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4 years ago

Which process is represented by the PV diagram?

Physics

1 answer:

Reika [66]4 years ago

3 0

Hi!

The answer would be A. Isobaric Process

<h3>Explanation:</h3>

Isobaric process is a process where the pressure inside a system remains unchanged. In the Pressure Volume graph given, you can see that the pressure (y axis) remains constant with an increasing volume ( x axis). An example of this would be heating a container with a movable piston. Now, the degree of pressure is dependent on the frequency of collisions of particles inside a system on the walls. If this frequency changes, the pressure changes (proportionally). In our example, heating a container with a movable piston results in the particles inside the container to gain kinetic energy and move faster, meaning an increased frequency of collisions (higher pressure), but at the system time the increase in pressure results in the piston being pushed outwards, causing the volume of the container to increase. This results in decreased frequency of collision of the particles with the walls of the container (lesser pressure). This results in the a zero net effect on the pressure.

Hope this helps!

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I need help with this problem

frosja888 [35]

Answer:

v = dr/dt = 7t x + 0 y + (- 4) z

a = dv/dt = 7 x + 0 y + 0 z

t = 0:

v = (-4) z

a = 7 x

Explanation:

v = dr/dt = 7t x + 0 y + (- 4) z

a = dv/dt = 7 x + 0 y + 0 z

t = 0:

v = (-4) z

a = 7 x

6 0

2 years ago

Ranking Task: The Electromagnetic Spectrum

shutvik [7]

The increasing order of the wavelength from left to right are as follows Gamma Rays < X-rays < UV rays < visible rays < infrared rays < microwaves < radio waves.

The electromagnetic spectrum consists of many waves which are made up of electric field and the magnetic field.

It is said that the electromagnetic wave behaves like particle as well as the wave.

The particle like pieces of the wave are called photons which contains the energy of the wave and the electromagnetic waves are primarily decided by the wavelength and the frequency of the wave.

The group of electromagnetic waves are known as electromagnetic spectrum.

The electromagnetic wave in the order of increasing wavelength are Gamma Rays < X-rays < UV rays < visible rays < infrared rays < microwaves < radio waves.

To know more about electromagnetic spectrum, visit,

brainly.com/question/23423065

#SPJ4

8 0

1 year ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

iris [78.8K]

Answer:

Explanation:

1. $V_{x}$ = $V_{0}$ * cos $\alpha$ ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. $V_{y}$ = $V_{0}$ * sin $\alpha$ ⇒ 16* sin32 ≈ 9.4 m/s

3. $y_{max}$ = $\frac{v_{0}^2*sin^2\alpha}{2g}$ = $\frac{16^2*sin^232}{2*9.8}$ (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

$y_{max}$ ≈ 3.6677+1.5 ≈ 5.2m

4. $x_{max}$ = $\frac{v_{0}^2*sin(2\alpha)}{g}$ = $\frac{16^2*sin(2*32)}{9.8}$ ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

3 0

3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. A uniform electric field

Brums [2.3K]

Answer:

the electric field direction should be in positive y axis

Explanation:

Lets assume that charge on particle is positive and it isequal to +q

First calculate the magnetic force on it

$F_B = q V\times B =qVBsin\theta$

for direction

use Right Hand Rule which will give the direction and by using his the direction will come towards negative y axis.

As given in the question that charge particle does not change their velocity so we need to apply electric field in such a way that electric force direction should be opposite to the magnectic field.

and magnitude should be same as magnectic force and also direcion of electric force depend on the direction of elecric field when charge is positive because electric force $F_E = qE$

Hence the electric field direction should be in positive y axis

3 0

4 years ago