uses of concave lens​

Carcinoma refers to a malignant neoplasm of epithelial origin or cancer of the internal or external lining of the body. Carcinomas, malignancies of epithelial tissue, account for 80 to 90 percent of all cancer cases. Epithelial tissue is found throughout the body.

Answer:

omparing the values ​​of the strung density, the one that is closest to carbon is number 2

Explanation:

In this exercise we are given the mass and volume of a body, we are asked to calculate the density, using the equation

          ρ = m / V

in the third and fourth column of the table is the density and the substance with the closest value

#    mass   volume  density   material

        (gr)      (cm³)       (gr/cm³)

1      6.95      4.0          1.74       magnesium

2     4.54      2.0          2.27       carbon

3     5.40      3.0          1.80        phosphorus

4    10.35      5.0         2.07        sulfur

When comparing the values ​​of the strung density, the one that is closest to carbon is number 2

Answer:Well toasting it makes it a solid so that’s a chemical change that happens! Hope this helped!

Answer:

The beaker that is matt black heats more quickly because black attracts more heat.

Explanation:

Three beakers are of identical shape and size, one beaker is pained matt black, one is dull white and one is gloss white . The beakers are filled with boiling water. The water in the matt black beaker cools more quickly.

The rate of cooling of an object is influenced by its surface properties and color. In this case, the matt black beaker will cool the water more quickly compared to the dull white and gloss white beakers.

The reason behind this lies in the concept of thermal radiation. Darker colors, such as matt black, absorb and emit thermal radiation more efficiently than lighter colors. When the boiling water is placed in the matt black beaker, the black surface absorbs a greater amount of thermal radiation from the water and its surroundings. This increased absorption accelerates the transfer of heat energy from the water to the beaker, leading to faster cooling.

On the other hand, the dull white and gloss white beakers reflect more thermal radiation, absorbing and emitting less heat. As a result, the water in these beakers cools at a slower rate compared to the matt black beaker.

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Answer:

The time for the rock to reach the bottom is 4.02 seconds.

Explanation:

Given;

mass of the rock, m = 5 kg

height of the rock fall, h = 79 meters

The time to drop to the given height is given by;

[tex]t = \sqrt{\frac{2h}{g} }[/tex]

where;

t is the time to fall to the bottom

g is acceleration due to gravity = 9.8 m/s²

[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s[/tex]

Therefore, the time for the rock to reach the bottom is 4.02 seconds.

Answer:

24g

Explanation:

mass of the product is 40g.

The mass of the first reactant is 16g

The mass of the second reactant  = ?

  Let the first reactant  = A

   Let the second reactant  = B

  Let the product  = C

Equation of reaction;

           A   +  B    →  C

           16g              40g

The mass of B must be  = 40 -16  = 24g

This question involves the concept of the orbital period.

The period of revolution of the second satellite is "3.95 x 10⁶ s".

First, we will consider the orbital period of the first satellite:

[tex]T_1=\sqrt{\frac{4\pi^2 r_1^3}{GM}}[/tex]

where,

Therefore,

[tex]1\ x\ 10^6\ s=\sqrt{\frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(M)}}\\\\M = \frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1\ x\ 10^6\ s)^2}\\\\M = 3.03\ x\ 10^{20}\ kg[/tex]

Now, we find out the orbital period of the second satellite:

[tex]T_2=\sqrt{\frac{4\pi^2 r_2^3}{GM}}[/tex]

where,

Therefore,

[tex]T_2=\sqrt{\frac{4\pi^2(2\ x\ 10^7\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3.03\ x\ 10^{20}\ kg)}}[/tex]

T₂ = 3.95 x 10⁶ s

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Answer:

I believe its C

Explanation:

Answer:

C-An object weighs about two times as much on Jupiter as on Neptune.

Explanation:

on edg

Answer:

B) 0.500 m/s West

brainlist answer

Answer:

Explanation:

given

Mass of cart= 150kg

mass of each wheel=45kg

mass of 4 wheels= 180kg

angle of the track=  17 ∘

distance of track= 27m

The height  of the tracl is calculated thus:

sin 17° = h / 27

h = sin 17*27

h=7.89m

"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "

1. Potential energy at top= (M+4m)gh

2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²

3. rotational energy of wheels at bottom= 4(1/2 Iω²)

The total is expressed as

(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) -------------1

we know that  I = mr² / 2

Put I= mr² / 2

(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)

(M+4m)gh = 1/2 (M+4m)v² +  m r² ω²

we know that  v²= r² ω²

(M+4m)gh = 1/2 (M+4m)v² +  m v²

(M+4m)gh = v² (M/2 + 2m + m)

(M+4m)gh = v² (M/2 + 3m)

v = √[(M+4m)gh / (3m + M/2)]

v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]

v=√25516.26/142.5

v=√179.06

v = 13.4 m/s

The correct option is A i.e.  Van der Waals forces.

what is Van der Waals forces?

