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____ [38]
4 years ago
6

A student has paper sitting at rest on their desk. Which of the following statements best describes this situation?

Physics
2 answers:
ValentinkaMS [17]4 years ago
8 0

C. The forces acting on the paper are balanced .

-There are several forces acting on your paper, but they balance each other.

nirvana33 [79]4 years ago
3 0

Answer: c

Explanation: hey xdd

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid
serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is r_x=140 km

Distance of astronaut From asteroid Y is r_y=581 km

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

F_x=F_y

\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}

cancel out the common terms we get

\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}

\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2

\frac{M_x}{M_y}=(\frac{140}{581})^2

\frac{M_x}{M_y}=0.05806\approx 0.0581

8 0
3 years ago
Waves begin to "feel bottom" when the depth of water is
LekaFEV [45]
I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.
8 0
3 years ago
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Identify the areas on the image where the force of repulsion is the least.
ArbitrLikvidat [17]

Each magnet has a north pole and a south pole. We know that, from having played with bar magnets in our childhood, that a magnet's north pole will repel another magnet's north pole and attract its south pole.

From this diagram it is easy to see that the two lower bar magnets not only repel each other, but they are quite attracted to each other since their north and south poles are close together.

Therefore the region between the lower two magnets has the least force of repulsion.

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3 years ago
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Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
3 years ago
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