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Elenna [48]
3 years ago
7

How did newton use creativity and logic in his approach to investigating light?

Physics
2 answers:
Mekhanik [1.2K]3 years ago
7 0
He used is logic to form a structure that could hold light a light bulb.
Fed [463]3 years ago
4 0
He tested light in two different ways to see which was more of his idea he wanted to approach. newton used his own logic and creativity to seeing how lights are used against a prism.
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Which part of a scientific argument must be supported by valid data? claim evidence reasoning theory
nikitadnepr [17]
This question is a bit ambiguous because all parts of a scientific argument must be supported by valid data. However, among the choices, the closest synonym to "valid data" would be evidence. Evidence is the body of facts or information that support  a given idea.
7 0
3 years ago
Read 2 more answers
Please hurry please
Ghella [55]

Answer:

0.765m

Explanation:

gravitational potential energy GPE = mass * acceleration due to gravity * height

Given

GPE = 30Joules

Mass m = 4kg

acceleration = 9.8m/s²

Required

Height h

From the formula

h = GPE/mg

h = 30/4(9.8)

h = 30/39.2

h = 0.765m\

Hence the height of the counter is 0.765m

4 0
3 years ago
If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial accelerat
Cerrena [4.2K]

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\   F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2}  }{(6.50*10^{-10} )^{2}  } \\F=5.46*10^{-10}N

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}

For(b) Acceleration of proton

a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}

3 0
3 years ago
A 0.15 kg meter stick is balanced with the pivot point at the 18 cm mark. A weight of 3.2N is hung from the shorter end. Where s
Pani-rosa [81]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

(0.15 x g)/100 = 0.0147N. per cm. length. 
<span>Torque short end = (0.0147 x 9) = 0.1323N/cm. </span>
<span>Torque long end = ((100 - 18)/2) x 0.0147 = 0.6027N/cm. </span>
<span>Difference = (0.6027 - 0.1323) = 0.4704N/cm. </span>
<span>(0.4704/3.2) = 0.147cm. from the 18cm. mark, = 17.853cm. from the 0 end of the stick.</span>
3 0
3 years ago
Read 2 more answers
Someone plz answer this quickly!!
Serjik [45]

Answer:

B. In the same group

Explanation:

Elements in the same group shared chemical properties.

8 0
4 years ago
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