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Andrew [12]
2 years ago
7

A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal le

ngth and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Physics
1 answer:
Vikki [24]2 years ago
3 0

Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What will the sp
zloy xaker [14]

Explanation:

if the elevator is moving upward with the constant speed the spring scale will read 18 N which is the mass of each of the two blocks attached by separate springs to the scale at opposite ends.

4 0
3 years ago
Danny lowers the sails on his boat. He paddles upstream at 19 km/hr. The current is still running downstream at 15 km/hr. What i
Molodets [167]

Answer:

4 km/hr

Explanation:

The computation of the actual velocity is shown below:

Because the path of its paddles is opposed to the current direction, the real velocity can be determined by deducting the current velocity to its velocity while paddling

So, the actual velocity is

= Upstream - downstream

= 19 km/hr - 15 km/hr

= 4 km/hr

As we can see it is in positive, so it is an upstream direction  

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A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning
zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

L_{i}=L_{f}

Here,  = the system's angular momentum before the collision

L_{i} = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here,  = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

4 0
1 year ago
Read 2 more answers
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