ east of north, as they pull the tanker a distance 0.800 km toward the north. Part A What is the total work done by the two tugboats on the supertanker?
Answers
Answer:
W = 1,049 10⁹ J
Explanation:
Work is defined by the relation
W = F. d = F d cos θ
where tea is the angle between the forces and the displacement.
The total work is the sum of the work of each tug.
Tug 1
W₁ = F d cos θ₁
the angle measured from the positive side of the x-axis is
θ₁ = 14 + 90 = 104º
tugboat 2
W₂ = F d cos θ₂
θ₂ = 14
we substitute
W = F d cos θ₁ + F d cos θ₂
W = F d (cos θ₁ + cos θ₂)
let's calculate
W = 1.80 10⁶ 800 (cos 104 + cos 14)
W = 1,049 10⁹ J
A group of particles of total mass 48 kg has a total kinetic energy of 320 J. The kinetic energy relative to the center of mass is 80 J. What is the speed of the center of mass?
Answers
320=.5(48)v^2+80
V=3.16
Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.6-V battery, the current from the battery is 2.07 A. When the resistors are connected in parallel to the battery, the total current from the battery is 8.98 A. Determine R1 and R2. (Enter your answers from smallest to largest.)
Answers
Answer:
When R1 = 2.193, R2 = 3.894
When R1 = 3.894, R2 = 2.193
Explanation:
We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.
Formula for resistance is;
R = V/I
R = 12.6/2.07
R = 6.087 ohms
Since R1 and R2 are connected in series.
Thus; R1 + R2 = 6.087 ohms
R1 = 6.087 - R2
We are also told that when they are connected in parallel, the current is 8.98 A.
Thus, R = 12/8.98
R = 1.403 ohms
Thus;
(1/R1) + (1/R2) = 1/1.403
Let's put 6.087 - R2 for R1;
(1/(6.087 - R2)) + (1/R2) = 1/1.403
Multiply through by 1.403R2(6.087 - R2) to get;
1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)
Expanding gives;
1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²
(R2)² - 6.087R2 + 8.54 = 0
Using quadratic formula, we have;
R2 = 2.193 ohms or 3.894 ohms
Thus,
R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894
R1 = 3.894 or 2.193
When R1 = 2.193, R2 = 3.894
When R1 = 3.894, R2 = 2.193
The optics of your visual system have a total refractive power of about +60 D—about +20 D from the lens in your eye and +40 D from the curved shape of your cornea. Surgical procedures to correct vision generally do not work on the lens; they work to reshape the cornea. In the most common procedure, a laser is used to remove tissue from the center of the cornea, reducing its curvature. This change in shape can correct certain kinds of vision problems.
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, he would be:________.
A. Nearsighted.
B. Farsighted.
C. Neither nearsighted nor farsighted.
Answers
Answer:
Farsighted
Explanation:
Farsightedness also known as hypermetropia is caused by the eye being too short(the eye shortens with advancing age) or the crystalline lines not being sufficiently convergent.
A farsighted person can see far objects but can not see nearby objects. His near point is now farther than the 25 cm near point of a normal eye. Images are formed some distance behind the retina.
This eye defect is corrected by the use of a converging lens to reduce the divergence of the rays entering the eye from an object.