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3 years ago

Which type of load generates a magnetic field?

Engineering

1 answer:

kozerog [31]3 years ago

5 0

Answer:

inductive loads

Explanation:

Inductive: When a changing current passes through an inductor, it induces a magnetic field around itself. Turning the inductor into a coil increases the magnetic field. Examples of inductive loads include transformers, fans, electric motors and coils.

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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

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

Is the COP of a heat pump always larger than 1?

Liono4ka [1.6K]

Answer:

Yes

Explanation:

Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.

We know that

COP of heat pump= 1 + COP of refrigeration

It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.

3 0

2 years ago

The underground cafe has an operating cash flow of $187,000 and a cash flow to creditors of $71,400 for the past year. During th

Serggg [28]

Answer:

cash flow to stockholders = $39,700

Explanation:

Operating cash flow = $187,000

cash flow to creditors = $71,400

Net working capital = $28,000

Net capital spending = $47,900

Cash flow to stockholders = ?

CFF = operating cash flow - net working capital - net capital spending

CFF = $187,000 - $28,000 - $47,900 = $111,100

CFF = cash flow to creditors + cash flow to stockholders

cash flow to stockholders = CFF - cash flow to creditors

cash flow to stockholders = $111,100 - $71,400 = $39,700

Hence $39,700 is the amount of the cash flow to stockholders for the last year.

3 0

3 years ago

Technician A says mismatching tires of the same size on a heavy vehicle will generally not affect ABS operation. Technician B sa

marysya [2.9K]

Technician A is correct

3 0

3 years ago