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3 years ago

At what value of angle between two vectors will the resultant of the two vectors be maximum?

Physics

1 answer:

yan [13]3 years ago

3 0

The resultant is maximum when the angle between the two vectors is zero

Explanation:

The formula to calculate the resultant of two vectors A, B is the following:

$R=\sqrt{A^2+B^2+2ABcos \theta}$

where

A is the magnitude of vector A

B is the magnitude of vector B

$\theta$ is the angle between the directions of A and B

This formula can be derived by applying the parallelogram law.

From the formula, we observe that:

- When $\theta=0^{\circ}$ (vectors A and B parallel), $cos \theta = 1$ . so the resultant becomes simply

$R=\sqrt{A^2+B^2 + 2AB}=\sqrt{(A+B)^2}=A+B$

- When $\theta = 90^{\circ}$ (vectors A and B perpendicular), $cos \theta = 0$ , so the resultant becomes

$R=\sqrt{A^2+B^2}$

- When $\theta=180^{\circ}$ (vectors A and B anti-parallel), $cos \theta = -1$ , so the resultant becomes

$R=\sqrt{A^2+B^2 - 2AB}=\sqrt{(A-B)^2}=|A-B|$

By looking at the three cases and at the formula, we see that the maximum value of the resultant is when the angle between the two vectors is zero, since in that case the resultant is simply obtained by adding the magnitudes of the two vectors.

Learn more about vector addition:

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What is the momentum of a 750kg car traculing at a velocity of 24m/s

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3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

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Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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