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Novay_Z [31]
4 years ago
12

Determine the farthest away a car can be at night so that your eyes can resolve the two headlights. Assume that the the diameter

of your pupil is 0.50 cm, the dominant wavelength emitted by headlights is 500 nm and car headlights are about 1.4 m apart. Express your answer with the appropriate units.
Physics
1 answer:
Zigmanuir [339]4 years ago
8 0

Answer:

The farthest distance is L_{max}=11.48 km

Explanation:

The angular resolution is mathematically represented as

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

Where Z is the distance between the car head lamp with a value Z = 1.4 m

L is the distance between the eye and the car

\lambda is the wavelength  with a value 500 nm = = 500 *10^{-9}m

d is the diameter of the eye with a value  d = 0.50 cm = = \frac{0.50}{100} = 0.005m

   The maximum distance L_{max} is given as

              L_{max} = \frac{(1.4) *0.005}{1.22 * (500 *10^{-9})}

                       = 11.48*10^{3}m

                        =11.48 km

             

rrrrrrr

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Bezzdna [24]

Answer:

F = 0.112 N

Explanation:

To find the magnitude of magnetic force on the wire, you use the following formula:

|\vec{F}|=|i\vec{L}\ X\ \vec{B}|=iLBsin\theta   (1)

L: length of the wire = 200cm = 0.2m

i: current in the wire = 30 A

B: magnitude of the magnetic field = 0.055 T

θ: angle between the directions of L and B = 20°

You replace the values of L, i, B and θ in the equation (1):

|\vec{F}|=(30A)(0.2m)(0.055T)sin(20\°)=0.112N

hence, the magnetic force on teh wire is 0.112N

3 0
3 years ago
After three half-lives, one-ninth of an original radioactive parent isotope remains, and eight-ninths has decayed into the daugh
BartSMP [9]

Answer:

B. False

The true statement will be:

After three half-lives, one-eighth of an original radioactive parent isotope remains, and seven-eighth has decayed into daughter isotopes.

Explanation:

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Thus after each successive half life the material is halved or half of it is decayed, which leads to a general formula:

Remaining Amount = 1/2^n of parent isotope

where, n is the number of half lives passed. So, for three half-lives:

Remaining Amount = 1/2^3 of parent isotope

<u>Remaining Amount = 1/8 of parent isotope</u>

Thus, the decayed amount will be:

Decayed Amount = (1-1/8) of parent isotope

<u>Decayed Amount = 7/8 of parent isotope</u>

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8 0
3 years ago
The eye of the Atlantic giant squid has a diameter of 3.50 × 10^2 mm. If the eye
Lunna [17]

Answer:

   q = 224 mm,   h ’= - 98 mm, real imagen

Explanation:

For this exercise let's use the constructor equation

        \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

       

where f is the focal length, p and q are the distance to the object and the image respectively.

In a mirror the focal length is

        f = R / 2

indicate us radius of curvature is equal to the diameter of the eye

       R = 3,50  10² mm

       f = 3.50 10² /2 = 1.75 10² mm

they also say that the distance to the object is p = 0.800 10³ mm

        1 / q = 1 / f - 1 / p

        1 / q = 1 / 175 - 1 /800

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to calculate the size let's use the magnification ratio

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in concave mirrors the image is real.

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8 0
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Answer:

<h2>C) internal combustion engine</h2>

Explanation:

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