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kakasveta [241]
3 years ago
11

Sometimes, a crackling sound is heard while taking off a sweater during winters. Explain

Physics
1 answer:
hoa [83]3 years ago
6 0

Answer:

it is somewhat frozen

Explanation:

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A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi
vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel N_1=2\ rev/s

angular speed \omega_1=2\pi N_1=4\pi\  rad/s

mass of wheel m_1=4.5\ kg

diameter of wheel d_1=0.30\ m=30\ cm

radius of wheel r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm

mass of clay m_2=2.8\ kg

the radius of the chunk of clay r_2=8\ cm

Moment of inertia of Wheel

I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2

Combined moment of inertia of wheel and clay chunk

I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2

Conserving angular momentum

\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi

Common frequency of wheel and chunk of clay is

\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s

5 0
3 years ago
My dad gifted me a calculator. I have observed that very small cells are used in a calculator. What are these cells called and w
Morgarella [4.7K]
The cells are called insibatss, I think I spelled that right but
7 0
3 years ago
f1=30Nandf2=10Nact on amass of 2kg if the cofficient of friction between the mass and the horizontal surface is 0.2 what is the
Alexus [3.1K]

The mass is moving with an acceleration of 13.299 m/s².

<h3>What is the straight forward meaning of acceleration?</h3>

The rate at which an object's velocity with respect to time changes is referred to as acceleration in mechanics. They are vector quantities, accelerations.

According to therefore mentioned statement,

the direction of the net force applied on an object determines its acceleration;

according to mentioned data;

F₁=30N

F₂=10N

Net force on object;

F=F₁+F₂

F=F₁y+F₂Cos30°(-x)+F₂ Sinx(-y)

F=3y+10×(√3/2)(-x)+10×(1/2)(-y)

F=ma

ma=-5√3/2(x)+25y

2a=-5√3x/2+25y

a=-5√3x+25y

magnitude(a)=√(-5√3/2)²+(25/2)²

     (a) =13.299m/s²

to know more about acceleration visit;

brainly.com/question/15295474

#SPJ9

6 0
2 years ago
What does soil structure refer to?
Zigmanuir [339]
A this refers to the make up of the soil
6 0
4 years ago
Read 2 more answers
A spring with a spring constant of 165 N/m is attached to a 2.0 kg mass and set into motion.
marin [14]

Explanation:

We have,

Spring constant of the spring, k = 165 N/m

Mass, m = 2 kg

It is required to find the period of the mass-spring system. For the spring mass system, the period is given by :

T=2\pi \sqrt{\dfrac{m}{k} }\\\\T=2\pi \sqrt{\dfrac{2}{165} }\\\\T=0.69\ s

The frequency of vibration is reciprocal of its time period. So,

f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.69}\\\\f=1.44\ Hz

So, the period of the mass-spring system is 0.69 s and frequency is 1.44 Hz.

3 0
4 years ago
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