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3 years ago

When an electrical current develops between two dissimilar types of metal in the cooling system, this is called

Engineering

1 answer:

elena-s [515]3 years ago

7 0

Answer:

Galvanic corrosion is caused by the contact of two dissimilar metals.

Explanation:

The driving force for the corrosion is the potential difference that develops between the two metals.

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What are the pros and cons of cooking with each type of heat source (solar vs fossil fuel)

aleksley [76]

Answer: battle between solar energy vs. fossil fuels, it might seem like the predominant resources on which the global economy depends oil, coal and natural gas will be completely phased out of existence in 2019. In reality, these resources still power most of the planet, while renewable resources like solar and wind only contribute some two to three percent of global energy capacity.

Explanation: this is all I know,but I'm not really smart, but I've tried my best to help out you guys

8 0

3 years ago

There is no charge at the upper terminal of the ele- ment in Fig. 1.5 for t 6 0. At t = 0 a current of 125e-2500t mA enters the

sergiy2304 [10]

Answer:

A) The charge accumulated on upper plate for t>0 is

$q(t)=50[1-e^{-2500t}]\mu C$

B) The total charge that accumulates at the upper terminal is 50μC

C) If the current is stopped at t = 0.5 ms then total charge stored on upper terminal is 35.67μC

Explanation:

Given that:

$I(t)=0 \quad \quad \quad \quad \quad t$

A) The charge that accumulates at the upper terminal for t > 0:

As we know

$q(t)=\int {I(t)} \, dt$

for t > 0

$q(t)=\int\limits^t_0 {I(t)} \, dt\\q(t)=\int\limits^t_0 {125e^{-2500t} mA} \, dt\\q(t)=(125\times 10^{-3})[\frac{e^{-2500t}}{-2500}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}+1]\\\\$

The charge accumulated on upper plate for t>0 is

$q(t)=50[1-e^{-2500t}]\mu C---(1)$

B) The total charge that accumulates at the upper terminal can be found by substituting t → ∞ in equation (1)

$q(t)=( 50\times 10^{-6})[1-e^{-2500(\infty)}]\\q(t)=(50\times 10^{-6})[1-0]\\q(t) =50\mu C$

C) If the current is stopped at t = 0.5 ms then

$q(t)=( 50\times 10^{-6})[1-e^{-2500(0.5\times10^{-3})}]\\q(0.5ms)=35.67\mu C$

7 0

4 years ago

Which statement is true about what will happen when the example code runs?1: main PROC2: mov edx,03: mov eax,404: push eax5: cal

saw5 [17]

Answer:

The last option is the correct answer. EDX will equal 40 on line 6

Explanation:

An element is pushed into stack at line no. 04 and call instruction at line no. 05 pushes a return address.

The return addrees is the instruction address followed by call function.In this scenario it is the address of line 6.

Now, the subroutine pops the return address from the stack,i,e. it pops 40 from the stack into edx and then puts the return address back in the stack, and returns.

Therefore, edx contains the value 40 when it returns from subroutine,

7 0

3 years ago

A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 kpsi. The bar is subjected to a steady torsio

Lerok [7]

Answer:

Factor of safety against static failure = 2.097

Factor of safety against fatique failure = 1.6

Explanation:

Bending stress = 12 kpsi

Se = 40 kpsi,

Sy = 60 kpsi,

Sut = 80 kpsi

See the attached file for explanation

6 0

3 years ago

The following is a list of metals and alloys:

viktelen [127]

Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

5 0

3 years ago