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boyakko [2]
3 years ago
12

Is a light sensitive screen​

Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

The retina is the light sensitive focusing screen. Inside, there's two types of fluid, one watery behind the lens, and further back, a thicker one that helps the eyeball hold its shape.

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In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init
liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,  

n  = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,  

U = − 0.061 J

The charge on an electron,  

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,  

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:  

q ₁ = n q

   = 1 ×10 ¹³ *  − 1.6 × 10 ⁻¹⁹

   = -1.6  × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6  × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and  q ₂ are the two charges.

r  is the distance between the charge and the point.

K  =  8.98755  ×  10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 =  (8.98755  ×  10 ⁹ * (-1.6  × 10⁻⁶)²) / r

r = (18.41 ×  10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0
3 years ago
Help me please!!!!!!!!!
beks73 [17]

limiting frictionl force

6 0
3 years ago
Calculate the kinetic energy of a 2kg ball moving at 5m/s
shepuryov [24]

Answer:

25

Explanation:

The kinetic energy is 25

8 0
3 years ago
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
Read 2 more answers
A damped LC circuit consists of a 0.17 μF capacitor and a 15 mH inductor with resistance 1.4Ω. How many cycles will the circuit
Dafna1 [17]

Answer:

number of cycles = 4.68 × 10⁴ cycles

Explanation:

     In damped RLC oscillation

voltage (V(t)) = V_o\ e^{-\dfrac{tR}{2L}}............(1)

given,

C = 0.17μF = 0.17 × 10⁻⁶ F

R = 1.4 Ω

L = 15 m H = 15 × 10⁻³ H              V(t) = V₀/2

From the equation (1)

\dfrac{V_0}{2} = V_0\ e^{-\dfrac{tR}{2L}}

 2 = e^{\dfrac{tR}{2L}}

taking log both side

ln ( 2 ) = \dfrac{tR}{2L}

t = \dfrac{2 L ln(2)}{R}

t = \dfrac{2 \times 15 ln(2)}{1.4}

t = 14.85 sec

time period

T= 2\pi \sqrt{LC}

T= 2\pi \sqrt{0.015 \times 0.17 \times 10^{-6}}

T = 3.172 × 10⁻⁴

number of cycle =\dfrac{t}{T}

                           =  \dfrac{14.85}{3.172 \times 10^{-4}}

  number of cycles = 4.68 × 10⁴ cycles

8 0
3 years ago
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