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3 years ago

Write the term related to reflection of light

1 answer:

8 0

(a) The light ray which is incidence (falls) on to a surface is called the Incident ray. (b) The light ray which is reflected (reflects) from a surface is called the Reflected ray. ... (d) The angle made by the Reflected ray with the normal at the point of Reflection is called a Angle of reflection.

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How far from the nucleus in angstroms (1 Å = 10-10 m) is the electron in a hydrogen atom if it has an energy of -8.72 * 10-20 J?

olganol [36]

Answer:

1200

Explanation:

7 0

3 years ago

Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi

zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

cos ( 360-30) = cos (-30) = x₁ / d

sin (-30) = y₁ / d

x₁ = d cos (-30)

y₁ = d sin (-30)

x₁ = 1.2 cos (-30) = 1,039 miles

y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

cos (270-20) = x₂ / d

cos (250) = y₂ / d

x₂ = 2.0 cos 250 = -0.684 miles

y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

cos (180 + 30) = x₃ / d

sin (210) = y₃ / d

x₃ = 1.6 cos 210 = -1.3856 miles

y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

cos (90 + 15) = x₄ / d

sin (105) = y₄ / d

x₄ = 2.6 cos 105 = -0.6729 miles

y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis (West-East direction)

X = x₁ + x₂ + x₃ + x₄

X = 1.039 -0.684 - 1.3846 - 0.6729

X = -1.7025 miles

Y = y₁ + y₂ + y₃ + y₄

Y = -0.6 -1.879 -0.8 +2.511

Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

D = √ (X² + Y²)

D = √ (1.7025² + 0.768²)

D = 1.8677 miles

let's use trigonometry to find the direction

tan θ = Y / X

θ = tan⁻¹ Y / x

θ = tan⁻¹ (0.768 / 1.7025)

θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

θ = 24.28º at South of West

4 0

3 years ago

Why neon is in the same group as helium?

guajiro [1.7K]

Neon and helium are both gases that fall under the Noble Gases section in the periodic table of elements.

Hope this helped to answer your question. :D

5 0

3 years ago

Calculate work done by the electrostatic force to move the charge with the magnitude of 1nC between two points 2cm spaced, along

timurjin [86]

Answer:

work done is = 0

Explanation:

given data

distance = 2 cm

potential = 1 V

charge with magnitude = 1 nC

to find out

work done by the electrostatic force

solution

we know that at equipotential surface is that surface which have equal potential at each every point that we say

work done will be

work done = ∫dw

∫dw = $\int\limits^v1_v2 {q} \, dv$

here q is charge

net work done = q ( v2 - v1 )

and

so v2 = v1 = 0

work done is = 0

3 0

4 years ago

A car stops in 150 m. If it has an acceleration of –7.0m/s2, what was the cars starting velocity?

Kobotan [32]

Answer:

We know that:

The car stops in 150 m, so this will be the final position.

The acceleration is -7.0m/s^2

I will assume that the initial position is zero.

Then we can write the movement equations as:

Acceleration:

A(t) = -7.0m/s^2

For the velocity, we integrate over time and get:

V(t) = (-7m/s^2)*t + V0.

Now we can integrate again over the time to get the position.

P(t) = (-3.5 m/s^2)*t^2 + V0*t + P0

Where P0 is the initial position,, ut it is equal to zero.

Then the equations that we need to use are:

V(t) = (-7m/s^2)*t + V0.

P(t) = (-3.5 m/s^2)*t^2 + V0*t

The car will reach a complete stop when:

The velocity is equal to zero

The position is equal to 150m

Then we have a system of equations:

0m/s = (-7m/s^2)*t + V0.

150m = (-3.5 m/s^2)*t^2 + V0*t

To solve this, first let's isolate one variable in one of the equations, i wil isolate V0 in the first equation:

V0 = (7m/s^2)*t

Now i can replace this into the second eqation:

150m = (-3.5m/s^2)*t^2 + (7m/s^2)*t^2

150m = (3.5m/s^2)*t^2

t = √( 150/3.5) seconds = 6.54 seconds.

And we know that the initial velocity is:

V0 = (3.5m/s^2)*t

then we can replace the value of t and get:

V0 = (3.5m/s^2)*6.54s = 22.89 m/s

3 0

3 years ago