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svetoff [14.1K]
3 years ago
14

What is the approximate wavelength of a light whose second-order dark band forms a diffraction angle of 15.0° when it passes thr

ough a diffraction grating that has 250.0 lines per mm?
Physics
1 answer:
Natalija [7]3 years ago
7 0
The answer to this would be 414 nm
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You attach a meter stick to an oak tree, such that the top of the meter stick is 1.471.47 meters above the ground. Later, an aco
Alexandra [31]

Answer:

The value is  h_a  =   1.712 \  m

Explanation:

From the question we are told that

   The  height of the top meter stick above the ground is  h_1  =  1.47 \  m

 

    The  time taken for the acorn to pass the length of the  stick is t =  0.281 \  s

Generally the height of the acorn at the point it is the same height with the metered stick is mathematically represented as  

     h =  h_m =  u_a  * t  +  \frac{1}{g} t^2

Here h_m is height of the meter stick and the value is 1 m (This because we are told in the question that the stick is 1 meter in length ( a meter stick))

So

      1  =  u_a  *  0.281  +  \frac{1}{9.8} (0.281)^2

=>   u_a  =  -2.2 \  m/s

Generally the velocity of the acorn just before passing the top of the meter stick is mathematically represented by a kinematic equation as

         u^2_a   =  u^2  + 2gs

here  u is zero since the acorn started from rest

So  

         (-2.2)  =  0  + 2 *  9.8 * s

         s  =  0.242 \  m

Generally the height of the acorn is

     h_a   =  h_1 + s

      h_a  =  0.242  +  1.47\

      h_a  =   1.712 \  m

       

       

     

3 0
4 years ago
How does a change in resistance affect the conductivity of a material?
max2010maxim [7]

Answer: the answer is B the lower the resistance,the higher the conductivity

Explanation:

Resistance is inversely proportional to conductivity,the lower the resistance the higher the conductivity,the metal silver has a very low resistance,thus it conductivity is high

4 0
4 years ago
Read 2 more answers
If the average frequency emitted by a 160 W light bulb is 5.00 1014Hz and 10.0 of the input power is emitted as visible light ap
bearhunter [10]

Answer:

The value is \frac{n}{t}  = 4.83 *10^{19} \  photons / s

Explanation:

From the question we are told that

   The power rating of the bulb is  P = 160 \ W

   The frequency is f =  5.00 *10^{14} \ Hz

   The percentage of the input power that is emitted as visible light is \eta =  10\% = 0.10

   

Generally the amount of power emitted as visible light is mathematically represented as

       P_l =  0.10 * P_i

=>  P_l =  0.10 *160

=>  P_l =  16 \ W

Generally the amount of energy emitted as light is mathematically represented as

        E = n *  h  *  f

Here n is the number of photon ,  h is the Planks constant with value h =  6.625*10^{-34} \  J\cdot s

Generally this power emitted as visible light is mathematically represented as

   P_l = \frac{E}{t}

=>  P_l = \frac{E}{t} = \frac{nhf}{t}

=>  \frac{n}{t}  = \frac{P_l }{hf}

=>  \frac{n}{t}  = \frac{16 }{6.625 *10^{-34}* (5.00*10^{14})}

=>  \frac{n}{t}  = 4.83 *10^{19} \  photons / s

4 0
3 years ago
When a sound source is moving away from an observer, the observer will hear a lower pitch. This is due to
Art [367]

a decrease in the frequency of the sound wave.

Explanation:

When a sound source is moving away from an observer, the observer will hear a lower pitch due to a decrease in the frequency of the sound wave. This is known as the Doppler's effect.

  • As the sound source moves closer to the observer, they hear a higher pitch in the reverse process.
  • A sound wave transmits vibrational energy from the source to the hearer.
  • A  sound wave has frequency, amplitude, velocity and wavelength just like other wave types.
  • The frequency of a wave is the number of waves that passes through a point at a time.
  • This is related to the the wavelength of a wave.
  • Waves of a high frequency are known to have short wavelengths and vice versa.
  • To hear a low pitch, the source moves away and the wavelength between successive crests and troughs on a wave increases considerably.

Learn more:

Sound waves brainly.com/question/2845448

#learnwithBrainly

6 0
4 years ago
Read 2 more answers
An alternating-current (AC) source supplies a sinusoidally varying voltage that can be described with the function v of t is equ
Marrrta [24]

Answer:

ω, the angular frequency of the source equals 377 rad/s

Explanation:

From the question, V(t) = V cosωt.

Now, ω = the angular frequency of the sinusoidal wave is given by

ω = 2πf where f = the frequency of the source = 60 Hz

So, the angular frequency of the source ,ω = 2π × the frequency of the source.

So, ω = 2πf

ω = 2π × 60 Hz

ω = 120π rad/s

ω = 376.99 rad/s

ω ≅ 377 rad/s

So, ω, the angular frequency of the source equals 377 rad/s

3 0
3 years ago
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