Van der Waals forces are weak intermolecular forces that are depending on atom or molecule distance. Interactions between uncharged atoms/molecules produce these forces.

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Answer:

C

Explanation:

Answer:

x is vertical and y is horizontal

Explanation:

Answer:

Explanation:

Initial kinetic energy of particle

= 1/2 m V²

= .5 x .208 x 1.26²

= .165 J

Work done by force = force x displacement

= .766 x .195

= .149 J

This energy will be added up .

Total final kinetic energy

= initial kinetic energy + work done on the particle

= .165 + .149 J

=  .314 J .

Answer:

(C) 9 mAs

Explanation:

The computation of the new mAs is shown below:

Let us assume the New mAs be x

For adjusting the factors of exposure basically we compared the old one with the new one i.e. presented below:

6 ÷ x = 4 ÷ 6

where

6 = old mAs

4 = old grid factor

x = new mas

6 = new grid factor

Now solve this above equation with the help of cross multiplication

4x = 36

x = 9 mAs

Here we used the 16:1 ratio grid

Hence, the correct option is C. 9 mAs

Answer:

[tex]\mu_s=0.51[/tex]

Explanation:

Coefficient of friction

The static coefficient of friction is a measure of the force needed to start moving an object from rest along a rough surface.

It's calculated as:

[tex]\displaystyle \mu_s=\frac{F_r}{N}[/tex]

Where Fr is the friction force and N is the normal force exerted by the surface over the object.

In the absence of any other vertical force, the normal is equal to the weight of the object:

[tex]N = W = m.g[/tex]

The box has a mass of m=80 Kg, thus the normal force is:

[tex]N = 80\ Kg * 9.8\ m/s^2[/tex]

N = 784 N

The student needs to push with a force of 400 N to make the box move. That is the force that barely outcomes the friction force. Thus:

[tex]F_r=400\ N[/tex]

Calculating the coefficient:

[tex]\displaystyle \mu_s=\frac{400}{784}[/tex]

[tex]\mathbf{\mu_s=0.51}[/tex]

Answer:

Symbol: I unit: A

Explanation:

the formula for current is I = v/r

I: Current

V: voltage

R; Resistance

Symbol and unit of measurement are used for the electric current are symbol: I; unit: A

"Electric Current is the rate of flow of electrons in a conductor. The SI Unit of electric current is the Ampere. Electrons are minute particles that exist within the molecular structure of a substance. Sometimes, these electrons are tightly held, and the other times they are loosely held."

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Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

[tex]E = hv[/tex]

[tex]E = \frac{hc}{\lambda}[/tex]

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

[tex]E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}[/tex]

[tex]E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})[/tex]

E = 4.96 x 10³ eV

B)

The energy of a particle at rest is given as:

[tex]E = m_{0}c^2[/tex]

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

[tex]E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\[/tex]

[tex]E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]

E = 4.19 x 10⁴ eV

C)

The energy of a particle at rest is given as:

[tex]E = m_{0}c^2[/tex]

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

[tex]E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\[/tex]

[tex]E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]

E = 3.73 x 10⁹ eV

A) The energy in electron volts for a particle with this wavelength if the particle is a photon is; .E = 4969.5 eV or 4.9695 keV

B) The energy in electron volts for a particle with this wavelength if the particle is an electron is; E = 23.58 eV

C) E = 0.003306 eV

A) The formula for the energy here is;

E = hc/λ

where;

h is planck's constant = 6.626 × 10⁻³⁴ J.s

c is speed of light = 3 × 10⁸ m/s

λ is wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Thus;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(0.25 x 10⁻⁹)

79.512 × 10⁻¹⁷ J

converting to eV gives;

E = (79.512 × 10⁻¹⁷)/(1.6 × 10⁻¹⁹)

E = 4969.5 eV or 4.9695 keV

B) Formula for the energy if the particle is an electron is;

E = h²/(2mλ²)

where m = 9.31 × 10⁻³¹ kg

E = (6.626 × 10⁻³⁴)²/(2 × 9.31 × 10⁻³¹ × (0.25 x 10⁻⁹)²)

