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igor_vitrenko [27]
3 years ago
7

Why does the lens need to be thicker for viewing nearby objects?

Physics
1 answer:
Maurinko [17]3 years ago
5 0

Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.

Explanation:

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
Trace the path of a ray emitted from the tip of the object toward the focal point of the mirror and then the reflected ray that
Crazy boy [7]

Answer: find the attached files for the answer

Explanation:

The reflected ray appears to have originated from the focal point. We should actually draw a vector from the focal point through the point where the incident ray hits the mirror but we shorten the vector so that its starting point is on the mirror, without changing its angle.

Please find the attached files for the solution

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3 years ago
if a ball with an original velocity of 0 is dropped from a tall structure and takes 7 Seconds to hit the ground what velocity do
krok68 [10]

a_y=\dfrac{v_y-v_{0y}}t\implies-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-0}{7\,\mathrm s}\implies v_y=-68.7\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
Brainliest if right
mixas84 [53]
They traveling at -0.37/ms^
3 0
3 years ago
Equation for pressure at a depth H inside a fluid PLSS URGENTT
Blababa [14]

Answer:

We begin by solving the equation P = hρg for depth h: h=Pρg h = P ρ g . Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.

4 0
3 years ago
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