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Katena32 [7]
2 years ago
14

a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by

a fan standing in the bleaches 30 meters above the height at which it was hit
Physics
1 answer:
lisov135 [29]2 years ago
4 0

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

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You are on an interstellar mission from the Earth to the 8,7 light-years distant star Sirius. Your spaceship can travel with 70%
marishachu [46]

Answer:

Time = 12.43 years

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Explanation:

Given the following :

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Density = 1 hydrogen atom/m^3

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A.) time it takes your spaceship to reach Sirius :

From the relation: Speed = (distance / time)

Time = distance / speed

Time = (70/100) × 1 light year

Distance = 8.7 light years

Time = 8.7 / 0.7 = 12.4285 years

Time = 12.43 years

B.) Mass of inter-stellar gas that collides with the spaceship can be calcuted by finding the product of the surface area of the cylindrical space ship and the mass of proton.

That is ;

surface area * mass of proton

Surface area of a cylinder = 2πrh + πr^2×Me

= 2πrh + Me×πr^2)

=( 2 × 22/7 × 3 × 25) + (22/7 × 3^2 × 1.673 * 10^-27)

= 2× 235.714 + 28.285) × 1.673 * 10^-27

= 471.428 + 47.31 * 10^-27

= 471.428 + 4.73 × 10^-26

= 4.71.43kg approximately

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2 years ago
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Answer:

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Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

v_o=\sqrt{\frac{GM}{r}}

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

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Answer:

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21.0+10.0=31.0 km

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