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Misha Larkins [42]
3 years ago
13

Violet light (410 nm) and red light (685 nm) pass through a diffraction grating with d = 3.33 x 10-6 m. What is the angular sepa

ration between them for m = 2?​
Physics
1 answer:
jeyben [28]3 years ago
7 0

Answer:10.03

Explanation:

right on acellus

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The temperature of water in a beaker is 45°C. What does this measurement represent?
Tasya [4]

Answer:

the average kinetic energy of water particles - C.

8 0
3 years ago
What is the wavelength of a radar signal that has a frequency of 27 GHz? The speed of light is 3 × 108 m/s. Answer in units of m
marusya05 [52]

Explanation:

speed of light= c

wave length= L

frequency= f

c=Lf → L= c/f → L= 3 × 10⁸/ 27 × 10⁹ → L = 1/90 ≈ 0.011 m

4 0
2 years ago
1. In a biology experiment the number of yeast cells is determined after 24 hours of growth at
Anna11 [10]

Answer:

In a biology experiment the number of yeast cells is determined after 24 hours of culture at different temperatures. a) Identify the independent and dependent variables in the experiment. b) Draw the graph of the

Explanation:

5 0
2 years ago
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

5 0
2 years ago
Read 2 more answers
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