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3 years ago

What are the prefixes for 1, 10, 1000, 1,000,000, .1, .01, .001, .000001

Engineering

1 answer:

Volgvan3 years ago

5 0

Prefix Symbol Multiplier Exponential

yotta Y 1,000,000,000,000,000,000,000,000 1024

zetta Z 1,000,000,000,000,000,000,000 1021

exa E 1,000,000,000,000,000,000 1018

peta P 1,000,000,000,000,000 1015

tera T 1,000,000,000,000 1012

giga G 1,000,000,000 109

mega M 1,000,000 106

kilo k 1,000 103

hecto h 100 102

deca da 10 101

1 100

deci d 0.1 10¯1

centi c 0.01 10¯2

milli m 0.001 10¯3

micro µ 0.000001 10¯6

nano n 0.000000001 10¯9

pico p 0.000000000001 10¯12

femto f 0.000000000000001 10¯15

atto a 0.000000000000000001 10¯18

zepto z 0.000000000000000000001 10¯21

yocto y 0.000000000000000000000001 10¯24

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Explanation:

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3 years ago

An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n= 1200

Sample mean : $\overline{x}=2.45$

Standard deviation : $\sigma=0.07$

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, $z=\dfrac{x-\mu}{\sigma}$ , the z-value corresponds to 2.39 will be :-

$z=\dfrac{2.39-2.45}{0.07}\approx-0.86$

z-value corresponds to 2.60 will be :-

$z=\dfrac{2.60-2.45}{0.07}\approx2.14$

Using the standard normal table for z, we have

P-value = $P(-0.86$

$=P(z$

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0

3 years ago

An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610

KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + $\frac{q}{h\pi Di}$

Ts = $30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000} }$

$Ts=30+\frac{294}{0.3928} \\$

$Ts =30+748.4\\Ts = 778.4$

The surface temperature is therefore 778.4°C if the cable is bare

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3 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

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3 years ago

An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred b

irinina [24]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Download docx

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3 years ago