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2 years ago

8

A solid cylinder of mass 12.0 kg and radius 0.250 m is free to rotate without friction around its central axis. If you do 75.0 J

of work on the cylinder to increase its angular speed, what will be its final angular speed if the cylinder starts from rest?

1 answer:

faltersainse [42]2 years ago

8 0

Answer:

20 rad/s

Explanation:

mass, m = 12 kg

radius, r = 0.250 m

Moment of inertia of cylinder, I = 1/2 mr²

I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2

Work done = Change in kinetic energy

Initial K = 0

Final K = 1/2 Iω²

W = 1/2 Iω²

ω² = 2W/ I = 2 x 75 / (0.375)

ω = 20 rad/s

Thus, the final angular velocity is 20 rad/s .

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Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads north, at 8 AM, and walks steadily at 2 mi

polet [3.4K]

Answer:

2.57 hours

Explanation:

Let t (hours) be the times it takes for Elsie to walk until they are 25 miles apart. Since Steve is 2 hours earlier, the time it takes for him is t + 2

Distance Steve covers to the North is $s_s = 2(t + 2)$

Distance that Elsie covers to the West is $s_e = 2.5t$

Distance between Steve and Elsie is

$\sqrt{s_s^2 + s_e^2} = \sqrt{(2(t+2))^2 + (2.5t)^2} = 25$

We can solve for t by raise the power on both sides to the 2nd

$(2(t+2))^2 + (2.5t)^2 = 25^2 = 625$

$4(t+2)^2 + 6.25t^2 = 625$

$4(t^2 + 4t + 4) + 6.25t^2 = 625$

$10.25t^2 + 16t - 609 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{-16\pm \sqrt{(16)^2 - 4*(10.25)*(-109)}}{2*(10.25)}$

$t= \frac{-16\pm68.74}{20.5}$

t = 2.57 or t = -4.13

Since t can only be positive we will pick t = 2.57 hours

8 0

2 years ago

You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac

zaharov [31]

Answer:

Explanation:

Given

Original Frequency $f=723\ Hz$

apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v_s^2-0=2\times 9.8\times s$

$s=\frac{12.8^2}{2\times 9.8}$

$s=8.35\ m$

5 0

2 years ago

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the

Ratling [72]

Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.

5 0

2 years ago

Read 2 more answers

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

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2 years ago

An antibaryon composed of two antiup quarks

Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have $+\frac{2}{3} e$ charge where as down, strange and bottom quarks have $-\frac{1}{3}e$ charge.

The antiparticle of up quark is antiup quark and has charge $-\frac{2}{3}e$ charge.

The antiparticle of down quark is antidown quark and has charge $+\frac{1}{3}e$ charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

$2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e$

Thus, antibaryon has -1e charge.

5 0

2 years ago