All categories

3 years ago

How would you increase the size of the base unit of length in the metric system

1 answer:

4 0

The size of metric units increases tenfold as you go up the metric scale. The decimal system works the same way: a tenth is 10 times larger than a hundredth; a hundredth is 10 times larger than a thousandth, etc.

You might be interested in

Use the image below to measure the blocks. Enter your measurement in centimeters accurate to the 0.1 cm.

SOVA2 [1]

Answer:

Pink block = 1.4 cm

Green block= 1.5 cm

Blue block= 1.7 cm

Explanation:

Hi, the ruler is in centimeters, from left to right each long line ( the ones with the number ) is a centimeter.

Since 10 millimeter equals 1 centimeter each short line is a millimeter (0.1 cm)

Knowing this, the measures of each block are.

Pink block = 1.4 cm

Green block= 1.5 cm

Blue block= 1.7 cm

Feel free to ask for more if needed or if you did not understand something.

6 0

4 years ago

Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at

andriy [413]

Answer:

A) $v_2 = 2016.80 ft/s$

B) $\Delta s = 0.006 Btu/lbm R$

Explanation:

Given data:

P-1 = 100 lbf/in^2

$T_1 = 500$ degree f

$V_1 = 100 ft/s$

$P_2 = 40 lbf/inc^2$

effeciency = 80%

from steady flow enerfy equation

$h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

$H_1 = 1278.8 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

for P1 = 40 lbf/in^2

$H_1 = 1193.5 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

exit enthalapy h_2

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

$0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm$

from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$ [1 Btu/lbm = 25037 ft^2/s^2]

$v_2 = 2016.80 ft/s$

b) amount of entropy

$\Delta s = s_2 - s_1$

$s_1 = 1.708 Btu/lbm -R$

at $h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2$

$s_2 is 1.714 Btu/lbm -R$

$\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R$

6 0

3 years ago

Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat

Len [333]

Answer:

my_mul:

.globl my_mul

my_mul:

//Multiply X0 and X1

// Does not handle negative X1!

// Note : This is an in efficient way to multipy!

SUB SP, SP, 16 //make room for X19 on the stack

STUR X19, [SP, 0] //push X19

ADD X19, X1, XZR //set X19 equal to X1

ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

CBZ X19, mult_eol

ADD X9, X9, X0

SUB X19, X19, 1

B mult_loop

mult_eol:

LDUR X19, [SP, 0]

ADD X0, X9, XZR // Move X9 to X0 to return

ADD SP, SP, 16 // reset the stack

BR X30

Explanation:

6 0

4 years ago

Which of these material properties determines how much heat you must supply to an object to raise it by a given temperature diff

Finger [1]

Answer:

specific heat

Explanation:

7 0

3 years ago

An inventor proposes an engine that operates between the 27 deg C warm surface layer of the ocean and a 10 deg C layer a few met

SpyIntel [72]

Answer:

Engine not possible

Explanation:

source temperature T1 = 300 K

sink temperature T2= 283 K

therefore, carnot efficiency of the heat engine

η= 1- T_2/T_1

$\eta= 1-\frac{T_1}{T_2}$

$\eta= 1-\frac{283}{300}$

= 0.0566

= 5.66%

claims of work produce W = 100 kW, mass flow rate = 20 kg/s

$Q=mc_p(T_1-T_2)$

$Q=20\times4.18(300-283)$

= 1421.2 kW

now $\eta= \frac{W}Q}$

now $\eta= \frac{100}{1421.2}$

=7%

clearly, efficiency is greater than carnot efficiency hence the engine is not possible.

8 0

4 years ago