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2 years ago

In which image below is the angle of refraction the greatest?

Physics

1 answer:

pentagon [3]2 years ago

5 0

Answer:

Explanation:

Angle of refraction is the angle made by refracted ray with the normal at the point of incidence . In this figure , refracted ray has been shown by line having arrow-head . Normal has been shown by broken line .

We observe that in figure D , angle made by refracted ray with normal is greatest . So figure D is the answer.

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An airplane is flying at 635 km per hour at an altitude of 35,000 m. What is its velocity?

Elden [556K]

Distance 350 Km

Time 1 hour

Velocity = 350 : 1 =

350Km/h

your answer is a

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3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

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3 years ago

How does mass affect the gravitational pull of the Sun, Earth, and the Moon?

madreJ [45]

Larger mass creates a stronger pull

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3 years ago

Read 2 more answers

A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop

postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination $(\theta )=30^{\circ}$

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

$F_{net}=0$

Force applied by cable $=mg\sin \theta$

$F=60\times 9.8\times \sin (30)=294 N$

work done $=F\cdot x$

$W=294\cdot 75=22.125 J$

(b)Power $=F\cdot v$

$P=294\cdot 2.4=705.6 W\approx 0.946 hp$

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3 years ago

Chemical properties of an acid include...

Cerrena [4.2K]

and C. corrosive, increases the concentration of hydrogen ions when added to water, forms hydrogen gas when it comes in contact with a metal, and formssalt and water when added to a base.

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3 years ago