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3 years ago

What are 5 uses of reflection?

2 answers:

8 0

Optomety-contact/corrective lenses
binoculars periscopes and refracting telescopes and refracting telescopes
mangnifying glass
knowledge of refraction creates the most sparkly diamonds
prisms to divdide white light into it's into it's component colours
certain optical illusions

Gre4nikov [31]3 years ago

3 0

<span>here u go!
fiber-Optics
Mirrors
Sonar
Radar
Study of seismic waves </span>

You might be interested in

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

An object is placed at 10.2 cm in front of a diverging lens with a focal length of -10.6 cm. What is the magnification

adoni [48]

Answer:

M= -0.51

Explanation:

After i calculated my v to be -5.2cm from the formula 1/f=1/v+1/u

Then m=v/u which is -0.51

7 0

4 years ago

A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent

Margarita [4]

Answer:

$F= 1333.767\,N$

Explanation:

The velocity of the swimmer just before touching the water is:

$v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}$

$v \approx 6.569\,\frac{m}{s}$

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

$(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J$

$F= 1333.767\,N$

6 0

3 years ago

Read 2 more answers

What occurs when salt is evenly mixed with water

zimovet [89]

The more you mix salt with water the more the salt will dissolve
The act of dissolving in chemistry describes a physical change such as change in shape or state. Its important to remember that physical change does not change the chemical state.

Hoped my answer helped!

8 0

4 years ago

Read 2 more answers

Number 38 please!!!

vodomira [7]

Answer:

D. 30 Ω

Explanation:

R₁ + R₂ = 33

-0.0005 R₁ + 0.005 R₂ = 0

Solve the system of equations for R₁.

-0.0005 R₁ + 0.005 R₂ = 0

0.005 R₂ = 0.0005 R₁

R₂ = 0.1 R₁

R₁ + R₂ = 33

R₁ + 0.1 R₁ = 33

1.1 R₁ = 33

R₁ = 30

3 0

3 years ago