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3 years ago

Please help me with this question

Physics

2 answers:

snow_lady [41]3 years ago

6 0

politics. Famous possible eg is of Werner Heisenberg in WW2. He delayed German attempt to build a nuclear bomb. US did build one ... hiroshima and nagasaki.

debbie may have got skilfully lucky by trial and error

Delvig [45]3 years ago

3 0

politics. Famous possible eg is of Werner Heisenberg in WW2. He delayed German attempt to build a nuclear bomb. US did build one ... hiroshima and nagasaki.

debbie may have got skilfully lucky by trial and error

You might be interested in

A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with m

MAXImum [283]

Answer:

Part(a): The final angular velocity is $\omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b): The ratio of the rotational energies is $\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$ ,showing the the energy of th system will decrease.

Explanation:

Part(a):

If ' $I$ ' be the moment of inertia of an object and ' $\omega$ ' be its angular velocity then the angular momentum ' $L$ ' of the object can be written as

$L = I \omega$

If ' $I_{1}$ ' and ' $I_{2}$ ' be the moment of inertia of the two cylinders and ' $\omega_{1}$ ' and ' $\omega_{2}$ ' be the initial angular velocity of the cylinders and ' $\omega_{1}{'}$ ' and ' $\omega_{2}^'}$ ' be their respective final angular velocity, then from conservation of angular momentum,

$I_{1} \omega_{1} + I_{2} \omega_{2} = I_{1} \omega_{1}^{'} + I_{2} \omega_{2}^{'}$

Given, $\omega_{1} = \omega_{i},~\omega_{2} = 0,~\omega_{1}^{'} = \omega_{2}^{'} = \omega_{f}$ . From the above expression

$&& I_{1} \omega_{i} = (I_{1} + I_{2}) \omega_{f}\\&or,& \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b):

Initial kinetic energy

$K_{i} = \dfrac{1}{2} I_{1} \omega_{i}^{2}$

and Final kinetic energy

$K_{f} = \dfrac{1}{2}(I_{1} + I_{2}) \omega_{f}^{2}$

Substituting the value of $\omega_{f}$ ,

$&& K_{f} = \dfrac{1}{2}(I_{1} + I_{2})\dfrac{I_{1}^{2}\omega_{i}^{2}}{(I_{1} + I_{2})^{2}} = \dfrac{1}{(I_{1} + I_{2})} \dfrac{1}{2}I_{1}\omega_{i}^{2} = \dfrac{1}{(I_{1} + I_{2})} K_{i}\\&\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$

The above expression shows that the ebergy of the system will decrease.

7 0

3 years ago

How much tension must a cable withstand to accellerate a 1400 kg car vertically upward at 0.70 m/s^2

MariettaO [177]

Answer:

Below

Explanation:

The cars upward accel adds to the gravity accel (think about how suddenly heavy you feel when the elevator starts to move upward)

F = tension = ma

= 1400 * ( .7 + 9.81) = 14714 N

7 0

2 years ago

when using parallax distance is calculated from the sun and not earth do you think this matters? explain your answer.

Elden [556K]

When using parallax, astronomers calculate distance from the sun and not earth to improve on the accuracy of their measurement, since parallax angle decreases as star distance increases.

<h3>What is parallax distance?</h3>

Parallax enables astronomers to measure the distances of far away stars by using trigonometry.

<h3>Why does astronomers measure parallax distance from sun?</h3>

As the distance of star increases, the parallax angle decreases, and great degree of accuracy is required for its measurement.

So taking a refence from the earth instead of the sun will impact the accuracy of their measurement.

Thus, when using parallax, astronomers calculate distance from the sun and not earth to improve on the accuracy of their measurement, since parallax angle decreases as star distance increases.

Learn more about parallax distance here: brainly.com/question/2128443

#SPJ1

7 0

1 year ago

At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead

Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

$n_{air}=1\\ n_{water}=1.33=4/3$

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

$\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3$

At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

$v_{water} = \frac{c}{n_{water}} = 0.75c$