A cataract is a clouding or opacity that develops in the eye's lens, often in older people. In extreme cases, the lens of the eye may need to be removed. What effect would this have on someone?

a. He would become nearsighted.
b. He would become farsighted.
c. He would become neither nearsighted nor farsighted.

Answers

Answer 1

Answer:

b. He would become farsighted.

Explanation:

A  cataract is defined as a medical condition where a person eyes becomes partially opaque and the person is not bale to see properly.

This is mainly caused due to aging or any injury in the eyes tissue which make up the lens of the eye.

It is the clouding of the lens of the eyes or opacity of the eyes. When treating cataract, in some cases the lens of the eyes are needed to be  removed. This may lead to person becoming far sighted.

Therefore, the correct option is (b).


Related Questions

Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 655 nm and a peak electric field magnitude of 1.5 V/m. 0.002984 W/m2 (b) an electromagnetic wave with an angular frequency of 6.5 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T. 1.19366E-6 W/m2

Answers

The intensity of the electromagnetic wave in terms of the electric field is 0.00298 W/m² and the intensity of the electromagnetic wave in terms of the magnetic field is 1.193x10⁻⁶  W/m².

The intensity of the electromagnetic wave is related to the electric field as well as to the magnetic field.    

a) Intensity of the electromagnetic wave for the electromagnetic field.

The intensity of the electromagnetic wave (I) in terms of the electromagnetic field is given by:

[tex] I = \frac{E^{2}*c*\epsilon_{0}}{2} [/tex]   (1)

Where:

c: is the speed of light = 3.00*10⁸ m/s  

E: is the magnitude of the electric field = 1.5 V/m

ε₀: is the permittivity of free space = 8.85*10⁻¹² C²/Nm²

Hence, the intensity of the electromagnetic wave (eq 1) is:

[tex] I = \frac{(1.5 V/m)^{2}*3.00 \cdot 10^{8} m/s*8.85 \cdot 10^{-12} C^{2}/(N*m^{2})}{2} = 0.00298 W/m^{2} [/tex]                                                                                          

b) Intensity of the electromagnetic wave for the magnetic field

We can calculate the intensity of the electromagnetic wave (I) in terms of the magnetic field with the following equation:

[tex] I = \frac{cB^{2}}{2\mu_{0}} [/tex]   (2)

Where:

B: is the magnitude of the magnetic field = 10⁻¹⁰ T

μ₀: is the vacuum permeability = 4π*10⁻⁷ m*T/A

Therefore, the intensity of the electromagnetic wave (eq 2) is:

[tex] I = \frac{3.00 \cdot 10^{8} m/s*(1\cdot 10^{-10} m*T/A)^{2}}{2*4\pi \cdot 10^{-7} T/A} = 1.193 \cdot 10^{-6} W/m^{2} [/tex]

Learn more about electromagnetic waves and magnetic and electric fields here: https://brainly.com/question/11647801?referrer=searchResults                                          

                                   

I hope it helps you!

A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.

Answers

Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.

Answers

Answer:

An object in either state of equilibrium must have no net force acting on it.

Explanation:

Answer: An object in either state of equilibrium must have no net force acting on it.

Explanation:

II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49. ( ) If the buckets are at rest, what is the tension in each cord? ( ) If the two buckets are pulled upward with an acceleration of 1.25 m/s by the upper cord, calculate the tension in each cord

Answers

Answer:

Here , mass of bucket ,m = 3.2 Kg

Now , let the tension in upper rope is T1

the tension in the middle rope is T2

a)

For lower bucket, balancing forces in vertical direction

T2 - mg = 0

T2 = mg

T2 = 3.2 *9.8

T2 = 31.36 N

tension in the middle rope is 31.36 N

For the upper bucket , balancing forces in vertical direction

T1 - T2 - mg = 0

T1 = T2 + 3.2 *9.8

T1 = 62.72 N

the tension in the upper rope is 62.72 N

B)

for a = 1.25 m/s^2

Using second law of motion ,for both the buckets

Fnet = ma

T1 - 2mg = 2m*a

T1 = 2*3.2*(9.8 +1.25)

T1 = 70.72 N

the tension in the upper rope is 70.7 N

Now , the lower bucket

Using second law of motion,

T2 - mg = ma

T2 = 3.2 * (9.8 + 1.25)

T2 = 35.36 N

the tension in the lower rope is 35.36 N

A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.​

Answers

F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a

==>   a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²

To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².

The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:

Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)

F[tex]_{horizontal}[/tex] = 150 N × cos(60°)

F[tex]_{horizontal}[/tex] = 150 N × 0.5

F[tex]_{horizontal}[/tex] = 75 N

Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)

F[tex]_{vertical}[/tex] = 150 N × sin(60°)

F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)

F[tex]_{vertical}[/tex] ≈ 129.9 N

Now, let's calculate the net force in the horizontal direction:

Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]

F[tex]_{net horizontal}[/tex] = 75 N - 15 N

F[tex]_{net horizontal}[/tex] = 60 N

Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:

F[tex]_{net horizontal}[/tex] = m × a

60 N = 20 kg × a

Now, solve for acceleration (a):

a = 60 N / 20 kg

a = 3 m/s²

So, the acceleration of the box is 3 m/s².