E = 37.726 × 10⁻¹⁹ J

Converting to eV gives;

E = (37.726 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)

E = 23.58 eV

C) Mass of alpha particle is; m = 6.64 × 10⁻²⁷ kg

E = h²/(2mλ²)

where m = 6.64 × 10⁻²⁷ kg

E = (6.626 × 10⁻³⁴)²/(2 × 6.64 × 10⁻²⁷ × (0.25 x 10⁻⁹)²)

E = 52.896 × 10⁻²³ J

Converting to eV gives;

E = (52.896 × 10⁻²³)/(1.6 × 10⁻¹⁹)

E = 0.003306 eV

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Answer:

Explanation:(1) a constitution is a supreme law of the land, (2) a constitution is a framework for government; (3) a constitution is a legitimate way to grant and limit powers of government officials. Constitutional law is distinguished from statutory law.

Answer:

3

Explanation:

Answer:

[tex] \frac{1}{2} \times m \times {v}^{2} \\ \frac{1}{2} \times m \times {2}^{2} \\ 2 \times m[/tex]

I dont the child mass...you should substitute that value to (m)

then you can get your answer

Answer:

Use

[tex]t = \: \sqrt{2h \div g } [/tex]

[tex]x = u \times \sqrt{2h \div g} [/tex]

Answer:

He concluded the fact that most alpha particles went straight through the foil is evidence for the atom being mostly empty space. A small number of alpha particles being deflected at large angles suggested that there is a concentration of positive charge in the atom.

Explanation:

Answer:

1) 0.0806 m/s

2) External forces don't act on the system during the jump

Explanation:

velocity of ball ( Vbf )= 15 m/s

mass of quarter back ( Mq )= 80 kg

mass of football ( Mb ) = 0.43 kg

A) How fast will be be moving backward just after releasing the ball

we can determine this speed with the use of conservation of momentum equation

= Mb ( Vbf - Vbi ) = Mq ( Vq )

where Vq = Mb ( Vbf - Vbi ) / Mq

                 = 0.43 ( 15 m/s - 0 ) / 80 kg

                 = 0.0806 m/s

B) External forces don't act on the system during the jump

Answer:

a) 0.3 m

b) 3.3 diopters

Explanation:

Given that

Object distance is 26cm from the eye.

Image distance is 105 cm

From the above, we will use the formula

1/f = 1/v + 1/u

where

f is the focal length of the lens,

v is the image distance,

u is the object distance.

Since the image is a virtual image, we will give our v a negative sign. So, on calculating, we have

1/f = 1/25 - 1/151

1/f = 0.0333

f = 1/0.0333

f = 30.03 cm

f = 0.3 m

b) The power of the needed corrective lens is the reciprocal of the focal length in metres;

1/0.3 = 3.3 diopters

This is usually in diopters

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

[tex]y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (1)

Where:

[tex]y(t)[/tex] - Position of the mass as a function of time, measured in meters.

[tex]A[/tex] - Amplitude, measured in meters.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]t[/tex] - Time, measured in seconds.

[tex]\phi[/tex] - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

[tex]k = \frac{m\cdot g}{\Delta y}[/tex] (2)

Where:

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]\Delta y[/tex] - Deformation of the spring due to gravity, measured in meters.

If we know that [tex]m=1.65\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta y = 0.260\,m[/tex], then the spring constant is:

[tex]k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}[/tex]

[tex]k = 62.237\,\frac{N}{m}[/tex]

If we know that [tex]A = 0.130\,m[/tex], [tex]k = 62.237\,\frac{N}{m}[/tex], [tex]m=1.65\,kg[/tex], [tex]x(t) = 0\,m[/tex] and [tex]\phi = 0\,rad[/tex], then (1) is reduced into this form:

[tex]0.130\cdot \cos (6.142\cdot t)=0[/tex] (1)

And now we solve for [tex]t[/tex]. Given that cosine is a periodic function, we are only interested in the least value of [tex]t[/tex] such that mass reaches equilibrium position. Then:

[tex]\cos (6.142\cdot t) = 0[/tex]

[tex]6.142\cdot t = \cos^{-1} 0[/tex]

[tex]t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s[/tex]

[tex]t \approx 0.255\,s[/tex]

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

The time taken for the spring to reach new equilibrium position is 0.26 s.

