Ask question

Ask question

All categories

3 years ago

13

A hollow spherical shell has mass 8.05kg and radius 0.215m . It is initially at rest and then rotates about a stationary axis th

at lies along a diameter with a constant acceleration of 0.895rad/s2 .
What is the kinetic energy of the shell after it has turned through 5.25rev ?

Express your answer with the appropriate units.

2 answers:

hram777 [196]3 years ago

4 0

Answer: The kinetic energy of the shell after it has turned through 5.25rev is 7.32J

Explanation: Please see the attachments below

Neporo4naja [7]3 years ago

3 0

Answer:

930.21J

Explanation:

The kinetic energy of a solid body that is rotating, is given by:

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

where I is the moment of inertia and w is the angular velocity.

The moment of inertia for a spherical shell is:

$I=\frac{2}{3}MR^2$

where M is the mass and R is the radius of the sphere. By replacing we have

$I=\frac{2}{3}(8.05kg)(0.215m)^2=0.55\ kgm^2$

To calculate w we have to use the equation

$\omega^2=\omega_0^2+2\alpha \theta\\\\\omega=\sqrt{2(0.895rad/s^2)(5.25rev*\frac{360\°}{1\ rev})}=58.16\frac{rad}{s}$

where we have taken w0=0 rad/s.

Finally, by replacing I and w we obtain:

$E_k=\frac{1}{2}I\omega^2 =\frac{1}{2}(0.55\ kgm^2)(58.16\frac{rad}{s})^2=930.21J$

HOPE THIS HELPS!

You might be interested in

How much time would be saved if the delicious chicken breast were thawed on the kitchen counter instead, given that room tempera

Art [367]

The time that would be saved if the delicious chicken breast were thawed on the kitchen counter instead, given that room temperature is around 69 F could be calculated by :
tf - ti = 5.7

hope this helps

5 0

3 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

3 years ago

What is the name of the specification that indicates the frequencies that are dedicated for data transmission over cable lines?

mihalych1998 [28]

The answer to this question is the term
DOCSIS. A DOCSIS or Data Over Cable Service Interface Specification is a telecommunication standard or interface where in an internet was being provided by the use of cables. The advantage of using DOCSIS is that the speed of the internet is faster using this kind of interface.

4 0

3 years ago

Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr

nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

$\frac{E}{t} = VI$

solving for t

$t = \frac{E}{VI}$

$t = \frac{m_w C_w \Delta T}{VI}$

$t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}$

t = 444.125 sec

5 0

4 years ago

Read 2 more answers

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0

3 years ago