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babymother [125]
3 years ago
10

With the simplified model of the eye, what corrective lens (specified by focal length as measured in air) would be needed to ena

ble a person underwater to focus an infinitely distant object? (Be careful-the focal length of a lens underwater is not the same as in air! Assume that the corrective lens has a refractive index of 1.62 and that the lens is used in eyeglasses, not goggles, so there is water on both sides of the lens. Assume that the eyeglasses are 1.79 cm in front of the eye

Physics
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

Please see the attached picture for the complete answer.

Explanation:

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a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. the centrip
qaws [65]
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity.  The direction of motion is part of velocity, and
the direction is constantly changing on a curve.

The centripetal force that keeps an object moving in a circle is

         Force  =  (mass of the object) · (speed)² / (radius of the circle)

         F  =  m s² / r

We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.

                                          F     =  m s² / r

Multiply each side by 'r':       F· r  =  m · s²

Divide each side by 'F':            r  =  m · s² / F    

We know all the numbers on the right side,
so we can pluggum in:

                      r  =       m       ·        s²      /     F

                      r  =  (1200 kg) · (20 m/s)² / (6000 N) .

I'm pretty sure you can finish it up from here.

                                      


5 0
3 years ago
Visible light occupies the majority of the electromagnetic spectrum? Ture or false
Novosadov [1.4K]
From what i can tell its false
5 0
3 years ago
Read 2 more answers
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
3 years ago
A wave has angular frequency 30 rad/s and wavelength 2.0 m. What are its (a) wave number and (b) wave speed
masya89 [10]

Answer:

a) 15 b) 60 i think is the answer

4 0
2 years ago
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

work done by gravity+work done by friction=Energy stored in Spring

mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

6 0
3 years ago
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