The given parameters;

The general wave equation is given as;

[tex]y(t) = A\ cos(\omega t \ + \phi)[/tex]

The angular frequency is given as follows;

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

The spring constant is calculated as;

[tex]F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{1.65 \times 9.8}{0.26} \\\\k = 62.2 \ N/m[/tex]

The angular frequency is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{62.2}{1.65} }\\\\\omega = 6.14 \ rad/s[/tex]

The time taken for the spring to reach new equilibrium position is calculated as follows;

[tex]y(t) = A \ cos(\omega t)\\\\0 = 0.13 \times cos(6.14t)\\\\6.14t = cos^{-1}(0)\\\\6.14t = 1.57 \ rad\\\\t = \frac{1.57 \ rad}{6.14 \ rad/s} \\\\t = 0.26 \ s[/tex]

Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.

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Answer: ionosphere

Explanation: aurora boreal is is caused by the ionosphere which is a part of the thermosphere

Answer:

centripetal and of constant magnitude

Explanation:

The acceleration of an object traveling on a circular path at constant speed is directed towards the center of the circle (centripetal) and of constant magnitude equal to the square of the object's speed divided by the radius of the circle.

Answer:

A. 7 s ≤ t < 10 s

Explanation:

From Definite Integral Theory and Mechanical Physics we remember that change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:

[tex]A_{1}+A_{2} = 0[/tex] (1)

Where:

[tex]A_{1}[/tex] - Change in position from [tex]t = 0\,s[/tex] and [tex]t = 3\,s[/tex], measured in meters.

[tex]A_{2}[/tex] - Change in position from [tex]t = 5\,s[/tex] and [tex]t = T[/tex], measured in meters.

By definitions of triangle, we expand the equation above:

[tex]\frac{1}{2}\cdot (3\,s)\cdot \left(1.2\,\frac{m}{s} \right) -\frac{1}{2}\cdot (T-5\,s)\cdot v = 0\,m[/tex]

[tex]1.8\,m-0.5\cdot T \cdot v +2.5\cdot v = 0[/tex]

And the time is now cleared:

[tex]1.8\,m+2.5\cdot v = 0.5\cdot T\cdot v[/tex]

[tex]T = \frac{1.8+2.5\cdot v}{0.5\cdot v}[/tex]

Where:

[tex]v[/tex] - Final velocity of the cart, measured in meters per second.

[tex]T[/tex] - Time, measured in seconds.

If we know that [tex]v = 1\,\frac{m}{s}[/tex], then the intended instant is:

[tex]T = \frac{1.8+2.5\cdot (1)}{0.5\cdot (1)}[/tex]

[tex]T = 8.6\,s[/tex]

The value does not coincide with the time from the graph.

If we know that [tex]v = 1.5\,\frac{m}{s}[/tex], then the intended instant is:

[tex]T = \frac{1.8+2.5\cdot (1.5)}{0.5\cdot (1.5)}[/tex]

[tex]T = 7.4\,s[/tex]

This value does not coincide with the time from the graph.

If we know that [tex]v = 1.1\,\frac{m}{s}[/tex], then the intended instant is:

[tex]T = \frac{1.8+2.5\cdot (1.1)}{0.5\cdot (1.1)}[/tex]

[tex]T = 8.273\,s[/tex]

This results offer a reasonable approximation, which that the correct answer is A.

The car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.

The change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:

A + A' = 0 ..................................................(1)

A is the change in position from t = 0 and t = 3 s, measured in metres.

A' is the change in position from t = 5 and t = T s, measured in metres.

As per the triangle law,

1/2 ×(3 s)(1.2) - 1/2 ×(T - 5)v = 0

1.8 - 0.5T(v) +2.5(v) =0

T = (1.8 + 2.5v)/ (0.5v)

Here, v is the final velocity of cart.

For v = 1 m/s, we have,

T = (1.8 + 2.5(1))/ (0.5(1))

T = 8.6 s

Since, the values do not coincide with the time from the graph, so taking another value, v = 1.5 m/s to obtain the time as,

T = (1.8 + 2.5(1.5))/ (0.5(1.5))

T = 7.4 s

This value does not coincide with the time from the graph.  If we know that , v = 1.1 m/s, then the intended instant is:

T = (1.8 + 2.5(1.1))/ (0.5(1.1))

T = 8.273 s

Thus, we can conclude that the car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.

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Answer:

High and low tides are caused by the Moon. The Moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the Moon and the side farthest from the Moon. These bulges of water are high tides.

Explanation: