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3 years ago

You need to design a photodetector that can respond to the entire range of visible light. True or False

Physics

1 answer:

Rasek [7]3 years ago

7 0

Answer: True

Explanation:

A photo detector that can respond to the entire rang of visible light can be design, it is true.

Photo detector is a device in an optical receiver which receives optical signals and convert it to electric signal. It is the key device position in front of the optical receiver.

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platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin

posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

$a=g=9.8 m/s^2$ is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s$

For the diver jumping from 10 m, s = 10 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s$

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

$s=ut+\frac{1}{2}at^2$

And since u = 0, it can be reduced to

$s=\frac{1}{2}at^2$

For the diver jumping from 5 m, s = 5 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s$

For the diver jumping from 10 m, s = 10 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s$

5 0

3 years ago

Which formation is one feature of karst topography?

steposvetlana [31]

Karst is a topography formed from the dissolution of soluble rocks such as limestone, dolomite, and gypsum. It is characterized by underground drainage systems with sinkholes and caves. It has also been documented for more weathering- resistant rocks, such as quartzite, given the right conditions.

8 0

3 years ago

Read 2 more answers

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

3 years ago

What is the speed of the animal from 0-20s? 50 m/s 0.4 m/s 20 m/s 2.5 m/s

Shkiper50 [21]

Answer:

2.5 m/s

Explanation:

The speed of the animal is given by the ratio between the distance travelled by the animal and the time elapsed:

$v=\frac{d}{t}$

where d is the distance travelled and t the time elapsed. Note that this quantity is also equal to the slope of the curve.

In the time interval 0-20 s, we have

d = 50 m - 0 m = 50 m

t = 20 s - 0 s = 20 s

So, the speed is

$v=\frac{50 m}{20 s}=2.5 m/s$

3 0

3 years ago

By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl

drek231 [11]

The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.

This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.

The more the distance between the charges, the less is the Electric Force.

The lesser the distance between the charges, the more is the Electric Force.

If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

Hence, the charge present determines a force to be attractive or repulsive.

Learn more about Coulomb Force here, brainly.com/question/15451944

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6 0

2 years ago