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Novay_Z [31]
3 years ago
8

Who was the first known contributor to the atomic theory? Select one: a. Thomson b. Bohr c. Dalton d. Democritus

Physics
1 answer:
ser-zykov [4K]3 years ago
6 0
It would be D Democritus
He contributed in around 460-370 BC
Whereas the others contributed much later from around 1700's - 1900's
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The equation of a progressive
Juliette [100K]

Answer:

Amplitude = 0.02m

Frequency = 640 Hz

Wavelength, λ = 0.5m

v = 320 m/s

Explanation:

Given the wave equation :

y=0.02 sin2π/0.5 (320t - x) where x and y are in

meters and t is in second

Comparing the above relation with the general wave equation :

y(x, t) = Asin2π/λ(wt - kx)

The amplitude, A = 0.02

From the equation :

2π/0.5 = 2π/λ

λ = 0.5 m

320t = vt

Hence, v = 320 m/s

Recall :

v = fλ

320 = f * 0.5

f = 320 / 0.5

f = 640 Hz

6 0
3 years ago
Which of the following items involve a wedge? Check all that apply A. A scissors blade B. A corkscrew C. An axe D. A wheelchair
Oksanka [162]

Answer:

Axe and Scissor blade

Explanation:

6 0
3 years ago
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Identify the two atmospheric layers that contain air as warm as 25◦ C
BigorU [14]

Answer:

troposphere and stratosphere

4 0
3 years ago
Cuando habla una persona, ¿por qué identificas que se trata de un hombre,una mujer o un niño​
Lubov Fominskaja [6]

Answer:

Lo ases por su voz

Explanation:

Creo ,,,,,,

4 0
3 years ago
A 4.0kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially
Kisachek [45]

a)

We use the formula :

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the values in:

4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f

Calculating this we get:

32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f

Rearrange for v2f:

v2f = \frac{32.0kg*m/s}{4.0kg}

This gives us 8.0 m/s as the final velocity of the second ball.

b)

Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.

This means we use the formula:

Ek = \frac{1}{2} *m*v^{2}+ \frac{1}{2} *m*v^{2} = \frac{1}{2} *m*v^{2} +  \frac{1}{2}*m*v^{2}

Substituting in values:

Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2

This simplifies to:

Ek= 128J + 0J = 0J + 128J

This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.

6 0
3 years ago
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