Se tiene una esfera de 100g, ¿Qué espacio ocupa?<br><br> La necesito, por favor!!!

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

2 years ago

Which scientific law is also known as the law of inertia? PLZ HELP ASAPPPPPPPPP!!!

Flauer [41]

Answer:

Newton's First Law of Motion

Explanation:

Without external forces acting on an object, the object tends to move at constant speed in a straight line. This property is referred to as inertia. Newton's first law states this natural observation.

4 0

3 years ago

Read 2 more answers

How does temperature change affect surface tension

svlad2 [7]

When the temperature increases, the intermolecular forces between the molecules of a liquid become weaker, and some bonds break easily. Thus as temperature increases, the surface tension of a liquid decreases.

7 0

2 years ago

What is the approximate weight of a 20-kg cannonball on earth?

disa [49]

The weight of an object is taken to be the force on the object due to gravity. The weight ( W ) is the product of the mass ( m ) of the object and the magnitude of the gravitational acceleration ( g ).
On Earth: g = 9.81 m/s²
m = 20 kg
W = m · g = 20 kg · 9.81 m/s² = 196.2 N

6 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

2 years ago

Se tiene una esfera de 100g, ¿Qué espacio ocupa? La necesito, por favor!!!