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Arte-miy333 [17]
3 years ago
14

How many forces are acting on a stationary raft floating in a swimming pool A1 B2 or C3

Physics
1 answer:
Hitman42 [59]3 years ago
3 0
There are at least two forces on it, and there could be more.

Vertical forces:
-- gravity, directed downward
-- buoyant force, directed upward
These two forces must be exactly equal, so that the net
vertical force on the raft is zero.  Otherwise, it would be
accelerating either up or down.

Horizontal forces:
We know that the net horizontal force on the raft is zero. 
Otherwise, it would be accelerating horizontally.

But we don't know if there are actually no horizontal forces
at all, or a balanced group of horizontal forces, that add up
to a net force of zero.
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A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo
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Volume of balloon =  1000 cm^3

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 The head of a normal person can be assumed as a sphere with radius 10 cm.

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  Approximate volume of head =\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3

 In the given options the closest value to the approximate volume is 1000 cm^3.

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2 years ago
9. A plane starts at rest & accelerates along the ground before takeoff. It
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  9.877 m/s^2

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A force of 10 N pressing on an area of 2m2. Calculate the pressure please answer fast it’s a test rn
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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
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Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' 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 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' 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 \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

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B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

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3 years ago
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