Convert 385k to temperature of

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

Explanation:

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant

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

Answer:

[tex]P=-0.2D[/tex]

Explanation:

From the question we are told that:

Far-point [tex]v=-5.04m[/tex]

Where

u=-\alpha

Generally the equation for Lens is mathematically given by

 [tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

 [tex]P=\frac{1}{-5.04}-\frac{1}{\alpha}[/tex]

 [tex]P=-0.2D[/tex]

Answer:

chemical potential energy​

Explanation:

A 9v battery comes in different formats, such that the most common one is the carbon-zinc and alkaline chemistry, so these are alkaline batteries (there are also rechargeable or lithium batteries, these also depend on chemical interactions).

These batteries "draw" the energy from chemical interactions of the materials inside of it, so the type of potential energy that is stored in a battery is actually chemical (regardless of the fact that the energy can be transformed into electrical energy later) the "potential" refers to how the energy is stored.

Then the correct option is chemical potential energy​

Answer:

Chemical Potential Energy

Explanation:

Hope this helps!!

Have a blessed day/night!! <33

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.

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.

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.

Solution :

a). B at the center :

     [tex]$=\frac{u\times I}{2R}$[/tex]

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

[tex]$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$[/tex]

So,

[tex]$\frac{I_1}{d_1}= \frac{I_2}{d_2}$[/tex]

[tex]$=\frac{16}{21}=\frac{I_2}{32}$[/tex]

[tex]$I_2=24.38 $[/tex] A

Therefore, the current in the outer wire is 24.38 ampere.

Answer:

(a) counter clockwise

(b) 24.38 A

Explanation:

inner diameter, d = 21 cm

inner radius, r = 10.5 cm

Current in inner loop, I = 16 A clock wise

Outer diameter, D = 32 cm

Outer radius, R = 16 cm

(a) The magnetic filed due to the inner wire is inwards to the plane of paper. According to the Maxwell's right hand thumb rule, the direction of magnetic field in outer wire should be outwards so that the net magnetic field is zero at the center.

So, the direction of current in outer wire is counter clock wise in direction.

(b) Let the current in outer wire is I'.

The magnetic field due to the inner wire is balanced by the magnetic field due to the outer wire.

[tex]\frac{ \mu 0}{4\pi}\times \frac{2 I}{r}=\frac{\mu 0}{4\pi}\times \frac{2 I'}{R}\\\frac{16}{10.5}=\frac{I'}{16}\\\\I' = 24.38 A[/tex]

Answer:

B = 1.03 10⁻⁸ T

Explanation:

For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish

This relational is expressed by the relation

           E /B = c

           B = E / c

let's calculate

            B = 3.10 / 3 10⁸

            B = 1.03 10⁻⁸ T

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

Answer:

https://youtu.be/ymHHdoCGJOU

Answer:

B. Community

Took science classes for 6 years now

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

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).

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.

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

Answer:

24m

Explanation:

you can use the formula

s=ut+1/2at²

s=0+1/2(3)(4)²

=1/2(3)(8)

=24m

I hope this helps

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.

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]

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.

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

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]

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I hope it helps you!

Question:- A scientist who studies insects, spiders, snails, and other bugs of an environment

Explanation:-

Entomologist word comes from two words Entomon and biologist

∑ 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.

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Answer:

pendulum, quartz

Explanation:

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

Answer:

53.8Joule

Explanation:

hope it is helpful

Answer:

approximate 153.1J

Explanation:

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

Answer:

effect on motation.effect on direction

Answer:

UY Scuti is slightly larger than VY Canis Majoris

Explanation:

These stars are millions of miles away and cannot be seen by the naked eye.

Beetlejuice is another large star that can be seen by the eye.

Explanation:

kE =1/2mv²

1/2(0.6×(3m/s)²

1/2(0.6×9m/s)

2.7J I think this is the answer

The average kinetic energy of the toy car is 2.7 J.

The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.

Given data;

Mass of car is,m= 600 g = 0.6 kg

d is the distance travelled = 6 m

T is the time travelled = 2 sec

The velocity of the car is found as;

v  = d /t

V = 6m / 2 sec

V = 3 m/sec

KE =1/2mv²

KE = 1/2 × 0.6 kg ×( 3 m/sec )²

KE = 2.7 J

Hence, the average kinetic energy of the toy car is 2.7 J.

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