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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 ?

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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Volume of sale (i.e., the number of parts sold) is a factorwhen determining which

nadya68 [22]

Answer: N has to be lesser than or equal to 1666.

Explanation:

Cost of parts N in FPGA = $15N

Cost of parts N in gate array = $3N + $20000

Cost of parts N in standard cell = $1N + $100000

So,

15N < 3N + 20000 lets say this is equation 1

(cost of FPGA lesser than that of gate array)

Also. 15N < 1N + 100000 lets say this is equation 2

(cost of FPGA lesser than that of standardcell)

Now

From equation 1

12N < 20000

N < 1666.67

From equation 2

14N < 100000

N < 7142.85

AT the same time, Both conditions must hold true

So N <= 1666 (Since N has to be an integer)

N has to be lesser than or equal to 1666.

3 0

3 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

What is the difference between filler and electrode in Welding? Can a filler be an electrode? Can an electrode be a filler? Why?

Vlada [557]

Explanation:

<u>Filler:</u>

Filler is the material rod is used when we are joining two material by using welding process.If thickness of work piece is more so it will become compulsory to provide some filler material for making the welding join to withstand high stresses.

<u>Electrode:</u>

Electrode is the element which is used to complete the electric circuit in welding .Some time electrode is connected with positive terminal and some time with negative terminal ,it depends on the requirement of welding process.In Tungsten inert gas welding electrode is connected negative terminal but on the other hand Metal inert gas welding electrode is connected with positive terminal.Electrode can be consumable non-consumable depends on the condition.

Yes electrode can be work as filler material ,in Metal inert gas welding wire is used as electrode as well as filler material.In Metal inert gas welding consumable electrode is used on the other hand Tungsten inert gas welding non-consumable electrode is used.In Tungsten inert gas welding if thickness of work pieces is less than 5 mm then no need to used any filler material but if thickness is more than 5 mm then we have to use filler material.

8 0

3 years ago

Military glorification through mosaics, relief carvings and triumphant arches are often associated with?

emmasim [6.3K]

Military glorification through mosaics, relief carvings and triumphant arches are often associated with roman architectural monument and is the correct choice.

<h3>What is a Monument?</h3>

This can be defined as a structure which is usually large and is used to commemorate the history of an influential person. This helps to serve as an example and also a reminder as to why the performance of good deeds are important

The use of military glorification through mosaics, relief carvings and triumphant arches were also often used by the Romans in other to commemorate war victories and the succession of a new ruler which is known as the emperor.

Read more about Roman monuments here brainly.com/question/15786258

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8 0

2 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago