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Rom4ik [11]
2 years ago
15

Ohm's law states that the current (I) in amps equals the voltage (E) in volts decided by the resistance (R) in ohm's. If you con

nected a 2 megohm resistor (2•10^6 ohms)across a 2.4 kilovolt voltage source (2.4•10^3 volts ) , what would be the currents amps ?
Engineering
1 answer:
Hitman42 [59]2 years ago
7 0

Answer:

1.14 * 10^-3  amperes

Explanation:

According to ohms law

V = IR   ---- ( 1 )

V = voltage  = 2.4 * 10^3 v

I = current  = ?

R = resistance =  2.1 * 10^6 Ω

from equation 1

I = V / R

 = ( 2.4 * 10^3 ) / (2.1 * 10^6 )

 = 1.14 * 10^-3  amperes

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The Danish scientist Niels Bohr proposed a model of the atom that is not totally accurate today but still contains components th
blagie [28]

Answer:

Option B

Planetary

Explanation:

In 1913, Niels Bohr proposed a model to explain the stability of orbits around the nucleus. Niels believed that light is emitted by an electron when the electron's energy changes. Bohr Atomic Model is sometimes called planetary model because it resembles a mini-solar system.

7 0
3 years ago
B. Is the “Loading Time” of any online application a functional or a non-functional requirement? Can the requirement engineers s
Oksi-84 [34.3K]

Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

Explanation:

non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

4 0
2 years ago
1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt
Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = \frac{width}{period}      ...............1

duty cycle = 1.5 × 10^{-3}

and applied voltage will be

applied voltage = duty energy × voltage    ...........2

applied voltage = 1.5 × 10^{-3} × 2

applied voltage = 3 mV

so current will be

current = \frac{applied\ voltage}{resistance}   ................3

current = \frac{3}{150}

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = \frac{1.5}{18.75*10^{-6}}

battery life = 80000 hours

battery life in year = \frac{80000}{8760}

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

4 0
3 years ago
What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A
ikadub [295]

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

8 0
2 years ago
Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
3 years ago
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