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Alinara [238K]
3 years ago
13

a camera with a 100mm lens can be used to focus objects from 6pm to infinity onto screen. how much must the lens be moved to foc

us on the extremities of this range
Physics
1 answer:
Romashka [77]3 years ago
3 0

Answer:

i = f = 0.1 m until the lens moves towards the screen 0.1 m

Explanation:

For this exercise let's use the constructor equation

         1 / f = 1 / o + 1 / i

     

Where f is the focal distance, or the distance to the object and "i" the distance to the image

It indicates that the focal distance is 100 mm (f = 100 mm), when an object is at infinity the image is formed at its focal length

     1 / f = 1 / inf + 1 / i = 1 / i

      i = f = 100 mm

At this point the screen is placed

For the shortest distance the lens has to move a little so the distance to the image is

        f = 100 mm = 0.1 m

        o = 6 pm = 6 10⁻¹² m

        i = 100 + x

        1 / f = 1 / o + 1 / 100+ x

        1 /( 0.10 + x) = 1 / f - 1 / o

        1 / 0.100+ x = 1/0.100 - 1/6 10-12

        1 / 0.100 + x = 10 - 10¹¹ = -10¹¹

        0.100 + x = -10⁻¹¹

        x = -10⁻¹¹ -10⁻¹

        x = -10⁻¹

        x = - 0.1 m

This negative distance indicates that the lens moves towards the screen 0.1 m

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Answer:

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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Kinetic energy is the energy that is possessed by an object that is moving. It is calculated by one-half the product of the mass and the square of the velocity of the object. We calculate as follows:<span>

For the truck,
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For the astronaut,
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