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adelina 88 [10]
3 years ago
9

Light passes through a pair of narrow slits with a 0.67-mm separation. It is found that the fourth bright fringe makes an angle

of 0.19° with the normal. What is the light’s wavelength in nanometers? Answer in nanometers

Physics
1 answer:
babunello [35]3 years ago
7 0

Answer:

The wavelength of the light is 555 nm.

Explanation:

according to Bragg's law..

n×λ = d×sin(θ)

n is the fringe number

λ is the wavelength of the light

d is the slit separation

θ is the angle the light makes with the normal at the fringe.

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3 years ago
A student adds two vectors of magnitudes 48 m and 22 m. What are the maximum and minimum possible values for the resultant of th
Julli [10]

Answer:

<em>Maximum=70 m</em>

<em>Minimum=26 m</em>

Explanation:

<u>Vector Addition </u>

Since vectors have magnitude and direction, adding them takes into consideration not only the magnitudes but also their respective directions. Two vectors can be totally collaborative, i.e., point to the same direction, or be totally opposite. In the first case, the magnitude of the sum is at maximum. Otherwise, it's at a minimum.

Thus, the maximum magnitude of the sum is 48+22 = 70 m and the minimum magnitude of the sum is 48-22= 26 m

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3 years ago
The metric unit of force is _____. kg kg/cm3 kg/cm2 N
ioda

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3 years ago
Read 2 more answers
a 28.7 KG sled is pulled forward with a 63 net force across the ground with MK equals 0.169 what is the acceleration of the sled
abruzzese [7]

Answer:

0.54\ \text{m/s}^2

Explanation:

F = Force on the sled = 63 N

m = Mass of sled = 28.7 kg

\mu_k = Coefficient of kinetic friction = 0.169

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

The force balance of the system is given by

F-\mu_k mg=ma\\\Rightarrow a=\dfrac{F-\mu_k mg}{m}\\\Rightarrow a=\dfrac{63-0.169\times 28.7\times 9.81}{28.7}\\\Rightarrow a=0.54\ \text{m/s}^2

The acceleration of the sled is 0.54\ \text{m/s}^2.

5 0
3 years ago
Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol
valentina_108 [34]

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

tan 16^0 = \dfrac{h}{x}

x =\dfrac{h}{tan 16^0}

now,

again applying

tan 20^0 = \dfrac{h}{300-x}

300-x=\dfrac{h}{tan 20^0}

300-3.49 h=2.75 h

6.24h = 300

h = 48.077 ft

3 0
3 years ago
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