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2 years ago

7. True or False? The positive effects of a new technology always outweigh its negative effects.

Engineering

2 answers:

Nadusha1986 [10]2 years ago

6 0

Answer:

True

Explanation:

The reason this is true is that technology has a greater good other than just our phones, technology is how we can create anything from new medicine to a new microwave. So the answer is true because technology has more positive effects than negative effects.

stich3 [128]2 years ago

3 0

False, there are negative side affects that have just as much if not more impact. One example of this is the effect on natural habit.

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Which statement most accurately describes Pascal's law?

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Answer:

La C

Explanation:

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3 years ago

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How buy airpods in my phone

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2 years ago

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Tech A says that horsepower is a measurement simply of the amount of work being performed. Tech B says that horsepower can be ca

snow_tiger [21]

Answer:

Tech B

Explanation:

Horsepower (hp) refers to a unit of measurement of power in respect of the output of engines or motors.

Horsepower is the common unit of power. It indicates the rate at which work is done.

The formula $\frac{rpm*T}{5252}$ , where rpm is the engine speed, T is the torque, and 5,252 is radians per second.

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6 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

before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t

Vsevolod [243]

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

5 0

3 years ago