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Alex73 [517]
3 years ago
15

Which scientist first discovered that white light is a mixture of a rainbow spectrum of light rays?

Physics
2 answers:
Zielflug [23.3K]3 years ago
5 0

c A ray of light is divided into its constituent colors by the first prism (left), and the resulting bundle of colred rays is reconstituted into white light by the second. Our modern understanding of light and color begins with Isaac Newton (1642-1726) and a series of experiments that he publishes in 1672.

isaac newton

katovenus [111]3 years ago
4 0

Answer:

The correct answer is Isaac Newton , option C

Explanation:

During the 1660’s, Newton, an English physicist, started some experiments with sunlight and prisms to know more about the optic field. He passed sunlight through a prism and separated the light beam into various colours. Once he did it, he discovered and understood that this could happened because every different ray of colour is diverged by the glassy prism by a different amount. Consequently, he could conclude that white light passes through a translucent medium, such as air, into another like glass, the components are glanced the first-time up t their colour and when they go back to air they do it again. This created the spectrum with the rainbow colours.  

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Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
What kind of radiation is emitted in the following nuclear reaction
miv72 [106K]

Beta emission is occurring in the given nuclear reaction.

Answer: Option B

<u>Explanation:</u>

In this equation, the reactant is the Thorium atom, which is reduced to palladium. As the atomic number get decreased by one, so an electron will be emitted. This process of emission of electrons by radiation or decaying the reactant nuclei to form a new product nuclei is termed as beta emission.

So, the electrons are generally termed as beta particles while the positrons are termed as positive beta particles. So this is a kind of radioactive reactions where the reactant changes to new element by releasing an electron and thus there is a change in the atomic number of the product by one.

4 0
3 years ago
An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam
Amanda [17]
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V =  the volume of the sample.
Then
W = ρVg
or
V =  W/(ρg)
   = (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
   = 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³
5 0
3 years ago
There is a circuit with 2 resistors in series and a cell of 3.0v.what is the resistance across the first resistor if the current
Marysya12 [62]

Answer:

The resistance of the  first resistor is, R₁ = 10 Ω

Explanation:

Given data,

The potential of the cell v = 3 V

The current through the first resistor, I = 0.2 A

The P.D across the first resistor, v₁ = 2.0 V

Let R₁, and R₂ be the resistance of the resistors,

Since the voltage across the series connection is,

                                   v = v₁ + v₂

Substituting the values

                                    3 V = 2 V + v₂

∴                                       v₂ = 1 V

Hence, the P.D across the second resistor is, v₂ = 1 V

The resistance of the first resistor is,

                                   R₁ = v₁ / I

                                        = 2 V / 0.2 A

                                         = 10 Ω

Hence, the resistance of the  first resistor is, R₁ = 10 Ω

6 0
3 years ago
Determine the magnitude of the force F1F1 component acting along the uu axis. Express your answer to three significant figures a
dybincka [34]

The first image below shows force F1 and the axes.

Answer: (F_{1})_{u}= 3.62 kN

Explanation: The second figure below express the parallelogram method to calculate the u component of force F1.

The <u>Parallelogram</u> <u>Method</u> is a method to determine resultant force and is applied as described in the question above.

With the three components, F_{1},(F_{1})_{u} and (F_{1})_{v} and angles, it can be used the <u>Law</u> <u>of</u> <u>Sines</u>, which states:

\frac{a}{sin\alpha} =\frac{b}{sin\beta} =\frac{c}{sin\theta}

i.e., there is a relation of proportionality between an angle and its opposite side.

For the triangle below:

\frac{u}{sin30} =\frac{F_{1}}{sin105}

u=F_{1}\frac{sin(30)}{sin(105)}

u=7\frac{0.5}{0.966}

u = 3.62

The magnitude of the component acting along the u-axis is 3.62kN.

5 0
3 years ago
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