brainly A person's eye lens is 2.9 cm away from the retina. This lens has a near point of 25 cm and a far point at infinity. What must the focal length of this lens be in order for an object placed at the near point of the eye to focus on the retina

[tex]F = 5.93×10^{13}\:\text{N}[/tex]

Explanation:

Given:

[tex]m_1= 2×10^{16}\:\text{kg}[/tex]

[tex]m_2= 4×10^{22}\:\text{kg}[/tex]

[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]

Using Newton's universal law of gravitation, we can write

[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]

[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]

[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]

Answer:

The correct answer is A. The sun is the energy source of the surface currents in the ocean

The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .

Surface currents are currents that are located in the upper feet of the ocean , they are simply how water moves from one place to another . Pattern of surface current are determined by wind direction .

Surface currents are formed by global wind system that are fueled by energy from the sun . Because of heating effect of sun , the earth's atmosphere gets warmed up . As we know , warm air is lighter then cool air , it rises up and create low pressure near the equator because of this wind causes surface currents the ocean .

hence , The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .

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Answer:

[tex]v_1=87.40m/s[/tex]

Explanation:

From the question we are told that:

Mass of arrow [tex]m=52g[/tex]

Mass of rock [tex]m_r=1.50kg[/tex]

Height [tex]h=0.47m[/tex]

Generally the equation for Velocity is mathematically given by

 [tex]v = \sqrt{(2gh)}[/tex]

 [tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]

 [tex]v= 3.035m/s[/tex]

Generally the equation for conservation of momentum is mathematically given by

 [tex]m_1v_1=m_2v_2[/tex]

 [tex]0.052kg * v = 1.5 * 3.03m/s[/tex]

 [tex]v_1=87.40m/s[/tex]

This question is incomplete, the complete question is;

A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answer:

the second moth is moving at 5.062 m/s

Explanation:

Given the data in the question;

Using doppler's effect

[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )

f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )

frequency reflected from the moth,

Now, moth is the source and the bat is the receiver

f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )

hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )

we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.

given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.

we substitute

55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[  ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )

1.02996 =  ( 343 + u₂ ) / ( 343 - u )

( 343 + u₂ ) = 1.02996( 343 - u )

343 + u = 353.27628 - 1.02996u

u + 1.02996u = 353.27628 - 343

2.02996u = 10.27628

u = 10.27628 / 2.02996

u = 5.062 m/s

Therefore, the second moth is moving at 5.062 m/s

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

Explanation:

Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]

The sum of the two vectors is

[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]

[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]

The difference between the two vectors can be written as

[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]

[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]

Answer:

PlROCA

Explanation:

The velocity of the ball is 100m/s

The first step is to write out the parameters;

The force used to kick the ball is 1000N

The mass of the ball is 0.8 kg

Time is 0.8 seconds

Therefore the velocity can be calculated as follows

F= Mv-mu/t

1000= 0.8(v) - 0.8(0)/0.8

1000= 0.8v- 0.8/0.8

Cross multiply both sides

1000(0.8) = 0.8v

800= 0.8v

divide both sides by the coefficient of v which is 8

800/0.8= 0.8v/0.8

v= 1000m/s

Hence the velocity is 1000m/s

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Answer:

d = 30kg/cm³

Explanation:

d = m/v

d = 1080kg/(3cm*4cm*3cm)

d = 30kg/cm³

Answer:

a) When moving body applies brake then velocity and acceleration would be in opposite direction

b) When body starts to increase velocity then velocity and acceleration would be in same direction

c) When body is circulating then velocity and acceleration would be perpendicular to each other

Explanation:

a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction

b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction

c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.

Explanation:

The equivalent capacitance of capacitors in parallel can be determined as

[tex]C_{eq} = C_1 + C_2 + C_3[/tex]

[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]

Answer:

where is the attachment

Explanation:

The velocity of B after the collision is obtained as -2.6 m/s.

Now we now that the  principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.

Thus;

(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)

-7 = 10 + 6.5v

-7 - 10 = 6.5v

v = -7 - 10 /6.5

v = -2.6 m/s

Hence, the velocity of B after the collision is obtained as -2.6 m/s.

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Answer:

Charge of the particle is 1 coulomb.

Explanation:

Force, F:

[tex]{ \bf{F=BeV}}[/tex]

F is magnetic force.

B is the magnetic flux density.

e is the charge of the particle.

V is the velocity

[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]

Answer:

The total surface area is "90 km²".

