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svetoff [14.1K]
4 years ago
9

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edg

e of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?
Physics
1 answer:
uysha [10]4 years ago
4 0

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and  laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

i=arctan(2.2/.90)

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2=  refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

1\times sin(67.8) = 1.33\times sin(r)

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

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The start of the Space Age throughout the 1950s and 1960s helped Florida's economy flourish. How did it continue to impact Flori
Inessa05 [86]

Answer:

A) After the Space Age ended, many people moved to other states and the economy began to decline.

Explanation:

After the space age ended, businesses slumped, when the space shuttle program ended those made redundant could not be re absorbed due to a lack of follow-on by NASA. The 2008 recession also had impacted businesses in the area. As such many moved to other states, which had the effect of a brain drain.

After Apollo, in the late 70's, there was a similar decline of economic activities as workers were laid off and where finding it hard to sell off their properties as the real estate prices where low.

7 0
3 years ago
One of Lex Luthor's henchman attacks Superman, shooting a rapid-fire stream of 3.3 g bullets at him at a rate of 112/min. The sp
igor_vitrenko [27]

Answer:

3.2451N

Explanation:

Mass of the bullet (m) = 3.3g = 3.3*10^{-3}Kg

Speed of the bullet (V)= 527m/s

Rate of bullet (r) = 112/min = 1.866\sec

We can calculate with this information the average acceleration of bullets

a=V*r = 527\frac{m}{s}\frac{1.866}{s} = 983.38m/s^2

The force is given by,

F=ma\\F=(3.3*10^{-3})*983.38m/s^2 = 3.2451N

That is just because he is Superman.

4 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
4 years ago
<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%5Csqrt%7B4%7D%20%3D87" id="TexFormula1" title="x^{2} +\sqrt{4} =87" alt="x^{
Alja [10]

Answer:

<h2><em><u>ᎪꪀsωꫀᏒ</u></em></h2>

➪x= √ 85

Explanation:

x²+√4 = 87

=> x²+2 = 87

=> x² = 87-2

=> x²= 85

=> x= √85

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