To know  more about acceleration

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Газ имеет объем 2 м³ при давлении 10° Па. Найдите обьем этого газа при изотермическом уменьшении давления в два раза.​

Answers

Answer:

sorry i didn't understand

how does laser works ?

Answers

Explanation:

Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.

Phân biệt các đặc điểm khác nhau giữa chất rắn, chất lỏng

Answers

Answer:

şen çal kapimi turkish drama

widely accepted scientific principles do not change. true or false​

Answers

Answer:

False

Explanation:

As technology advances and new evidence is found which either contradicts or supports accepted scientific principles, the principles are susceptible to change.

.
A mass of 8.72 kg gains 446 J of gravitational potential energy. To what height was it lifted?

Answers

Answer:

[tex]\boxed {\boxed {\sf 5.22 \ m}}[/tex]

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is calculated using the following formula:

[tex]E_P=mgh[/tex]

Where m is the mass, g is the acceleration due to gravity, and h is the height.

The object has a mass of 8.72 kilograms. Assuming this occurs on Earth, the acceleration due to gravity is 9.8 meters per second squared. The object gains 446 Joules of potential energy.

Let's convert the units of Joules. This makes the process of canceling units simpler later on. 1 Joule is equal to 1 kilogram meter squared per second squared. The object gains 446 J, which is equal to 446 kg *m²/s².

EP= 446 kg*m²/s²m= 8.72 kg g= 9.8 m/s²

Substitute the values into the formula.

[tex]446 \ kg*m^2/s^2 = 8.72 \ kg * 9.8 \ m/s^2 *h[/tex]

Multiply on the right side of the equation.

[tex]446 \ kg*m^2/s^2 = 85.456 kg*m/s^2 *h[/tex]

We are solving for the height, so we must isolate the variable h. It is being multiplied by 85.456 kg*m/s². The inverse operation of multiplication is division, so we divide both sides by this value.

[tex]\frac{ 446 \ kg*m^2/s^2}{85.456 kg*m/s^2} = \frac{85.456 kg*m/s^2 *h}{85.456 kg*m/s^2}[/tex]

[tex]\frac{ 446 \ kg*m^2/s^2}{85.456 kg*m/s^2} =h[/tex]

The units of kg*m/s² cancel, leaving meters as our unit.

[tex]\frac{ 446 }{85.456 } \ m =h[/tex]

[tex]5.2190601011 \ m =h[/tex]

The original measurements of mass and potential energy have 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredths place. The 9 in the thousandths place to the right tells us to round the 1 up to a 2.

[tex]5.22 \ m \approx h[/tex]

The object was lifted to a height of approximately 5.22 meters.

Define hydropower or hydroelectric power ?
No Spam..

Answers

[tex]\:[/tex]

Hydroelectric power, also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.

Answer:

Hydroelectric power/hydropower -  electricity produced by a hydraulic source, specifically energy generated falling or flowing water

A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 106 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 32.8°. How wide is the river?

Answers

Answer:

x = 68.3 m

Explanation:

tan 32.8 = x / 106

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a distance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?

Answers

Answer:

[tex]D_l=d[/tex]

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

[tex]V_{e1}=V_{e2}[/tex]

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

[tex]D=\frac{at^2}{2}[/tex]

Where

Acceleration is given as

[tex]a=\frac{V_o}{2d}[/tex]

And

Time

[tex]T=\frac{d}{v_0}[/tex]

Therefore horizontal displacement towards the left is

[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]

[tex]D_l=d[/tex]

Observe: Air pressure is equal to the weight of a column of air on a particular location. Air pressure is measured in hectopascals (hPa). Note how the air pressure changes as you move Station B towards the center of the high-pressure system.

a. What do you notice?
b. Why do you think this is called a high-pressure system?

Answers

Answer:

A. When moving towards a high pressure center the pressure values ​​increase in the equipment

B. This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

A political campaign manager must decide whether to emphasize television advertisements or letters to potential voters in a reelection campaign. Describe the production function for campaign votes.
A. Campaign managers produce campaign votes.
B. Reelection campaigns produce campaign votes.
C. Television advertisements and campaign votes produce letters to potential voters.
D. Television advertisements and letters to potential voters produce reelection campaigns.
E. Television advertisements and letters to potential voters produce campaign votes.
How might information about this function (such as the shape of the isoquants) help the campaign manager plan strategy?
A. If the marginal rate of technical substitution of television advertisements for letters to potential voters is constant, then the campaign manager should use a combination of the two inputs.
B. If television advertisements and letters to potential voters are perfect complements, then the campaign manager should use them in fixed proportions.
C. If television advertisements and letters to potential voters are perfect substitutes, then the campaign manager should use them in fixed proportions.
D. If the isoquant curves for television advertisements and letters to potential voters are convex, then the campaign manager should use only the cheaper input per vote.
E. If the isocost lines for television advertisements and letters to potential voters are convex, then the campaign manager should use a combination of the two inputs.

