Ask question

Ask question

All categories

3 years ago

12

The rotational inertia of a collapsing spinning star changes to 2/5 its initial value. what is the ratio of the new rotational k

inetic energy to the initial rotational kinetic energy (kenew / kei)?

1 answer:

Alexandra [31]3 years ago

4 0

Let K.E $_{i}$ be the intial K.E and K.E $_{new}$ be new K.E,

I $_{new}$ = $\frac{2}{5} I _{i}$

K.E $_{i} = \frac{I_{i} w^{2} }{2}$ --- (i)
K.E $_{new} = \frac{I_{new}w^{2} }{2}$

Therefore,
K.E $_{new} = \frac{2I_{i}w^{2} }{5*2}$ ----(ii)

Dividing i and ii,

$\frac{K.E_{new} }{K.E_{i} } = \frac{2}{5}$

You might be interested in

High school and collage students answers only. <br> Are all compounds molecules? Why or why not?

Shkiper50 [21]

All compounds are molecules but not all molecules are compounds. A molecule is two atoms joined together. A compound is two different atoms joined together.

5 0

3 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

Why water in earthern pot remain cool in summer

hjlf

Answer:

Since in summer, the eastern side do not face the sunlight and hence the water in eastern pot remain cool in summer.

8 0

3 years ago

Read 2 more answers

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

$W_{all}=Change\ in\ kinetic\ energy\ of\ system$

We know that kinetic energy of moving mass given as

$KE=\dfrac{1}{2}mv^2$

Now by pitting the values

$KE=\dfrac{1}{2}mv^2$

$KE=\dfrac{1}{2}\times 0.011\times 29^2$

KE= 4.25 J

So the work done on the ball is 4.25 J

8 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$

$\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}$

Removing $y_o$ and dividing by t (t different of zero)

$\displaystyle 0=v_osin\theta-\frac{gt}{2}$

Then we find the total flight as

$\displaystyle t=\frac{2v_osin\theta}{g}$

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

$\boxed{t=24.5/2=12.25\ seconds}$

4 0

3 years ago