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Nadya [2.5K]
3 years ago
7

A ball of mass 0.35 kg hangs straight down on a string of length 13 cm. It is then swung upward, keeping the string taut, until

the string makes an angle of 70° with respect to the vertical. Find the change in the gravitational potential energy of the Earth-ball system.

Physics
1 answer:
Anni [7]3 years ago
6 0

Answer:

PE=0.29J

Explanation:

According to the description, there is a angle and in point swung upward of 70°

So,

Y=13*10^{-2}*cos(70) \\Y=0.0444m

Appling the equation of Potential Energy we have,

PE=mgh\\PE=(0.35)(9.8)(0.13-0.0444)\\PE=0.29J

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Una rueda que tiene 15 cm de radio, realiza 64 vueltas en 16 seg. Calcula: Periodo Frecuencia Velocidad angular Velocidad lineal
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Answer:

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Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

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m2·kg·s-2

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