A is 588N.
b) When she reaches her terminal speed, 10 seconds into the dive, she is no longer accelerating, so the net force on her is zero.
Think of it this way: If the net force were not zero she would continue to accelerate.
c) She is no longer accelerating.
Her acceleration is zero.
40% of the Earth's surface is covered by continental crust.
The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = (
-
) / (
+
)
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
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Answer:
700 mL or 0.0007 m³
Explanation:
P₁ = Initial pressure = 2 atm
V₁ = Initial volume = 350 mL
P₂ = Final pressure = 1 atm
V₂ = Final volume
Here the temperature remains constant. So, Boyle's law can be applied here.
P₁V₁ = P₂V₂

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³