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4 years ago

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what optical properties do holograms depend on? Please give explanations for answers and/or sources please! I need this for my s

cience project!

1 answer:

Stels [109]4 years ago

7 0

A hologram is a physical structure that diffracts light into an image. The term ‘hologram’ can refer to both the encoded material and the resulting image. A holographic image can be seen by looking into an illuminated holographic print or by shining a laser through a hologram and projecting the image onto a screen. Hope this helps! Oh and here's a good source to use just don't forget to credit the sources you use.

http://file.scirp.org/Html/70316_70316.htm

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Which of the outer planets would you most likely visit? Why? What would you see on your visit?

Tanzania [10]

I would like to visit Pluto because i want to see what a Dwarf planet would look like, i would like to see what kind of minerals are in the planet its self..

Brainliest answer?

6 0

3 years ago

Read 2 more answers

The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo

notka56 [123]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t = 0.00075 s

The impulsive force is given by the formula:

$F=\frac{mv}{t}$

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

$F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N$

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

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2 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

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3 years ago

What colors are in a 24 pack of crayola crayons?

Xelga [282]

Blue puprle white gold organs yellow red gray brown green pink black tan light blue

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4 years ago

lozanna [386]

The force of gravity produces acceleration in all C. freely falling objects and this is known as acceleration due to gravity

Explanation:

A body is said to be in free fall when there is only one force acting on the body: the force of gravity.

Gravity is a force that acts downward, i.e. towards the Earth's centre.

If we are near the Earth's surface, the magnitude of the force of gravity on a body is given by

$F=mg$

where:

m is the mass of the body

g is known as the acceleration of gravity , whose value near the Earth's surface is $9.8 m/s^2$ ).

We can apply Newton's second law on an object in free-fall, to find its acceleration. In fact, we have:

$F=ma$

where F is the force acting on the body and a is its acceleration.

Solving for the acceleration,

$a=\frac{F}{m}$

And substituting F,

$a=\frac{mg}{m}=g=9.8 m/s^2$

Therefore, every object in free-fall accelerates at $9.8 m/s^2$ towards the ground.

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3 years ago