All categories

3 years ago

Romeo sets off in his spaceship tp marry juliet who lives on the planet lovelon

Physics

1 answer:

Rus_ich [418]3 years ago

4 0

False? I'm not sure what you're asking.

You might be interested in

What is the metric unit for time?

spin [16.1K]

The base unit of time in the metric and SI system is the second.

5 0

4 years ago

The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi

lukranit [14]

Answer:

a. volume of gas: (decreases)

b. temperature of gas: (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

4 0

2 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

Download pdf

7 0

3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago

Different color ca

777dan777 [17]

Answer: The color of candy cane

Explanation:

An independent variable is the Variable that the scientist changed in order to change the dependent variable. The scientist changes the color of the candy cane to test the rate at which they dissolve.

4 0

3 years ago