Answer:
By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.
Explanation:
Answer:children burn calories to being a student
Explanation:That mean when a children getting ready to go to high school
Answer:v nxfgdjngdnmgndjfnncnfndngndsnbxzmnfn
Explanation:
The wave property which is independent of all other properties is THE VELOCITY OF A WAVE.
The velocity of a wave is defined as the distance moved by a cyclic motion per unit time. The velocity of a wave is determined by the properties of the medium through which it moves; it does not not depend on the properties of the wave itself.
Answer:
Let's investigate the case where the cable breaks.
Conservation of angular momentum can be used to find the speed.
![\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\](https://tex.z-dn.net/?f=%5Cvec%7BL%7D_1%20%3D%20%5Cvec%7BL%7D_2%5C%5C%5Cvec%7BL%7D_1%20%3D%20m%5Cvec%7Bv_0%7D%20%5C%5C%5Cvec%7BL%7D_2%20%3D%20I%5Cvec%7B%5Comega%7D%5C%5C)
The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>
![I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2](https://tex.z-dn.net/?f=I%20%3D%20I_%7Bprojectile%7D%20%2B%20I_%7Bball%7D%5C%5CI%20%3D%20mr%5E2%20%2B%20mr%5E2%5C%5CI%20%3D%202mr%5E2)
where r is the length of the cable.
<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is
![F = \frac{mv^2}{r}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
The relation between linear velocity and the angular velocity is
![v = \omega r](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20r)
So,
![F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bm%28%5Comega%20r%29%5E2%7D%7Br%7D%20%3D%20m%5Comega%5E2%20r%20%3D%20300%5C%5Cmv_0r%20%20%3DI%5Comega%5C%5C%5C%5C%5Comega%20%3D%20mv_0r%2FI%5C%5C300%20%3D%20m%28%5Cfrac%7Bmv_0r%7D%7BI%7D%29%5E2r%20%3D%20m%28%5Cfrac%7Bmv_0r%7D%7B2mr%5E2%7D%29%5E2r%20%3D%20m%28%5Cfrac%7Bv_0%7D%7B2r%7D%29%5E2r%20%3D%20%5Cfrac%7Bmv_0%5E2r%7D%7B4r%5E2%7D%20%3D%20%5Cfrac%7Bmv_0%5E2%7D%7B4r%7D%5C%5C300%20%3D%20%5Cfrac%7B0.8v_0%5E2%7D%7B4r%7D%5C%5C1500%20%3D%20v_0%5E2%2Fr%5C%5Cv_0%20%3D%20%5Csqrt%7B1500r%7D)
As can be seen, the maximum velocity for the projectile without breaking the cable is
, where r is the length of the cable.