Ask question

Ask question

All categories

3 years ago

8

Can you answer the question in the picture

1 answer:

adelina 88 [10]3 years ago

3 0

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. A free-body diagram is a special example of the vector diagrams that were discussed in an earlier unit. These diagrams will be used throughout our study of physics.

You might be interested in

Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass fl

amid [387]

It appears that your answer contains either a link or inappropriate words. Please correct and submit again! error

Had to screenshot the solution check attached

5 0

4 years ago

1. A car engine transfers 3000J in 20 seconds. What is the power generated by the engine?

Scorpion4ik [409]

Answer:

150 W

Explanation:

Power is energy per unit time. The power generated by the engine is ...

(3000 J)/(20 s) = 150 J/s = 150 W

The engine generated 150 watts.

3 0

3 years ago

PLZ HELP ME!! Can someone plz give me a summary on engineering in 3-5 sentences.

myrzilka [38]

Engineering is about building things, learning how to build things and it being fun

4 0

3 years ago

Consider the following set of processes, with the length of the CPU burst given in milliseconds:Process Burst Time PriorityP1 2

Iteru [2.4K]

Answer and Explanation:

The answer is attached below

6 0

3 years ago

Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per

Cloud [144]

Answer:

a) $Re_{D} = 111896.745$ , b) $Re_{D} = 1.119\times 10^{-7}$

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

$Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}$

Let assume that density and dynamic viscosity at 25 °C are $997\,\frac{kg}{m^{3}}$ $0.891\times 10^{-3}\,\frac{kg}{m\cdot s}$ , respectively. Then:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 111896.745$

b) The result is:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 1.119\times 10^{-7}$

6 0

4 years ago