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4 years ago

What are plane mirrors ?

Physics

1 answer:

forsale [732]4 years ago

3 0

Explanation:

Mirrors consist of reflecting surfaces that reflect light.

Reflection is a phenomenon of light wave (but also of other types of waves) in which a ray of light hits a surface, and then it bounces back into the original medium at a certain angle.

The direction of the reflected ray is determined by the law of reflection:

The incident ray, the reflected ray and the normal to the surface all lie in the same plane
The angle of reflection is equal to the angle of incidence (where both angles are measured between the ray and the normal to the surface)

A plane mirror is a type of mirror consisting of a straight surface. As a result, light incident perpendicular to the surface is reflected back exactly in the opposite direction.

The image formed by a plane mirror is:

- The same size as the object

- Virtual (it is located behind the mirror)

- Laterally inverted

- Upright

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Naming covalent compounds <br> P4S5

Andrew [12]

Answer:

what the heck is sakurfa

Explanation:

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3 years ago

At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/

Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0

3 years ago

Water changes from a liquid to a gas in the process of evaporation. Please select the best answer from the choices provided T F

const2013 [10]

The correct answer is T.

6 0

4 years ago

Read 2 more answers

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

Now we can solve for $v$ . We find that

$v=20.8\,\dfrac{\mathrm m}{\mathrm s}$

In the case that the ratio of accelerations is actually

$\dfrac{a_2}{a_1}=1.66$

we would instead have

$\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)$

in which case we would get a velocity of

$v=24.4\,\dfrac{\mathrm m}{\mathrm s}$

6 0

3 years ago

onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re

solmaris [256]

Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA = 1.00 * 10⁻⁴ * 1

Number of moles of HA = 1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺ + A⁻

If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴

Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left

Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴

Number of mols of HA left = 8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ = 1.6 * 10⁻⁵/V

To calculate the dissociation constant

$k_{a} = [H^{+} ][A^{-} ]/[HA]$

$k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L$

3 0

3 years ago