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rusak2 [61]
3 years ago
8

How dose silicon shape our technological reality?

Physics
1 answer:
vichka [17]3 years ago
3 0
Silicon is used in the manufacturing of microchips and is also used in motherboards. We use it in our daily lives. It is a high tech element. 
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1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

8 0
3 years ago
A magnetic field would be produced by a beam of
kumpel [21]
3 protons should be your answer
4 0
3 years ago
If yall friend me ill give lots of points like this in the future
Ne4ueva [31]

Answer:

ok......................

3 0
2 years ago
A 2011 Porsche 911 Turbo S goes from 0-27 m/s in 2.5 seconds. What is the car's acceleration?
Natalka [10]

Answer:

-10.8m/s^2

Explanation:

a=change in velocity/change in time

-27 m/s/2.5=10.8m/s^2

or if its not negative

27m/s/2.5=10.8m/s^2

3 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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