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3 years ago

The first transistor, made of gold, plastic, and germanium (a metallic

Physics

1 answer:

Citrus2011 [14]3 years ago

3 0

The first transistor made of gold, plastic and germanium was about the size of adult's fingernail.

Option C

<u>Explanation:</u>

The first transistor made of gold, plastic and germanium was invented in Bell laboratories. It is termed as point contact transistor. As it is made like a pointed arrow with both the sides covered with layer of gold foil. The germanium is used at the tip, just like the base and the gold foil ends as collector and emitter.

The size of this transistor is about the size of adult's fingernail. It is very small in size and it was one of its kind. Due to this small size and the working capacity by the point contact, it is termed as point contact transistor.

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calculate the force between two objects that have masses of 20 kg and 100 kg separated by a distance of 2.6 m

olga55 [171]

The gravitational force between the objects is $1.97\cdot 10^{-8} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between the objects

For the two objects in this problem:

$m_1 = 20 kg\\m_2 = 100 kg$

And their distance is

r = 2.6 m

So, the gravitational force between them is

$F=\frac{(6.67\cdot 10^{-11})(20)(100)}{2.6^2}=1.97\cdot 10^{-8} N$

Learn more about gravitational force:

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brainly.com/question/12785992

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3 years ago

A football player kicks a field goal from a distance of 45 m from the goalpost. The football is launched at a 35° angle above th

daser333 [38]

Answer: try searching It up on go0gle

Explanation:

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2 years ago

Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar ( $a_{D}$ ), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities ( $v_{D}$ , $\omega_{BD}$ ), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations ( $a_{D}$ , $\alpha_{BD}$ ), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$ (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$ (2)

<h3>Accelerations</h3>

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$ (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$ (4)

If we know that $\theta = 60^{\circ}$ , $\gamma = 19.889^{\circ}$ , $r_{BD} = 10\,in$ , $\omega_{AB} = 16\,\frac{rad}{s}$ , $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$ , then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

$v_{D}+3.402\cdot \omega_{BD} = -41.569$ (1)

$9.404\cdot \omega_{BD} = -24$ (2)

$v_{D} = -32.887\,\frac{in}{s}$ , $\omega_{BD} = -2.552\,\frac{rad}{s}$

<h3>Accelerations</h3>

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$ (3)

$-9.404\cdot \alpha_{BD} = -687.264$ (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$ , $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

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2 years ago

What is the distance covered by a Freely falling object 5 seconds after being dropped ? After 6 seconds?

mario62 [17]

This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:

D = 1/2 A T²

Distance = (1/2) (acceleration) (time²)

The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)

The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .

In 5 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (5 sec)²

D = (4.9 m/s²) (25 sec²)

D = 122.5 meters

In 6 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (6 sec)²

D = (4.9 m/s²) (36 sec²)

D = 176 meters

5 0

3 years ago

How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires

Alina [70]

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

t is the time

Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs = 1,000,000/3600

1,000,000secs = 277.78 hours

Hence it will take 277.78 hours for the charge to flow through the diode

6 0

3 years ago