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4 years ago

2. Who were the first philosophers to propose that atoms existed? What time period was this?

Physics

1 answer:

Rus_ich [418]4 years ago

7 0

Answer:

"Greek Origins" were the first philosophers to propose that atoms existed.

Explanation:

The idea that all matter is made up of tiny, indivisible particles, or atoms, is believed to have originated with the Greek philosopher Leucippus of Miletus and his student Democritus of Abdera in the 5th century B.C. (The word atom comes from the Greek word atomos, which means “indivisible.”)

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What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.

Rasek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The tension is $T = 440 \ N$

The length of the wire is $l = 0.600 \ m$

The mass is $m = 5.60 \ g = 0.0056 \ kg$

Generally the frequency is mathematically represented as

$f = \frac{1}{2l} * \sqrt{\frac{T}{\frac{m}{l} } }$

=> $f = \frac{1}{2 * 0.600 } * \sqrt{\frac{440}{\frac{0.0056}{0.600} } }$

=> $f = 181 \ Hz$

4 0

3 years ago

ayo 20 m long is heated from a temperature of 5 degree celsius to 55 degree celsius if the change in length is 0.020 m calculate

Lera25 [3.4K]

Answer:2 x 10^-5K^-1

Explanation:

The solution is in the attached file

6 0

3 years ago

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement

eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

$a(t)=5+4t-2t^2$

Also, $a=\dfrac{dv}{dt}$

$v=\int\limits {a.dt}$

$v=\int\limits {(5+4t-2t^2).dt}$

$v=5t+2t^2-\dfrac{2}{3}t^3+c$

At t = 0, $v(0)=3\ m/s$ . So, c = 3

$v=5t+2t^2-\dfrac{2}{3}t^3+3$

Also, $v=\dfrac{ds}{dt}$ , s is the position

$s=\int\limits {v.dt}$

$s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}$

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'$

At t = 0, $s(0)=10\ m$ . So, c' = 10

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10$

At t = 4 s

$s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10$

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

5 0

3 years ago

When the shuttle bus comes to a sudden stop to avoid hitting a dog, it decelerates uniformly at 4.5 m/s2 as it slows from 9.3 m/

saw5 [17]

V=u+at
V= speed 0 m/s
u = initial speed 9.3 m/s
a = acceleration - 4.5 (negative)
0=9.3-4.5t
t=9.3/4.5= ? :) seconds

3 0

3 years ago

A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t

Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

F = mg sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

W = F d = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

W = K = 1/2 mv ^2

Solve for v :

v = √(2W/m) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0

3 years ago