Answer:
La tensión es 85.3 N.
Explanation:
Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:

Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.
En donde:
m: es la masa del objeto = 200 g = 0.200 kg
: es la aceleración centrípeta
La aceleración centrípeta viene dada por:

En donde:
ω: es la velocidad angular del objeto = 3 rev/s
r: es el radio = 1.20 m
Entonces, la tensión es:

Por lo tanto, la tensión es 85.3 N.
Espero que te sea de utilidad!
Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Answer:
C
Explanation:
i could be wrong but it seems the most logical
The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of
950 N = (2300 kg) <em>a</em>
<em>a</em> = (950 N) / (2300 kg)
<em>a</em> ≈ 0.413 m/s²
Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of
<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)
<em>v</em> = -0.75 m/s
which means the wind slows the boat down to a velocity of 0.75 m/s westward.
It's 3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.
Does that tell you anything ?
Why not ?