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3 years ago

Why the dust particles undergoes small random movements? Plz help!!?

Physics

1 answer:

Rashid [163]3 years ago

8 0

Brownian motion<span> or pedesis is the </span>random motion<span> of particles suspended in a fluid </span>

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A ball starts from rest and rolls down a hill at a constant acceleration of 5 m/s2. If it travels for 8 m how fast is it going i

makkiz [27]

Hi there!

We can use the following kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = final velocity (? m/s)
vi = intial velocity (0 m/s)

a = acceleration (5 m/s²)

d = displacement (8 m)

Plug in the givens and solve.

$v_f^2 = 0 + 2(5)(8)\\\\v_f = \sqrt{80} = \boxed{8.944 \frac{m}{s}}$

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2 years ago

Can you see gas in a bottle

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Depends on what type of gass

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3 years ago

Read 2 more answers

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0

3 years ago

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

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3 years ago

An unstretched spring has a length of 0.30 m. When the spring is stretched to a total length of 0.60 m, it supports traveling wa

IceJOKER [234]

Explanation:

Below is an attachment containing the solution

7 0

3 years ago