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2 years ago

What would be the required voltage of an energy source in a circuit with a current of 10.0 A and a resistance of 11.0 Ω?

Engineering

1 answer:

Tju [1.3M]2 years ago

8 0

Answer:

110 V

Explanation:

V = IR

V = (10.0 A)(11.0 Ω) = 110 volts

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The parts of a feature control frame are the tolerance value, the datum references, and the

Elan Coil [88]

Answer:

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3 years ago

An air conditioning system is to be filled from a rigid container that initially contains 5 kg of saturated liquid at 24° Celsiu

gtnhenbr [62]

And air-conditioning system is to be filled for my ridge the containerBut that internally contains 5 kgDetermine the final quality of the arm 134

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2 years ago

On the position time curve, if the slope of a tangent at a point is positive, that means:

Gnoma [55]

Answer:

C. the object is moving forward

Explanation:

A positive slope means position is increasing when time is increasing. Generally, increasing position is "moving forward."

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3 years ago

Read 2 more answers

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

3 years ago

A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k

goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = $\frac{ln\frac{r2}{r1}}{2 \pi *k * L}$ .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

heat loss is change in temperature divide thermal resistance

Q = $\frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}$

Q = $\frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }$

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0

3 years ago