A raindrop of mass 57.40 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, 17.50 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.

To calculate the weight of the raindrop after it reaches terminal velocity, we first need to determine the mass of the raindrop. Given that the raindrop has a mass of 57.40 mg, we need to convert it to kilograms before calculating the weight.

1 milligram (mg) = 1/1000 gram (g)
1 gram (g) = 1/1000 kilogram (kg)

Therefore, the mass of the raindrop (m) in kilograms is:

m = 57.40 mg * (1 g / 1000 mg) * (1 kg / 1000 g)
m = 57.40 * 1/1000 * 1/1000 kg
m = 57.40 * 10^-6 kg

Now that we have the mass of the raindrop, we can calculate its weight using the formula:

Weight (W) = mass (m) * acceleration due to gravity (g)

Given that g = 10 m/s^2, we can substitute the values into the formula:

W = m * g
W = 57.40 * 10^-6 kg * 10 m/s^2
W = 57.40 * 10^-5 N

Therefore, the weight of the raindrop after it reaches terminal velocity but before it reaches the ground is 57.40 * 10^-5 N.

To calculate the weight of the raindrop after it reaches terminal velocity and before it reaches the ground, we need to find the gravitational force acting on it.

The weight of an object can be calculated using the formula:

Weight = mass * gravitational acceleration

Given:
Mass of the raindrop = 57.40 mg = 57.40 * 10^-6 kg
Gravitational acceleration (g) = 10 m/s^2

Weight = mass * g

Weight = (57.40 * 10^-6) kg * 10 m/s^2

Weight = 0.574 * 10^-4 kg * 10 m/s^2

Weight = 0.00574 N

Therefore, the weight of the raindrop after it reaches terminal velocity and before it reaches the ground is 0.00574 N.