Ask question

Ask question

All categories

Aleksandr-060686 [28]

2 years ago

8

Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but

with opposite angular velocities ~ω and −~ω. what is the total rotational kinetic energy kr of this system of two disks? 1. 1 2 i ω 2 2. zero 3. none of these

1 answer:

valentina_108 [34]2 years ago

5 0

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

$K=\frac{1}{2}I \omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

$K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2$

For disk 2:

$K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2$

so the total energy is

$K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2$

So, none of the options is correct.

You might be interested in

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

USPshnik [31]

No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.

7 0

2 years ago

Read 2 more answers

(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45

Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m$

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0

2 years ago

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

6 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f

ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c = 25 ms = 25 × 10⁻³ s

The initial temperature $T_1$ = 273 K

The final temperature $T_2$ = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume = 0.4 mV/K × 473 K = 189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2 + (189.2 - 109.2)( 1 - $\mathbf{e^{-t/c}}$ )mV

y(t) = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 × 189.2 mV

= 170.28 mV

170.28 mV = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

170.28 mV - 109.2 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

61.08 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

0.7635 mV = ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

t = 1.44 × 25 × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = $\dfrac{189.2-170.28 }{189.2}$

The error fraction = 0.1

The error fraction = 10%

8 0

2 years ago