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dlinn [17]
2 years ago
10

The most common type of mirage is an illusion that light from faraway objects seem to be reflected by a pool of water that is no

t really there. Mirages are generally observed in deserts, when there is a hot layer of air near the ground. The index of refraction of air is lower at higher temperature. What physical phenomena explains how mirages are formed? Give a brief explanation of the phenomena as your LPE.
Physics
1 answer:
olga nikolaevna [1]2 years ago
5 0

Answer:

Mirage is an optical illusion due to the refraction of light as it travels through different layers of air at different temperature.

Explanation:

On a clear hot afternoon, the sun heats up the ground, and the ground heats up the air immediately above it. The result is that the air directly above the ground is hotter than the air above it, leaving the air layers with different refractive indies. Since refractive index decreases with increase in temperature, the air immediately above the ground bends the light that travels through it from the sky, directly into our eyes. The image seen is perceived to be reflected from the ground, but in an actual sense is a direct image of the sky, giving one the impression of seeing a pool of water on the ground.

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g 2. The _____ spans the distance from the _____ to the location of the applied force. moment arm; pivot point moment of inertia
Alik [6]

Answer:

The correct answer to the following question will be Option A (moment arm; pivot point).

Explanation:

  • The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
  • The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
  • The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.

The other three choices are not related to the given situation. So that option A is the appropriate choice.

7 0
2 years ago
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

6 0
2 years ago
A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the
Elis [28]

Answer:

a)R = 171μΩ

b)E = 1.7 *10^{-4} V/m

c)R_{2} = 1.16 *10^{-4}Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder  =  1.1 cm

resistivity depends on the distance x from the left

p(x)=a+bx^2 ............(i)

using equation

R = \frac{pl}{a}

let dR is the resistance of thickness dx

dR =\frac{p(x)dx}{a}

where p(x) is resistivity  l is length

a is area

\int\limits^R_0 {dR}  =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx  \\.........................(2)

after integration

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}  ...............(3)

it is given p(0) = a = 2.25 * 10 ^{-8}Ωm

p(L) = a + b(L)^2  = 8.5 * 10 ^{-8} Ωm

8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\

(here * stand for multiplication )

on solving we get

b = 2.78* 10^{-8} Ωm

put each value of a  and b and r value in equation 3rd we get

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}

R = 1.71 * 10^{-4}Ω

R = 171μΩ

FOR (b)

for mid point  x = L/2

E = p(x)L

for x = L/2

p(L/2) = a+b(L/2)^2

for given current  I = 1.75 A

so electric field

 

E = \frac{[a+b(L/2)^2]I }{\pi  r^2}

by substitute the values

we get;

E = 1.7 *10^{-4} V/m

(here * stand for multiplication )

c ).

75 cm means length will be half

 that is   x =  L/2

integrate  the second equation with upper limit  L/2  

Let resistance is R_{1}

so after integration we get

R_{1}  =  \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}

substitute the value of a , b and L we get

R_{1} = 5.47 * 10 ^{-5}Ω

for second half resistance

R_{2} =  R- R_{1}

R_{2}  = 1.7 *10^{-4} -5.47 *10^{-5}

R_{2} = 1.16 *10^{-4}Ω

(here * stand for multiplication )

5 0
2 years ago
A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
vodka [1.7K]

Answer:

v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

F=qE

F_p=F_e

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

4 0
2 years ago
A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force
boyakko [2]

Answer:

Explanation:

D

7 0
1 year ago
Read 2 more answers
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