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mars1129 [50]
2 years ago
11

White light passes through a diffraction grating and forms rainbow patterns on a screen behind the grating. For each rainbow,

Physics
1 answer:
dlinn [17]2 years ago
7 0

Answer:

The answer is a) The red side is on the right, the violet side on the left and d) The red side is farthest from the center of the screen, the violet side is closest to the center

A rainbow drawing is attached to the attached file.

Explanation:

A rainbow is an optical and meteorological phenomenon that consists of the appearance of a multicolored arc of light, originated when sunlight is broken down into the visible spectrum. It is an arc composed of an arc of colors, with red towards the outside (angle of incidence of 42°) and violet towards the inside (angle of incidence of 40°).

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You might be interested in
If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times t
Tasya [4]

Answer:

F = K Q1 Q2 / R^2       force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

8 0
1 year ago
10. Convert the following:<br> a. 37.4 mL into ML<br> b. 689 km/hr into m/s<br> c. 34.5 m² into mm²
Snezhnost [94]

A. When we convert 37.4 mL to ML, the result obtained is 3.74×10¯⁸ ML

B. When we convert 689 km/hr to m/s, the result obtained is 191.39 m/s

C. When we convert 34.5 m² to mm², the result obtained is 3.45×10⁷ mm²

<h3>A. How to convert millimeters (mL) to megaliter (ML)</h3>
  • Volume (mL) = 37.4 mL
  • Volume (ML) =?

1 mL = 1×10¯⁹ ML

Therefore,

37.4 mL = 37.4 × 1×10¯⁹

37.4 mL = 3.74×10¯⁸ ML

Thus, 37.4 mLis equivalent to 3.74×10¯⁸ ML

<h3>B. How to convert 689 km/hr to m/s</h3>

Conversion scale

3.6 Km/hr = 1 m/s

Therefore,

689 km/hr = 689 / 3.6

689 km/hr = 191.39 m/s

Thus, 689 km/hr is equivalent to 191.39 m/s

<h3>C. How to convert 34.5 m² to mm²</h3>

Conversion scale

1 m² = 1×10⁶ mm²

Therefore,

34.5 m² = 34.5 × 1×10⁶

34.5 m² = 3.45×10⁷ mm²

Thus, 34.5 m² is equivalent to 3.45×10⁷ mm²

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

6 0
1 year ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a
zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2

b)

Also , by equation of motion :

s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m

Hence , this is the required solution .

6 0
3 years ago
Which of the following is MOST useful to scientists in measuring the size of asteroids?
Alenkasestr [34]

Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

5 0
3 years ago
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