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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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Need help with 1 A and 1 B

Bingel [31]

I hate this class .....

8 0

3 years ago

A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc

blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0

3 years ago

Read 2 more answers

A wedge with a mechanical advantage of 0.78 is used to raise a house corner from its foundation. If the output force is 7500 N,

Zigmanuir [339]

Answer:

5850

Explanation:

MA = Force out/ Force in

.78 = out/ 7500

.78 times 7500 = input

6 0

2 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

7 0

3 years ago

How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?

DiKsa [7]

<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>

4 0

3 years ago

Read 2 more answers