A raindrop of mass 61.60 mg (that is milli- gram) falls in quiet air (no wind). Due to the air resistance force, the drop has a maximum velocity called the terminal velocity of magnitude, 11.00 m/s, which is reached high above ground. What would be in N the weight of the rain drop after it reaches terminal velocity and before it reaches ground? Use g = 10 m/s2 and take the upward direction as positive.
You use of the term weight is confusing to me. THe raindrop if falling through the air. If by weight, you mean the force the air is pushing against the drop, and thereby the Earth is pushing against the raindrop, then the answer is at constant velocity mass*g
Frankly, your use of Weight her bothers me, as it is being likely used in some specific problem writers thought.
Consider. You stand on a scale and weigh 110 lbs. You get into an airplane, and jump out. Whether or not you are at constant velocity, the Earth is pulling you with a force of 110lbs. If air is resisting that motion, the acceleration decreases, but the Earth is still pulling on you with 110 lbs. Gravity does not change. If at constant velocity, then air is pushing on you at 110lbs, but the air is also pushing on the Earth by 110 lbs, so net force is still zero between Earth, you, and the air.
To find the weight of the raindrop after it reaches terminal velocity and before it reaches the ground, we need to consider the force equilibrium.
Weight (W) = Mass (m) * Acceleration due to gravity (g)
Given:
Mass (m) = 61.60 mg = 0.06160 g = 0.06160 * 10^(-3) kg
Acceleration due to gravity (g) = 10 m/s^2
To find the weight of the raindrop, we need to determine its mass. The mass remains constant regardless of the velocity. Therefore, we can use the given mass directly.
Weight (W) = m * g
Plugging in the values:
W = 0.06160 * 10^(-3) kg * 10 m/s^2
Simplifying, we calculate:
W = 0.000616 N
Therefore, the weight of the raindrop after it reaches terminal velocity and before it reaches the ground is 0.000616 N.