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Anika [276]
2 years ago
6

Some lenses are shaped with one flat side and one spherically-shaped side. This shape is designed to focus parallel light rays o

nto a single point. In a few sentences, explain how the spherical shape of the lens' surfaces causes parallel light rays to focus on a single point. (Assume the light is travelling through air into a lens with an index of refraction greater than that of air.) g
Physics
1 answer:
Komok [63]2 years ago
7 0

Answer:

Some lenses are used to focus light to a pre-defined point based on the amount of curvature of their surfaces.

In a piano design convex, some surfaces are flat while others has positive lenses (biconvex)

Explanation:

Solution

These lenses are applied to pay attention to light in a  point pre-defined  based on the amount of curvature of their surfaces.

For that of a plano-convex design, one surface has a positive curve and for biconvex lenses, both surfaces are positively curved while the other  remains flat.

when used practically, plano-convex lenses are most commonly used where the object being imaged is far apart from lens.

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A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it
Lisa [10]

Answer:

Vertical distance=  3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time=  0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

3 0
2 years ago
Juan was wearing a bright red shirt in a very dark room. What color did his shirt appear to the people with him in the room? A)
ikadub [295]
It would appear black.   

Hope I helped.  
8 0
2 years ago
Read 2 more answers
A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
tankabanditka [31]

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

mass of book = 0.305 Kg

so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

5 0
2 years ago
Which statement is not a good practice when working inside a computer case?
lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

3 0
2 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
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