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Gala2k [10]
2 years ago
10

A red water balloon is thrown horizontally from the top of a bridge. At the same instant, a yellow water balloon is dropped off

the bridge from the same height. Compare the time of fall for the two balloons. Assume air resistance is negligible.
Physics
1 answer:
Advocard [28]2 years ago
7 0

The time of fall for the two balloons is equal.

Here the conditions given resembles a free fall. Since the air resistance is negligible, gravitational force is the only force acting on both the balloons which means the two balloons are falling in a vacuum. In a vacuum, if two objects of different mass are thrown from the same height, both objects will reach the ground at the same time.

Free fall is nothing but a condition in which an object falls solely under the influence of gravity.

Therefore, the time of fall for both yellow and red water balloons is equal no matter the weight where air resistance is negligible.

To know more about free fall

brainly.com/question/13299152

#SPJ4

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
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Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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3 years ago
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Answer:

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Explanation:

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Which is the 2nd closest star to earth??
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Answer:

Proxima Centauri

Explanation:

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1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su
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Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

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Donde:

v_{o} - Velocidad inicial, en metros por segundo.

v - Velocidad final, en metros por segundo.

g - Aceleración gravitacional, en metros por segundo al cuadrado.

t - Tiempo, en segundos.

Si sabemos que v_{o} = -75\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} y t = 26\,s, entonces la velocidad final del mango es:

v = v_{o}+g\cdot t

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v = -329.982\,\frac{m}{s}

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

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