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goldfiish [28.3K]

3 years ago

15

value of the refractive index of lens is 2.5 The curved surfaces are having The radius of curvature 10 cm and -12cm Out The foca

l length of the lens respectively find out the focal length of the lens?

2 answers:

Hoochie [10]3 years ago

3 0

Answer:

4

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Savatey [412]3 years ago

3 0

Answer:

16.5

Explanation:

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In the early 1900s, it was proposed that the law of conservation of mass should be simultaneously considered with the law of con

svet-max [94.6K]

Answer:

B. After a photon of light is absorbed by certain metals, electrons are found to be ejected from the metals. This is because the energy contained in the massless photon is used to eject an electron with mass out of the metal.

Explanation:

Before, in the early days, it was proposed to form a combined theory by joining the theory of conservation of mass and the theory of conservation of energy and form a combined theory of conservation of mass-energy. It was done to explain a particular theory of $\text{photoelectric effect}$ .

The $\text{photoelectric effect}$ is the emission of the electrons form the surface of a metal when light energy strikes on it. Here, in this phenomenon, both mass and energy is conserved.

When the light strikes a metal surface, electrons gets ejected from the surface. The energy of the photon is used to eject the electron form the metal surface.

7 0

2 years ago

Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

3 0

3 years ago

Your neighbor is riding her bike around the block. When she slows down and turns a corner, what changes about her?.

Ann [662]

Answer:

nothing

Explanation:If you ride a bike around the block and return to the exact point where you started, your displacement is zero.

By definition, displacement involves changing an object from its original position. No matter how far or for how long a body moves, if it returns to the position it started from, it has not been displaced at all. This means that the body has zero displacements.

4 0

2 years ago

A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring

ANEK [815]

Answer:

29856.521 N/m

Explanation:

$m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m$

5 0

4 years ago

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago