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3 years ago

Brownian motion is due to:

Physics

1 answer:

My name is Ann [436]3 years ago

8 0

Answer:

Explanation:

Brownian motion is the random movement of particles in a fluid due to their collisions with other atoms or molecules. ... Brownian motion takes its name from the Scottish botanist Robert Brown, who observed pollen grains moving randomly in water. He described the motion in 1827 but was unable to explain it.

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Greg dropped a coin from the h meter tall building. If it takes t seconds to reach the ground, determine the position of the coi

My name is Ann [436]

consider the motion of the coin in vertical direction :

Assuming down direction as positive and top of building as origin

Y₀ = initial position of displacement of the coin at the top of building = 0 m

Y = final position of displacement of the coin as it hits the ground = h

t = time taken to hit the ground

a = acceleration = acceleration due to gravity = g

v₀ = initial velocity at the top of the building = 0 m/s

using the equation

Y = Y₀ + v₀ t + (0.5) a t²

h = 0 + 0 t + (0.5) g t²

h = (0.5) g t² eq-1

consider the motion of the coin for time "t/2"

Y₀ = initial position of displacement of the coin at the top of building = 0 m

Y' = final position of the coin after time "t/2"

t' = time of travel = t/2

a = acceleration = acceleration due to gravity = g

v₀ = initial velocity at the top of the building = 0 m/s

using the equation

Y' = Y₀ + v₀ t' + (0.5) a t'²

Y' = 0 + 0 t + (0.5) g (t/2)²

Y' = (0.5) gt²/4

using eq-1

Y' = h/4 since h = (0.5) g t²

8 0

4 years ago

Trial.

olga nikolaevna [1]

Answer:

Explanation:

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4 0

3 years ago

A 250 g air-track glider is attached to a spring with springconstant 4.0 N/m. Th damping constant due to air resistance is0.015

vaieri [72.5K]

Answer:

33.33 seconds

Explanation:

$N=\dfrac{1}{e}N_0$

$N_0$ = Initial length pulled = 20 cm

b = Damping constant = 0.015 kg/s

k = Spring constant = 4 N/m

m = Mass of glider = 250 g

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.25}{4}}\\\Rightarrow T=1.57079632679\ s$

Using exponential decay formula

$N=N_0e^{\dfrac{-bt}{m}}$

Final amplitude = Initial times decay

$\dfrac{1}{e}0.2=0.2e^{\dfrac{-0.015t}{2\times 0.25}}\\\Rightarrow 0.2=0.2e^{\frac{-0.015t}{2\cdot \:0.25}+1}\\\Rightarrow e^{\frac{-0.015t}{2\cdot \:0.25}+1}=1\\\Rightarrow \ln \left(e^{\frac{-0.015t}{2\cdot \:0.25}+1}\right)=\ln \left(1\right)\\\Rightarrow \left(\frac{-0.015t}{2\cdot \:0.25}+1\right)\ln \left(e\right)=\ln \left(1\right)\\\Rightarrow \frac{-0.015t}{2\cdot \:0.25}+1=\ln \left(1\right)\\\Rightarrow -\frac{0.015t}{0.5}=-1\\\Rightarrow -0.000225t=-0.0075\\\Rightarrow t=33.33\ s$