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UNO [17]
1 year ago
5

What is the value of the equivalent resistance for the three resistors connected in series?

Physics
1 answer:
Harman [31]1 year ago
7 0

The value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

To find the answer, we have to know more about the equivalent resistance.

<h3>What is meant by equivalent resistance?</h3>
  • equivalent resistance is the total value of the resistance connected in a circuit.
  • If n resistors are connected in series, then the equivalent resistance will be,

                R_E=R_1+R_2+..........+R_n

  • In our question we have three resistors. Thus, the equivalent resistance will be,

               R_E=R_1+R_2+R_3

Thus, we can conclude that, the value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

Learn more about the equivalent resistance here:

brainly.com/question/11603204

#SPJ4

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Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0
3 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
How can you verify the archimedes principle?​
Ipatiy [6.2K]

Answer:

It is found that W1 - W2 loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes' principle.

6 0
3 years ago
With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons
Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0
3 years ago
A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample
Musya8 [376]

Answer:

remain 1s60

Explanation:

I took away sample

7 0
2 years ago
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