Finding velocity when onlu a mass and acceleration is given

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To find the velocity of the instruments released from the weather balloon after 10 seconds, the initial velocity is assumed to be 0 m/s. The acceleration calculated from the upward force of 98 N and the mass of 5.0 kg is 1.96 m/s². Using the equation V(sub f) = V(sub i) + at, with t set to 10 seconds, the final velocity can be determined. It’s important to clarify the net force acting on the instruments after accounting for gravity, which affects the overall acceleration. The discussion emphasizes understanding the relationship between force, mass, and acceleration to solve for velocity at the moment of release.
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Homework Statement


The instruments attached to a weather balloon have a mass of 5.0kg.The balloon is released and exerts an upward force of 98 N on the instruments. I have found that the acceleration is 1.96 m/s/s by F=ma.
After the balloons have accelerated for 10 seconds, the instruments are released. What is the velocity of the instruments at the moment of their release?


Homework Equations


V(sub f)^2=V(sub i)^2 + 2ad
I have tried this but I don't know either velocity or the distance.
or would it be:
V(sub f) = V(sub i) +at

The Attempt at a Solution


V(sub f)^2= 0^2 + (2)(1.96m/s^2)(d)
or
V(sub f)= 0 + (1.96m/s^2)(10s)
 
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Be careful, that's not the acceleration. How did you figure out what the acceleration was?

Once you know the acceleration, you also know that

v = v_0 + at

It's assumed to be released from rest, so v_0 = 0. It's a simple matter of letting t = 10 sec. Remember, the question is quite simple since it only asks for what happens at the release.
 
Oh okay. I looked at the calculator wrong for the decimal. So the initial velocity is 0?
 
The force exerted by the balloon on the instruments is 98N, but what is the net force after accounting for gravity?
 
is it 10?
 
How did you arrive at that? E.G. which equation did you use?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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