Explanation:

Given:

Power from solar installations,

= 27 GW

Other natural installations,

= 740 GW

Intensity,

[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]

%n,

= 30%

Now,

⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]

then,

⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]

           [tex]=90 \ GW[/tex]

As we know,

⇒ [tex]I=\frac{P}{A}[/tex]

by substituting the values, we get

[tex]1000=\frac{90\times 10^9}{A}[/tex]

    [tex]A = \frac{90\times 10^9}{10^3}[/tex]

        [tex]=90\times 10^6[/tex]

        [tex]=90 \ km^2[/tex]

Answer:

Distance from sun to Neptune = 4.495E9 km

Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec

That is from sun to Neptune time fof light = 250 min

Time for light to travel from sun to earth is about 8 min

So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and

250 min / 8 min is roughly 30

The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:

          t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s

depending on the relative distance of the two planets

given parameters

to find

They  velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships

             v = d / t

             t = d / v

where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)

We look in the tables for the distances and the rotation periods around the sun

                           distance ( m)         period (s)

Sun Neptunium     4.50 10¹²             5.2 10⁹

Sun - Earth             1.5    10¹¹              3.2 10⁷

With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:

Maximum approach

positions relative distance from the dos Plantetas is

         Δd = [tex]x_{Neptuno - Sum} - x_{Earth - Sum}[/tex]d  

         Δd = 4.50 10¹²  - 1.5 10¹¹

         Δd = 43.5 10¹¹ m

the time it takes for Neptune's light to reach Earth is

        Δt = [tex]\frac{ 43.5 \ 10^{11} }{3 \ 10^8}[/tex]  

        Δt = 14.5 10³ s

        Δt = 1.45 10⁴ s

We reduce to hours

        Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h

Maximum away

         Δd = [tex]x_{Neptune - Sum} + x_{Neptune-Sum}[/tex]  

         Δd = 4.50 10¹² + 1.5 10¹¹

         Δd = 46.5 10¹¹

The time is

         Δt = [tex]\frac{46.5 \ 10^{11}}{ 3 \ 10^8}[/tex]  

         Δt = 15.5 10³

         Δt = 1.55 10⁴ s

We reduce to hours

         Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h

In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:

          t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s

depending on the relative distance of the two plants

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Explanation:

OGLE-2005-BLG-390Lb.

PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...

Neptune. Surface Temperature: 72 Kelvin. ...

Uranus. Surface Temperature: 76 Kelvin. ...

Saturn. Surface Temperature: 134 Kelvin. ...

Jupiter. Image Courtesy: NASA. ...

OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown

Answer: D. 0 m

Explanation:

Concept:

Here, we need to know the concept of displacement.

Displacement is defined to be the change in position of an object.

The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

STEP ONE: the boy walks from point C to point D (a distance of 50 m)

C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D

                                              50 m

STEP TWO: the boy walks from point D to point C (a distance of 50 m)

D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C

                                               50 m

STEP THREE: find the displacement

The boy started with point C

The boy ended with point C

He did not change his position throughout the journey.

Therefore, his displacement is 0 m.

Hope this helps!! :)

Please let me know if you have any questions

Answer:

Ecologist.

Your answer is Ecologist.

(Ecologist) is a scientist who studies the whole environment as a working unit.

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.

Answer:

We have that the spring constant is mathematically given as

[tex]k=2.37*10^{11}N/m[/tex]

Generally, the equation for angular velocity is mathematically given by

[tex]\omega=\sqrt{k}{m}[/tex]

Where

k=spring constant

And

[tex]\omega =\frac{2\pi}{T}[/tex]

Therefore

[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]

Hence giving spring constant k

[tex]k=m((\frac{2 \pi}{T})^2[/tex]

Generally

Mass of earth [tex]m=5.97*10^{24}[/tex]

Period for on complete resolution of Earth around the Sun

[tex]T=365 days[/tex]

[tex]T=365*24*3600[/tex]

Therefore

[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]

[tex]k=2.37*10^{11}N/m[/tex]

In conclusion

The effective spring constant of this simple harmonic motion is

[tex]k=2.37*10^{11}N/m[/tex]

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Answer:

[tex]E=35921.96N/C[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.321mm[/tex]

Charge Density [tex]\mu=0.100[/tex]

Distance [tex]d= 5.00 cm[/tex]

Generally the equation for electric field is mathematically given by

[tex]E=\frac{mu}{2\pi E_0r}[/tex]

[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]

[tex]E=35921.96N/C[/tex]

Answer:

杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音

Explanation:

莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd

Answer:

B velocity

Explanation:

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex]  Rounding to the correct number of sig fig's to simplify:

[tex]v=\sqrt{400+2.0*10^2}[/tex] to get

v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

Answer:

No. As we gain mass the force of gravity on us does not decrease