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Radda [10]
2 years ago
6

Light with a wavelength of about 510 nm is made to pass through a diffraction grating. The angle formed between the path of the

incident light and the diffracted light is 12.0° and forms a first-order bright band.
The number of lines per mm in the diffraction grating, rounded to the nearest whole number, is _______ lines per mm.

Formula??
Physics
1 answer:
xxMikexx [17]2 years ago
5 0
First, we use the relationship between the spacing of the openings, the wavelength of light, the order of the maxima or minima and the angle of diffraction which is:
d*sinθ = nλ, where d is the spacing, θ is the angle, n is the order and λ is the wavelength. We will use the wavelength in mm so that we may obtain the spacing in mm (510 x 10⁻⁶ mm). Solving for d,
d = (1)(510 x 10⁻⁶) / sin(12)
d = 2.45 x 10⁻³ mm
Using this, we may calculate openings per mm,
1 / 2.45 x 10⁻³ 
= 408 lines per mm.
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Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the
Yuki888 [10]

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

W_a = F_a d

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

W_b = F_b d

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

F_a = F_b

and so

W_a=W_b

3 0
2 years ago
The power that a student generates when walking at a steady pace of vw is the same as when the student is riding a bike at vb =
Nadya [2.5K]

Answer:

1/9

Explanation:

Sorry, I don't know if this is right, but here is what I did. We are ignoring potential energy because we assume that the student is walking and biking on level ground. Power = W/T, W = Mechanical Energy, or just Kinetic for this case. So P_{w}=\frac{E_{w} }{T}, and P_{b} =\frac{E_{b} }{T}. Ew = \frac{1}{2} mv_{w}^{2}, and Eb =\frac{1}{2} m(3v_{w})^{2}. Put Ew over Eb. the 1/2's cancel, the m's cancel, and you are left with \frac{v_{w} ^{2} }{9v_{w}^{2} }. Finally, this simplifies after cutting out the vw^2's to 1/9.

6 0
2 years ago
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens woul
Brilliant_brown [7]

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

7 0
2 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

Thus from eqn (2), the total charge is given by:

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

5 0
2 years ago
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