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3 years ago

What is approximate resistance between P and Q? a) 0.5 b)0.8 c)2.0 d)2.2 e)3.6

Physics

1 answer:

Rufina [12.5K]3 years ago

8 0

Answer:

0.5 Ohms

Explanation:

We note that the node Q is also between the resistors of 1ohm and 2ohms.

We note that the node P is also between the resistors of 2ohms and 3ohms.

Thus, all these resistors are in parallel, beween nodes P and Q

1/Re=1/R1+1/R2+1/R3

1/Re=1/1+1/2+1/3=(6+3+2)/6=11/6 [ohm^(-1)]

Re=6/11=0.54ohms

Rounding to the tenth: Re=0.5 ohms

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Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

From the left we have:

r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

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3 years ago

Calculate the pressure of water in a will if the deep of the water is 10 m

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Answer:

98,000 pa

Explanation:

The formula for water pressure is as follows:

$pressure = pgh$

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The density of water is 1000kg/m3, the gravitational field strength is 9.8, and the height is 10. Substituting in these values:

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2 years ago

Brainliest if correct Question 8 of 10

kramer

Answer:

Explanation:

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2 years ago

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Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo

kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

$W=F*d*cos(\theta)$

here we have the work done by Paul and the friction force, so:

$W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J$

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Now the change of energy is:

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3 years ago

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A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.

Galina-37 [17]

Answer:

Change in specific internal Energy $=250\ \rm Btu/lb$

Explanation:

Given:

Mass of the gas, m=0.4 lb
Initial pressure and volume are $p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\$

Final pressure and temperature are $p_1=480\ \rm lbf/in^2$
Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

$p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\$

So the final volume of the gas is calculated.

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$w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\$

According to the first law of thermodynamics we have

$Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\$

So the Specific Internal Change is given by

$\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb$

So the specific Change in Internal energy is calculated.

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3 years ago