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3 years ago

8

Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the fina

l rotational kinetic energy, Kf, of the two spinning disks?'

1 answer:

ikadub [295]3 years ago

4 0

Answer:

See attached picture.

Explanation:

See attached picture for explanation.

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Question 11 of 11 | Page 11 of 11

KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

T = 2π/ω

By substitution,

<em>T = 2π(R +h)√[(R + h)/GM] </em>

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

3 0

3 years ago

Can you get cool powers.

Morgarella [4.7K]

Answer: Sadly no I wish

Explanation:

3 0

3 years ago

What a neurology professor does = _______ brains

Natasha_Volkova [10]

That’s really easy ask your teacher and also peace happy

3 0

3 years ago

An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is

Ulleksa [173]

Answer:

1. a $W_t=746.63 J$

b $E_p=212.415 J$

c $W_n=183.96J$

2. $T_e=99.71J$

Explanation:

a). The work done by the tension is:

$W_t=T*dt$

$dt=\frac{5.1}{cos(17)}$

$W_t=140N*\frac{5.1}{cos(17)}$

$W_t=746.63 J$

b). The work done potential of gravity

$E_p=m*g*h$

$h=5.1*sin(17)$

$E_p=8.5kh*9.8*5.1*sin(30)$

$E_p=212.415 J$

c). The work done by the normal force

$W_n=N*d_n$

$d_n=5.1*sin(30)=2.55$

$W_n=8.5kg*9.8*cos(30)*2.55$

$W_n=183.96J$

2. The increase in thermal energy is:

$T_e=F*d$

$F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)$

$F_k=19.5N$

$T_e=19.55*5.1m$

$T_e=99.71J$

4 0

3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

8 0

4 years ago