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3 years ago

1. Which diagram shows the correct pattern of light rays through a concave lens?

Physics

1 answer:

vaieri [72.5K]3 years ago

6 0

The one of and eye at some point the light rays will cross and produce an upside down image

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Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also

ozzi

Answer:

a) $B_{max} = 1.784*10^{-12}$

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

$V_m$ = 160 V

a) $V = V_msin(2\pi ft)$

Where $f = 60$ Hz and $V_m = 160$ V

$E =V/d= \frac{V_msin(2\pi ft)}{d}$

For $r = R$

$A = \pi R^2$

Since $\Phi_E = EA$

$\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}$

From Ampere's Law:

$\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl}$ where $I_{encl}=0$

So at $r = R$ ,

$B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}$

For maximum B, cos(2πft) = 1. Hence,

$B_{max}=\frac{\mu_0\epsilon_0\pi fRV_m}{d}=\frac{4\pi*10^{-7}*8.85*10^{-12}*\pi*60*0.025*160}{4.7*10^{-3}}=1.784*10^{-12}$ T

b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}$

From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}$

The plot is given in the attachment.

5 0

4 years ago

What is the third most abundant gas in the atmosphere.

Nataly_w [17]

Answer:

Argon

Explanation:

Argon, an inert gas, is the third most abundant gas in the atmosphere.

7 0

2 years ago

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

mass of the block, = M
coefficient of static friction in section 1, = $\mu_s_1$
angle of inclination of the plane, = θ

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago

Mars moves in an elliptical orbit around the sun, and the mass of mars is much less than the mass of the sun. at the instant sho

Bess [88]

I believe the answer would be that the system potential energy will increase; while the Mar's angular momentum will remain the same.
Mars rotates on its axis, completing one revolution every 24.6 hours. The axis of mars is tilted at 25 degrees and 12 minutes relative to its orbital plane about the Sun. This produces seasons on the surfaces of Mars, similar to the seasons on Earth.

4 0

3 years ago

A kind of foliated metamorphic rock a. coal b. limestone c. marble d. slate

djyliett [7]

Limestone is a kind of Foliated Metamorphic rock.

4 0

4 years ago

Read 2 more answers