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laiz [17]
3 years ago
5

2. A child observes a caterpillar walking on a window sill. The caterpillar walks 18 cm to the left, then 6 cm to

Physics
1 answer:
motikmotik3 years ago
8 0

Answer:

D = 8 cm

Explanation:

It is given that, the caterpillar walks 18 cm to the left, then 6 cm to  the right, then 10 cm to the left.

We need to find the displacement of the caterpillar. The attached figure shows the path traveled by the caterpillar. A is the initial position, then it reaches B and finally it reaches C.

Let right side is positive and left is negative. We know that,

Displacement = final position - initial position

= C - A

= -10-(-18)

= 8 cm

Hence, the displacement of the caterpillar is 8 cm.

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Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the
polet [3.4K]

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

\rm \theta is the angle formed between the path of the incident light and the diffracted light = 9. 2°

λ is the wavelength of the light=490nm=4.9

N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\  d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm

The number of lines per mm is found as;

\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm

Hence the number of lines per mm in the diffraction grating is 326.

To learn more about diffraction grating refer to the link;

brainly.com/question/1812927

5 0
2 years ago
An experiment yielded the above temperature and time information. what is the freezing point of the material in this experiment
Dmitry [639]
The answer is 0 degrees Celsius (0°C). It will be where the line flat lines the first time.  The second time would be the boiling point.  An experiment yielded the above temperature and time information. The freezing point of the material in this experiment if the material is a solid at time zero is 0 degrees Celsius (0°C) .
7 0
3 years ago
The human activities in two locations are described below:
Bogdan [553]
Air pollution level is higher in Location B, because poisonous fumes are produced when coal burns.
5 0
3 years ago
Read 2 more answers
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
irakobra [83]

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>

So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.

3 0
3 years ago
Read 2 more answers
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

5 0
3 years ago
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