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Ilya [14]
2 years ago
6

A ball is thrown horizontally from a cliff at a speed of 10 m/s. you predict that its speed 1 s later will be slightly greater t

han 14 m/s. your friend says it will be 10 m/s. show who is correct.
Physics
2 answers:
Veseljchak [2.6K]2 years ago
6 0
The horizontal component of speed remains constant at 10 m/s.
After 1 s, a vertical component of speed = gt = 9.8(1) = 9.8 m/s is added to the horizontal speed.
The size of the total speed of the ball after 1 s = √(10²+9.8²) = √196.04 ≈ 14.0 m/s <=
You are correct.
Mice21 [21]2 years ago
4 0

Answer:

Speed is slightly greater than 14 m/s.          

Explanation:

The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.

Using first equation of motion, after 1 s, vertical velocity will be:

v = u +at

v=0+(9.8 m/s²)(1 s) = 9.8 m/s

Horizontal velocity = 10 m/s

Net velocity:

V= \sqrt{(9.8)^2+(10)^2}=\sqrt{196.04}=14.001 m/s

Speed is slightly greater than 14 m/s.

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Un resorte se alarga 5 cm bajo la acción de una fuerza de 39,2 N. ¿Cuál es la constante del resorte? Si ahora la fuerza es 68,6
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Answer:

k=784 N/m

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Explanation:

Usando la ley de Hook tenemos:

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Solving it for k we have:

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k=\frac{39,2}{0,05}

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Usando la misma ecuación y sabiendo k tenemos:

\Delta x=\frac{F}{k}

\Delta x=\frac{68,6}{784}

\Delta x=8,8 cm

Espero esto te ayude!

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