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3 years ago

How did dalton improve the atomic theory more than 2000 years after democritus's hypothesis about atoms?

1 answer:

6 0

<span>Dalton improved the atomic theory by establishing that elements are made of atoms & that all atoms of an element are identical.</span>

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If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob

Natalija [7]

Answer:

Explanation:

Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:

$T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m$

By using the $y=A\sin(\omega t) \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}$

The mean position is the position when <em>y</em> = 0, so:

$\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi$ rad/s

and $v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi}$ in centimeters (cm).

4 0

3 years ago

The planet Jupiter revolves around the Sun in a period of about 12 years (3.79 × 108 seconds). What is its mean distance from th

givi [52]

As we know that gravitational force of Sun will provide the centripetal force to all planets

so here we will have

$F_g = F_c$

$\frac{GMm}{r^2} = m\omega^2 r$

here we know that

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{3.79 \times 10^8}$

$\omega = 1.66 \times 10^{-8}$

now from above equation

$\frac{GM}{r^2} = \omega^2 r$

$\frac{GM}{\omega^2} = r^3$

$\frac{(6.67 \times 10^{-11})(1.99\times 10^{30})}{(1.66\times 10^{-8})^2} = r^3$

$r = 7.8 \times 10^{11} m$

7 0

4 years ago

An originally stationary car with a mass of 1500 kg which velocity of 15 m/s, five second after starting. What is the car’s acce

gladu [14]

15=0+a*5
a=3m/s^2

F=1500*3=4500N

4 0

3 years ago

What kind of friction exists between solid objects moving in water?

OLga [1]

In water, it's drag. Exactly the same as air resistance
only hundreds of times greater.

5 0

3 years ago

why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level

Genrish500 [490]

Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.

5 0

3 years ago