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3 years ago

Describe the shape of a convex lens and explain what it does to light.

1 answer:

4 0

Convex lenses are thicker at the centers than the edges, they are known as the converging lenses. Rays of light that pass through the lens are brought closer together (they converge). When rays of light that are parallel pass through a convex lens they are refracted, the refracted rays converge at one point called the principal focus.

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The 500 series bullet train travels between Tokyo and Hakata, Japan. If it is heading north with a momentum of 13,194,098.64 kg

Hitman42 [59]

Answer:

The mass of the 500 series bullet train is 181.4 tonnes.

Explanation:

The momentum is given by:

$p = mv$

Where:

m: is the mass =?

v: is the velocity = 261.8 km/h

p: is the momentum = 13194098.64 kg*m/s

By solving the above equation for "m" we have:

$m = \frac{p}{v} = \frac{13,194,098.64 kg*m/s}{261.8 km/h*\frac{1000 m}{1 km}*\frac{1 h}{3600 s}} = 181.4 tonnes$

Therefore, the mass of the 500 series bullet train is 181.4 tonnes.

I hope it helps you!

5 0

2 years ago

What is the value of sin r and sin i in physics

taurus [48]

Answer:

★The second law of refraction

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a light of given colour and for a given pair of media. This law is also called Snell's law of refraction. If 'i' is the angle of incidence and 'r' is the angle of refraction then, Sin i/Sin r = constant

This constant value is called the refractive index of the second medium with respect to the first.

7 0

3 years ago

A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?

gayaneshka [121]

F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.

3 0

2 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago

What is impulse measured in

MrMuchimi

Impulse = Force * time

5 0

3 years ago