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olasank [31]
2 years ago
5

Consider a diffraction grating of width 5cm with slits of width 0.0001 cm separated by distance of 0.0002cm 1:what is the corres

ponding grating element 2:how many orders will be observed at a wavelength of 0.000055 cm 3:calculate the dispersion in the different orders
Physics
1 answer:
sashaice [31]2 years ago
3 0

Answer:

a) d= 0.0002cm

b) m=3

Explanation:

From the question we are told that:

Width of diffraction grating W_g= 5cm

Width of silt w_s 0.0001 cm

Distance d= 0.0002cm

a)

Generally the grafting element is the distance b/w  the silts

d= 0.0002cm

Therefore the grafting element is

d= 0.0002cm

b)

Generally the equation for diffraction state is mathematically given by

 dsin\theta=m\lambda

Since

 sin\theta \leq 1

Therefore

 m \leq \frac{d}{\lambda}

 m \leq \frac{0.0002}{0.000055}

 m \leq 3.6

Therefore orders observed with \lambda is

 m=3

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Answer:

89.53    North

Explanation:

Total N-S displacement

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1 year ago
To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal
tatiyna
Answer:  Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
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The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
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At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
 Therefore the force that opposes the centrifugal force becomes
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5 0
2 years ago
A swing has a period of 10 seconds. What is its frequency ?
kow [346]

Frequency = 1 / (period)

Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.


4 0
2 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
2 years ago
Define anterior and posterior in correlation to the body. (ANATOMY)
tatyana61 [14]

Answer:

it should be right it's from go.ogle hm!!!

Explanation:

Anterior or ventral - front (example, the kneecap is located on the anterior side of the leg). Posterior or dorsal - back (example, the shoulder blades are located on the posterior side of the body). Medial - toward the midline of the body (example, the middle toe is located at the medial side of the foot).

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