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Pavlova-9 [17]
3 years ago
13

What did isaac Newton believe light was made of

Physics
2 answers:
adelina 88 [10]3 years ago
8 0
Sir Isaac Newton, held the theory that light was made up of tiny particles<span>. In 1678, Dutch physicist, Christiaan Huygens, believed that light was made up of </span>waves<span>vibrating up and down </span>perpendicular<span> to the direction of the light travels, and therefore formulated a way of visualising wave propagation.</span>
STALIN [3.7K]3 years ago
5 0

Answer:

PARTICLE

Explanation:

As per the theory proposed by the Newton for the propagation of light he assumed that light is made up of PARTICLEs

He said that light is made up of small rigid particles which behaves elastically. When light shows reflection then this is an example of 100% elastic collision where these particles will strike the surface and rebound with same speed at same angle with the normal

So this theory explains the propagation of light in particle form and hence this theory was known as particle theory of light

So as per Newton's explanation light is made up of PARTICLES

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When a bat uses echolocation to determine the distance to an insect, it sends out a sound wave and waits to see how long the sou
ZanzabumX [31]

Answer:

Explanation:

Distance travelled by sound in going to target( insect )  and returning back = 2d ,d is distance of target .

time t = .06 s

speed = distance / time

344.9 = 2d / .06

d = 10.35 m

3 0
3 years ago
Need help with these please
Montano1993 [528]

Answer:

Explanation:

Which number all or one?

6 0
3 years ago
Read 2 more answers
What is the specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene v
vazorg [7]

The specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene vapor at 45 c and 1.27 atm absolute pressure will be 0 kJ/mol.

<h3>What is specific enthalpy and how was it calculated in the question?</h3>

A thermodynamic system has a property called enthalpy (H). It is calculated by the sum of the internal energy (U) of the thermodynamic system and the product of its volume (V) and pressure (p). The SI Unit is Joule (J).

Equation:

H = U+pV

The specific enthalpy of vapor can be defined as the amount of energy spent in order to transform a liquid substance into its vapor or gaseous form. The SI Unit is kJ/mol.

In the above question, the formula to be used is

P1/P2 = (Δ Hvap)/R)(1/T2-1/T1)

T1 & P1 --> the starting temperature & pressure respectively (= 1.27 atm and 45c),

T2 & P2 --> the final temperature & pressure respectively (= 0.7 atm and 45c),

R --> the real gas constant i.e. 8.314kJ/mol and

ΔHvap --> The specific enthalpy of vaporization.

Putting the values in the equation;

1.27/0.7=(ΔHvap/8.314)(1/45-1/45)

Hence as after subtracting the equation becomes 0, our final answer also comes out to be ΔHvap= 0 kJ/mol.

To know more about specific enthalpy, visit:

brainly.com/question/16244647  

#SPJ4

6 0
2 years ago
What would be the answer for this and how?
Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

k = \frac{F}{\Delta x}\\

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\

For a load of 30 N, applying Hooke's Law:

\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\

Hence, the correct option is:

<u>B. 6 cm</u>

7 0
3 years ago
A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)
natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

\phi = 22.90°

8 0
3 years ago
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