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3 years ago

A person finds it easier to lift an anchor through the water than lift it into the boat through the air. Why?

Physics

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

because underwater a force called buoyant force of upthrust applies in the objects in water in the upward direction

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A sound wave has a frequency of 425 Hz. What is the period of this wave?

Serhud [2]

Period is 1/frequency
1/425 = 2.353ms

8 0

3 years ago

Calculate the wavelength of a photon having 3.26 x 10^-19 joules of energy

bekas [8.4K]

The energy of a photon is given by:
$E=hf$
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon, $E=3.26 \cdot 10^{-19} J$ , so we can rearrange the equation to calculate the frequency of the photon:
$f= \frac{E}{h}= \frac{3.26 \cdot 10^{-19}J}{6.6 \cdot 10^{-34}Js}=4.94 \cdot 10^{14}Hz$

And now we can use the following relationship between frequency f, wavelength $\lambda$ and speed of light c to find the wavelength of the photon:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{4.94 \cdot 10^{14} Hz}=6.07\cdot 10^{-7} m=607 nm$

8 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

Which of the following changes would make a heat engine less efficient

DIA [1.3K]

A heat engine would be less efficient due to many factors
For instance, a heat engine is more efficient when it uses in cold weather because there is a greater temperature difference ( Carnot Efficient )
A heat engine could be less efficient because of friction
Hope it helps I am a beginner

8 0

3 years ago

You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j

Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

$E= \frac{1}{2} kx^2$