Ask question

Ask question

All categories

3 years ago

11

It's nighttime, and you've dropped your goggles into a 3m deep swimming pool. If you hold a laser pointer 1.0m abovve the edge o

f the pool, you can illuminate the goggles if the laser beam enters the water 2.0m from the edge. How far are the goggles from the edge of the pool?

1 answer:

adoni [48]3 years ago

3 0

Answer:

x_total = 3.07 m

Explanation:

Let's analyze the situation presented, in this case we have the data to find the angle of incidence and with the law of refraction we can find the angle of refraction.

Let's start looking for the angle with which the laser pointer reaches the water, let's use trigonometry

tan θ₁ = 1.0 /2.0

θ₁ = tan⁻¹ 0.50

θ₁ = 26.57º

now let's use the law of refraction to find the angle of refraction (in water)

n₁ sin θ₁ = n₂ sin θ₂

the refractive index of air is n₁ = 1 and that of water n₂ = 1.33

sin θ₂ = $\frac{n_1}{n_2} \ sin \theta_1$

sin θ₂ = $\frac{1}{1.33} \ sin \ 26.57$

θ₂ = sin⁻¹ 0.3363

θ₂ = 19.65º

now we can find the distance from the entry point to the water to the lenses

tan θ₂ = x₂ / 3

x₂ = 3 tan θ₂

x₂ = 3 tan 19.65

x₂ = 1.07 m

the total distance from the edge of the pool is

x_total = 2 + x₂

x_total = 2 + 1.07

x_total = 3.07 m

You might be interested in

A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction

Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

Index refraction = 1.47

(a). We need to calculate the speed

Using formula of speed

$n = \dfrac{c}{v}$

Where, n = refraction index

c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

$1.47=\dfrac{3\times10^{8}}{v}$

$v=\dfrac{3\times10^{8}}{1.47}$

$v= 2.04\times10^{8}\ m/s$

(b). We need to calculate the wavelength

Using formula of wavelength

$n=\dfrac{\lambda_{0}}{\lambda}$

$\lambda=\dfrac{\lambda_{0}}{n}$

Where, $\lambda_{0}$ = wavelength in vacuum

$\lambda$ = wavelength in medium

Put the value into the formula

$\lambda=\dfrac{650\times10^{-9}}{1.47}$

$\lambda=442\times10^{-9}\ m$

Hence, The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

3 0

3 years ago

The fastest pitched baseball was clocked at 46 m/s. assume that the pitcher exerted his force (assumed to be horizontal and cons

Zielflug [23.3K]

Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance

<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,

Solving Eq.1 for acceleration,

</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²

<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058

F = 153.4 N</span>

8 0

3 years ago

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.

4 0

4 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

2 years ago

Your high beam headlights illuminate the road in front of you for __________ feet. A. 150 B. 450 C. 650

ASHA 777 [7]

Answer:

B. 450 feet

Explanation:

Due to the angle at which high beam headlights illuminate, they can illuminate the road for about 450 feet.

8 0

3 years ago