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LUCKY_DIMON [66]

2 years ago

5

By adding "-once", one can form the noun form of the word "organize" is that true or false?

2 answers:

garri49 [273]2 years ago

8 0

Answer:

False, mark brainlist?

Explanation:

denpristay [2]2 years ago

7 0

Answer:

<h2>False </h2>

Explanation:

The noun form of organize is just adding letter r

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A piston-cylinder apparatus has a piston of mass 2kg and diameterof

iragen [17]

Answer:

M =2.33 kg

Explanation:

given data:

mass of piston - 2kg

diameter of piston is 10 cm

height of water 30 cm

atmospheric pressure 101 kPa

water temperature = 50°C

Density of water at 50 degree celcius is 988kg/m^3

volume of cylinder is $V = A \times h$

$= \pi r^2 \times h$

$= \pi 0.05^2\times 0.3$

mass of available in the given container is

$M = V\times d$

$= volume \times density$

$= \pi 0.05^2\times 0.3 \times 988$

M =2.33 kg

6 0

3 years ago

10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20

vazorg [7]

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

7 0

2 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$

$\Delta W = 2.043\times 10^{9}\,J$

$\Delta W = 2.043\times 10^{6}\,kJ$

The change in chemical energy from gasoline is:

$\Delta E = \frac{\Delta W}{\eta}$

$\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}$

$\Delta E = 6.81\times 10^{6}\,kJ$

The changes in gasoline consumption is:

$\Delta m = \frac{\Delta E}{L_{c}}$

$\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }$

$\Delta m = 154.772\,kg$

$\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }$

$\Delta V = 209.151\,L$

Lastly, the money saved is:

$\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )$

$\Delta C = 217.517\,USD$

4 0

3 years ago

People with skills and training in areas such as marketing or accounting are an important part of the manufacturing industry.

adelina 88 [10]

Answer:

true

Explanation:

8 0

2 years ago

Technologies that allow for instant worldwide communication include

hram777 [196]

Answer:

mobile phones and internet access

Explanation:

I got it right on my quiz

7 0

2 years ago

Read 2 more answers