Answer:
showm
Explanation:
Consider a dipole having magnetic moment 'm' is placed in magnetic field
then the torque exerted by the field on the dipole is
![\tau = m\times B](https://tex.z-dn.net/?f=%5Ctau%20%3D%20m%5Ctimes%20B)
![\tau=mBsin\alpha](https://tex.z-dn.net/?f=%5Ctau%3DmBsin%5Calpha)
Now to rotate the dipole in the field to its final position the work required to be done is
![U=\int \tau d\alpha](https://tex.z-dn.net/?f=U%3D%5Cint%20%5Ctau%20d%5Calpha)
![U=\int mBsin\alpha d\alpha](https://tex.z-dn.net/?f=U%3D%5Cint%20mBsin%5Calpha%20d%5Calpha)
![U= -mBcos\alpha](https://tex.z-dn.net/?f=U%3D%20-mBcos%5Calpha)
![U=-\vec{m\times \vec{B}}](https://tex.z-dn.net/?f=U%3D-%5Cvec%7Bm%5Ctimes%20%5Cvec%7BB%7D%7D)
Minimum energy mB is for the case when m is anti parallel to B.
Minimum energy -mB is for the case when m is parallel to B.
Answer:
98N and 147N
Explanation:
We have the following information:
![m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N](https://tex.z-dn.net/?f=m%3D50kg%5C%5C%5Cmu_s%20%3D0.4%5C%5C%5Cmu_k%20%3D%200.2%5C%5CF%3D140N)
We can find the static fricton force as follow,
![F=\mu_s * N](https://tex.z-dn.net/?f=F%3D%5Cmu_s%20%2A%20N)
Where N is the normal force (mg)
![F=0.3*50*9.8\\F=147N](https://tex.z-dn.net/?f=F%3D0.3%2A50%2A9.8%5C%5CF%3D147N)
Static friction force at 147N is greater than the force applied hence body does not move.
![F=\mu_k N = 0.2*50*9.8= 98N](https://tex.z-dn.net/?f=F%3D%5Cmu_k%20N%20%3D%200.2%2A50%2A9.8%3D%2098N)
Explanation:
<h2>Yes!</h2>
<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>
Answer:
![\Delta \theta = 56\,rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%2056%5C%2Crad)
![\Delta \theta \approx 3208.564^{\circ}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%5Capprox%203208.564%5E%7B%5Ccirc%7D)
Explanation:
El ángulo barrido en el intervalo de tiempo dado es (The covered angle in the given time interval is):
![\Delta \theta = \omega \cdot \Delta t](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Comega%20%5Ccdot%20%5CDelta%20t)
![\Delta \theta = \left(14\,\frac{rad}{s} \rjght)\cdot (4\,s)](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Cleft%2814%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%5Crjght%29%5Ccdot%20%284%5C%2Cs%29)
![\Delta \theta = 56\,rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%2056%5C%2Crad)
![\Delta \theta \approx 3208.564^{\circ}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%5Capprox%203208.564%5E%7B%5Ccirc%7D)