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2 years ago

Clarify the concept of refrection index with example

1 answer:

3 0

The refractive index is used to measure how much light bends as it goes from one medium to another the refractive index of air =1 as light doesn’t bend in it

When looking into a glass of water with a straw the straw may appear bent or morphed as the water has a different refractive index to air (and the light is passing from air to glass of water)

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A balloon is a sphere with a radius of 5.0 m. the force of air against the walls of the balloon is 45 n. what is the air pressur

Mazyrski [523]

The air pressure inside the balloon is: 0.1432 Pa

The formulas and procedures that we will use to solve this problem are:

a = 4 * π * r²
P = F/a

Where:

a = area of the sphere
r = radius
π = mathematical constant
P = Pressure
F = Force
a = surface area

Information about the problem:

r = 5.0 m
F = 45 N
1 Pa = N/m²
1 N = kg * m/s²
a=?
P=?

Using the formula of the sphere area we get:

a = 4 * π * r²

a = 4 * 3.1416 * (5.0 m)²

a = 314.16 m²

Applying the pressure formula we get:

P = F/a

P = 45 N/314.16 m²

P = 0.1432 Pa

<h3>What is pressure?</h3>

It is a physical quantity that expresses the force applied on the area of a surface.

Learn more about pressure at: brainly.com/question/26269477

#SPJ4

7 0

1 year ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

<u>Now the net force</u>

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why

lutik1710 [3]

It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.

Hopr it helps :)

4 0

2 years ago

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

3 0

3 years ago

John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

KIM [24]

The distance is 30 km and the displacement is 22.4 km North East

7 0

2 years ago

Clarify the concept of refrection index with example​

Clarify the concept of refrection index with example