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Dima020 [189]
3 years ago
7

A scientist uses a camera to study the stars.

Physics
2 answers:
Anika [276]3 years ago
6 0
It will be number 3 hope that help
Mars2501 [29]3 years ago
4 0

Answer:

It would most likely be 3!

Explanation:

A scientist taking photos of the stars would be using those photos as a sort of documentation. The only reason to record this is for data or research purposes, meaning that 1 and 2 are incorrect. While 4 is a possibility, it is a fruitless endeavor, as even if you managed to count all the stars in the night sky at one point, it would completely discount every star not visible from that part of the globe, making the data false/pointless. Again, this is just my educated assumption, but I do think the answer is 3. Good luck :D

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A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of th
lapo4ka [179]

Answer:

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

Explanation:

By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:

K =U_{g} (1)

Where:

K - Translational kinetic energy, measured in joules.

U_{g} - Gravitational potential energy, measured in joules.

By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:

\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h

v = \sqrt{2\cdot g\cdot h} (2)

Where:

m - Mass, measured in kilograms.

v - Initial speed, measured in meters per second.

g - Gravitational acceleration, measured in meters per square second.

h - Maximum height of the ball, measured in meters.

If we know that g = 9.807\,\frac{m}{s^{2}} and h = 5\,m, then the initial speed of the ball is:

v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}

v\approx 9.903\,\frac{m}{s}

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

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