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Ber [7]
3 years ago
7

Who was the man who lived from 460B.C. 370B.C. and was among the first to suggest the idea of atoms?

Physics
2 answers:
Jlenok [28]3 years ago
6 0
C Democritus

That's the Answer
Basile [38]3 years ago
4 0

Answer:

The correct answer is C. The man who lived from 460 BC to 370 BC and was among the first to suggest the idea of atoms was Democritus.

Explanation:

Democritus was a Greek philosopher, astronomer, mathematician and traveler. He is counted among the presocrats and was known as "the laughing philosopher". Together with his teacher Leucippus, he was the founder of atomism. According to it, the cosmos consisted of emptiness (the non-existent) and bodies (atoms, the existence) in an infinite space. Those bodies are eternal and can move. Democritus wrote a lot: sixty titles are known. They were about ethics, philosophy of nature, mathematics, music, poetry and technical themes.

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Think of three questions that will help you clarify your understanding of how of the factors that have contributed to the rise i
fenix001 [56]

Answer:

the glucose in air will solidify and it will stop the process.

it will cause hot air

6 0
3 years ago
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A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r
katen-ka-za [31]

Answer:

  • k = 167.33 N/m

Explanation:

  • The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
  • v = c = λ*f

        where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =  

        4.92*10⁷ m.

        Solving for f, we get the frequency of the radio waves:

        f = 6.1 Hz

  • Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by  a fixed relationship between the spring constant k and the mass m, as follows:

       \omega_{o}^{2} =\frac{k}{m}  (1)

  • Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

       \omega = 2*\pi *f (2)    

  • We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
  • k = 4*\pi ^{2}*f^{2} *m = 4*\pi ^{2} * (6.1Hz)^{2} * 0.114 kg = 167.33 N/m

3 0
3 years ago
An object has a coefficient of static friction of 0.3 and a normal force of 30 N. Find the force of static friction.
gladu [14]

Answer:

9N

Explanation:

static friction=normal force x coefficient of static friction

so static friction =30N x 0.3= 9N

7 0
3 years ago
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What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

<u>s= 200 m</u>


3 0
3 years ago
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Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
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