Answers

Answer:

First answer - (E)

Second answer - (B)

Explanation:

The trade-off here is between TV ADVERTISEMENTS and LETTERS TO POTENTIAL VOTERS. The campaign manager for the candidate who is running for reelection, is trying to decide which of the two factors he should use more of or emphasize. The production function for campaign votes can be simplified as

TVAD + LTPV = CV

This is the production function for campaign votes.

PART A

Describe the production function for campaign votes (in words).

ANSWER: (E)

Television advertisements and (or 'plus') letters to potential voters, produce (or 'equal') campaign votes.

PART B

How might information about this function (such as the shape of the isoquants) help the campaign manager plan strategy?

ANSWER: (B)

If television advertisements and letters to potential voters are perfect complements (complements are goods or actions that 'must' go together or be used together) then the campaign manager should use them in fixed proportions (e.g. in a ratio of 50:50).

which characteristic of nuclear fission makes it hazardous?

Answers

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

Q.3. The equivalent resistance across AB is:
(a)1
(c)2
(b)3
(d)4

Answers

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

B = 0.013(-j) T

Explanation:

Given that,

The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]

Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]

We need to find the magnitude and the direction of the magnetic field. At equilibrium,

[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]

The velocity is in +z direction, force in +x direction, then the field must be in -y direction.

herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran

Answers

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

Which of the following is not true about Triton, the large moon of Neptune? It is more reflective than Earth's Moon. It is larger than Earth's Moon. It is in a retrograde orbit. It has a thin atmosphere. It has nitrogen geysers.

Answers

Answer:

Triton is the largest of Neptune's 13 moons. It is unusual because it is the only large moon in our solar system that orbits in the opposite direction of its planet's rotation―a retrograde orbit. ... Like our own moon, Triton is locked in synchronous rotation with Neptune―one side faces the planet at all times.

One of the asteroids, Ida, looks like an elongated potato. Surprisingly it has a tiny (compared to Ida) spherical moon! This moon called Dactyl has a mass of 4.20 × 10^16 kg, and a radius of 1.57 × 10^4 meters, according to Wikipedia. Ida has a radius of 3.14 x 10^4 meters.
Find the acceleration of gravity on the surface of this little moon.

Answers

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

[tex]g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}[/tex]

g = 0.0114 m/s²

Yea, gonna need some help. Thanks

Answers

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

[tex]h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t[/tex]

where,

v₀ = initial speed = 110 ft/s

Therefore,

[tex]110 = 32t\\\\t = \frac{110}{32}\\\\[/tex]

t = 3.48 s

How much energy is stored in a spring that is compressed 0.650m if the spring constant is 725N/m?

Answers

Answer:

53.8Joule

Explanation:

hope it is helpful

please mark it as brainliest

Answer:

approximate 153.1J

Explanation:

W= 1/2k(x^2) = 1/2x725x(0.650)^2 = 153.15625 (J)

If car A passes car B, then car A must be
A. accelerating at a greater rate than car B.
B. moving faster than car B, but not necessarily accelerating
C. accelerating
D. moving faster than car B and accelerating more than car B

Answers

Answer:

B. moving faster than car B, but not necessarily accelerating

Explanation:

Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.

two objects A and B vertically thrown up with velocities 80m/s and 100m/s at two sec interval.where and when will they meet each other?

Answers

Answer:

hcbvdgsyyvjusvbxjxu usbsbhsi

Explanation:

ysggsghxuxgscsixigdvgsibxhdhshshjf

A bus starts from rest and accelerates at 1.5m/s squared until it reaches a velocity of 9m/s .the bus continues at this velocity and then deccelerate at -2m/s squared until it comes to stop 400m from it's starting point. how much time did the bus takes to cover the 400m?​

Answers

Answer:

23s

Explanation:

s=ut+1/2at^2

the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore

400=0t+1/2(1.5)t^2

400/0.75=0.75t^2/0.75

t^2=√533.33

t=23s

I hope this helps and sorry if it's wrong

1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.

Answers

Answer:

ΔT = 0.017 °C

Explanation:

According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:

Change in Internal Energy = (0.85)(Kinetic Energy)

[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]

where,

ΔT = increase in temperature = ?

v = speed of block = 4 m/s

C = specific heat capacity of copper = 389 J/kg.°C

Therefore,

[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]

ΔT = 0.017 °C

Chase grew up wanting to wear his sister's dresses over his brother's pants and button up shirts. When Chase turns 18, he decides to begin living as woman, though he's still only sexually attracted to women. He decides he doesn't want to undergo surgery. Chase is

Answers

Explanation:

she is a woman, i dont understand why youre still using he/him after she comes out

which watch is more preferable for the measurement of time among pendulum, quartz and atomic watch

Answers

Answer:

pendulum, quartz

Explanation:

19 point please please answer right need help

Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?

Answers

Explanation:

We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as

[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]

For the hanging mass [tex]m_2,[/tex] we can write NSL as

[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]

We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get

[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]

or

[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]

[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]

Using this value for the acceleration on Eqn(2), we find that the tension T is

[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:=24.7\:\text{N}[/tex]

Other Questions
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