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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 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equili
PolarNik [594]
The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
E=U_{max}= \frac{1}{2}k(x_{max})^2

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
E=K_{max}= \frac{1}{2}m(v_{max})^2

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
U_{max}=K_{max}
\frac{1}{2}k(x_{max})^2 =  \frac{1}{2}m(v_{max})^2
v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s
3 0
3 years ago
A hiker walks 15km due east then heads due north for 8km . What is the direction of the resultant vector?
4vir4ik [10]

his displacements are given as

d_1 = 15 km East

d_2 = 8 km North

now we can find the direction of displacement by using the concept

\theta = tan^{-1}\frac{\deta y}{\delta x}

\theta = tan^{-1}\frac{8}{15}

\theta = 28.1 degree

magnitude of displacement is given as

d = \sqrt{x^2 + y^2}

d = \sqrt{15^2 + 8^2}

d= 17 km

<em>so the displacement is 17 km at angle 28.1 degree North of East</em>

8 0
3 years ago
100 POINTSSSS PLEASE HELP
lara31 [8.8K]
Wood, wind, sunshine,geothermal energy, biomass and water 
4 0
3 years ago
Read 2 more answers
At a given time of​ day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4​-f
katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

tan\theta =\frac{4}{2.8}

let the height of tree be h

as \thetawill remain same thus

tan\theta =\frac{h}{24.64}

\frac{4}{2.8}=\frac{h}{24.64}

h=35.2 ft

8 0
3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

6 0
3 years ago
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