<span> For any body to move in a circle it requires the centripetal force (mv^2)/r.
In this case a ball is moving in a vertical circle swung by a mass less cord.
At the top of its arc if we draw its free body diagram and equate the forces in radial
direction to the centripetal force we get it as T +mg =(mv^2)/r
T is tension in cord
m is mass of ball
r is length of cord (radius of the vertical circle)
To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So
minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s
In the second case the speed of ball at top = (2*3.285) =6.57 m/s
Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom
we get velocity at bottom as 9.3m/s.
Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force
T-mg=(mv^2)/r
We get tension in cord T=13.27 N</span>
No. The moon always keeps the same side facing us. Its rotation and revolution periods are equal.
Answer:
The magnetic field will be
, '2d' being the distance the wires.
Explanation:
From Biot-Savart's law, the magnetic field (
) at a distance '
' due to a current carrying conductor carrying current '
' is given by
![\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Coverrightarrow%7BB%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI%7D%7B4%20%5Cpi%7D%7D%20%5Cint%20%5Cdfrac%7B%5Coverrightarrow%7Bdl%7D%20%5Ctimes%20%5Chat%7Br%7D%7D%7Br%5E%7B2%7D%7D%7D)
where '
' is an elemental length along the direction of the current flow through the conductor.
Using this law, the magnetic field due to straight current carrying conductor having current '
', at a distance '
' is given by
![\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}](https://tex.z-dn.net/?f=%5Clarge%7B%5Coverrightarrow%7BB%7D%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI%7D%7B2%20%5Cpi%20d%7D)
According to the figure if '
' be the current carried by the top wire, '
' be the current carried by the bottom wire and '
' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the
symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by
symbol.
Given
and ![\large{I_{B} = 12.5 A}](https://tex.z-dn.net/?f=%5Clarge%7BI_%7BB%7D%20%3D%2012.5%20A%7D)
Therefore, the magnetic field (
) at 'P' due to the top wire
![B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}](https://tex.z-dn.net/?f=B_%7Bt%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI_%7Bt%7D%7D%7B2%20%5Cpi%20d%7D)
and the magnetic field (
) at 'P' due to the bottom wire
![B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}](https://tex.z-dn.net/?f=B_%7Bb%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI_%7Bb%7D%7D%7B2%20%5Cpi%20d%7D)
Therefore taking the value of
the net magnetic field (
) at the midway between the wires will be
![\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T](https://tex.z-dn.net/?f=%5Clarge%7BB_%7BM%7D%20%3D%20%5Cdfrac%7B4%20%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%7D%7B2%20%5Cpi%20d%7D%20%28I_%7Bt%7D%20-%20I_%7Bb%7D%29%20%3D%20%5Cdfrac%7B2%20%5Ctimes%2010%5E%7B-7%7D%7D%7Bd%7D%20%3D%20%5Cdfrac%7B41.4%20%5Ctimes%2010%20%5E%7B-4%7D%7D%7Bd%7D%7D%20T)
Kinetic Energy would have to be the answer. Pretty sure.
Answer:
C.) The same as one
Explanation:
Because the Ferris wheel has two bars at the side that keep it in place so it doesn't roll away. The Ferris wheel only moves in one place. If the ferris wheel did roll then you would have a greater than one. If it did roll it would be dangerous.