In physics grade 10 number 8 in the pic how do we do it thanks in advance

What is the only component of scalar quantities?

Bad White [126]

Answer:

Scalar quantities have a size or magnitude only and need no other information to specify them. Thus, 10 cm, 50 sec, 7 litres and 3 kg are all examples of scalar quantities.

Explanation:

7 0

2 years ago

Find the final velocity

Verizon [17]

Answer: v = 4.4 m/s

Explanation:

In the absence of friction, the total mechanical energy will be constant

KE₀ + PE₀ = KE₁ + PE₁

0 + mg(6) = ½mv₁² + mg(5)

½mv₁² = mg(6 - 5)

v = √(2g(1)) = 4.4 m/s

3 0

2 years ago

Boyle’s law states that the volume of a gas is inversely proportional to its pressure if the

Romashka [77]

I believe the answer is A, temperature and number of particles are constant. Correct me if i'm wrong :)

4 0

3 years ago

Read 2 more answers

If a force is 100 N and is pointing 37 degrees north of east. (a) Draw a diagram of this force. (b) Draw the force's x and y com

sasho [114]

Answer

given,

force = 100 N

Point 37 degrees north of east

a) and b) part is shown in the diagram attached below.

c) to find the x and y component of the force

x- component of the force

$F_x = F cos \theta$

$F_x = 100\times cos 37^0$

$F_x = 79.86 N$

y- component of the force

$F_y = F sin \theta$

$F_y = 100\times sin 37^0$

$F_y = 60.18 N$

3 0

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.

PolarNik [594]

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:

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

7 0

3 years ago