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3 years ago

An integrated circuit is a ___.

1 answer:

5 0

HI buddy!

An integrated circuit(IC) is a semiconductor water on which thousands of tiny resistors, transistors are fabricated.

Remember that an integrated circuit are the heart and brains of most circuits. They are the keystone of modern electronics.

I hope this helps!

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Which of the following is the correct relationship among the acceleration, velocity, and position?

Damm [24]

Answer:1

Explanation:

s=displacement

v=velocity of particle

a=acceleration of particle

acceleration can be written as rate of change of velocity

so $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

multiply and divide by $ds$

$a=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{ds}{ds}$

$a=\frac{\mathrm{d} v}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}$

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ads=vdv$

option 1 is correct

5 0

3 years ago

Read 2 more answers

Which statement best describes the difference between acceleration and velocity?

Sloan [31]

C. is the answer because acceleration is the change in velocity in time while velocity is speed with a direction

8 0

3 years ago

The deepest well from which water can be pumped, by creating a vacuum on the top end of the pipe

Ivenika [448]

Can you further elaborate this isn't making much sense my mans

7 0

3 years ago

A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo

cestrela7 [59]

Answer:

Volume of balloon = 1000 cm^3

Explanation:

The head of a normal person can be assumed as a sphere with radius 10 cm.

Volume of sphere $=\frac{4}{3} \pi r^3$ , where r is the radius.

We have approximate radius = 10 cm.

Approximate volume of head $=\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3$

In the given options the closest value to the approximate volume is 1000 cm^3.

So, volume of head = Volume of balloon = 1000 cm^3

4 0

3 years ago

A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.

zzz [600]

Answer:

0.0257259766982 m

Explanation:

$P_2$ = Atmospheric pressure = 101325 Pa

$d_1$ = Initial diameter = 1.5 cm

$d_2$ = Final diameter

$\rho$ = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

$P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa$

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m$

The diameter of the bubble is 0.0257259766982 m

8 0

3 years ago