The answer is C, has negative acceleration.
The answer is True. The amount force exerted by any object is directly proportional to its mass. This means that our planet is exerting more gravitational force to Angelina, and Angelina is also exerting a gravitational force on our planet directly proportional to her mass. Angelina is actually falling towards the center of the earth,and also our planet is also moving towards Angelina, but it seems negligible with respect to Angelina.Our Sun is so massive that it held our planet in its orbit because of its gravitational force.
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)
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13 g —> 0.013 kg
KE = 1/2(m)(v)^2
KE = 1/2(0.013)(8.5)^2
KE = 0.47 J