<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
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The angular velocity of the orbit about the sun is:
w = 1 rev / year = 1 rev / 3.15 × 10^7 s
Now in 1 rev there is 360° or 2π rad, therefore:
w = 2π rad / 3.15 × 10^7 s
To convert in linear velocity, multiply the rad /s by the
radius:
v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles
<span>v = 18.55 miles / s = 29.85 km / s</span>
Centripetal acceleration = (speed)² / (radius) .
Force = (mass) · (acceleration)
Centripetal force = (mass) · (speed)² / (radius) .
= (11 kg) · (3.5 m/s)² / (0.6 m)
= (11 kg) · (12.25 m²/s²) / (0.6 m)
= (11 · 12.25) / 0.6 kg-m/s²
= 224.58 newtons. (about 50.5 pounds)
That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by. It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second. If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
Answer:
a = 15.1 g
Explanation:
The relation between mass and acceleration is given by :
If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂
So,
So, the car's acceleration would be 15.1 g.
Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²
