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Korolek [52]
1 year ago
13

a nearly monochromatic beam of light with an average wavelength of 514 nm is incident normally on a diffraction grating 1.30 cm

wide with 110 lines/cm. what is the smallest wavelength difference (in nm) this grating can resolve in the second order diffraction with this beam?
Physics
1 answer:
muminat1 year ago
8 0

smallest wavelength difference (in nm) this grating can resolve in the second order diffraction with this beam is 8 degree

The formula for diffraction grating: Obviously, d = \frac {1} { N }, where N is the grating constant, and it is the number of lines per unit length. Also, n is the order of grating, which is a positive integer, representing the repetition of the spectrum so in this case the wavelength difference can be calculated as  514 nm/ 1.30 X 110 =3.59 X2 =approx 8.Wavelengths between 100 nm and 10 m correspond to the optical regime, where grating application is most prevalent. In that situation, the groove density can range from a few tens to a few thousand grooves per millimeter, similar to echelle gratings.

to learn more about wavelength

brainly.com/question/13533093

#SPJ4

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The burner on a stove is 325° f. given that the burner emits electromagnetic radiation as a blackbody, what is the maximum wavel
lesantik [10]
To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin. 

C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C

C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K

λmax = 2898/435.78 = <em>6</em><em>.65 μm</em>

5 0
3 years ago
Which state of matter has the greatest distance between the individual particles?
Maurinko [17]
Your answer is c steam because steam is a gas...
4 0
4 years ago
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A proton traveling with speed 2 × 105 m/s in the -y direction passes through a region in which there is a uniform magnetic field
stepladder [879]

Answer:

\vec{E} =  1.2\times 10^5(-i)

Explanation:

Given that

Speed ,v= 2 x 10⁵ m/s ( - y direction)

B= 0.6 T (- z direction)

The resultant force on the proton given as

\vec{F}=q.\vec{E}+ q.(\vec{v}\times \vec{B})

F= m a

For uniform motion acceleration should be zero.

F = 0

0=q.\vec{E}+ q.(\vec{v}\times \vec{B})

0=\vec{E}+ (\vec{v}\times \vec{B})

0=\vec{E}+2\times 10^5(-j) \times 0.6(-k)

\vec{E} =- 2\times 10^5(-j) \times 0.6(-k)

\vec{E} =-1.2\times 10^5(i)

\vec{E} = 1.2\times 10^5(-i)

Electric filed should be apply in the negative x direction.

5 0
3 years ago
A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
3 years ago
A mass of 1 kg is moving in a circle of radius 3.3 meters. What is the liner velocity v m/s that would give a Centripetal Force
Ghella [55]

Answer:

The linear velocity of the object is 8.71 m/s.

Explanation:

Given;

mass of the object, m = 1 kg

radius of the circle, r = 3.3 meters

centripetal force, F = 23 N

Centripetal force is given by;

F_c = \frac{mv^2}{r}\\\\

where;

v is the linear velocity of the object

F_c = \frac{mv^2}{r}\\\\mv^2 = F_cr\\\\v^2 = \frac{F_cr}{m}\\\\v= \sqrt{\frac{F_cr}{m}} \\\\v= \sqrt{\frac{23*3.3}{1}}\\\\v = 8.71 \ m/s

Therefore, the linear velocity of the object is 8.71 m/s.

4 0
3 years ago
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