Question

A boy on board a cruise ship drops a 30. 0 gm marble into the ocean. If the resistive force proportionality constant is 0. 500 kg/s. What is the terminal speed of the marble in m/s?.

Answer 1

The terminal speed of the marble is 0.588 m/s.

Calculation:

We know that,

F = mg                                                   ......(1)

where,

F = force

m = mass

g = acceleration due to gravity

Also,

v = F/k                                                       ......(2)

where,

v = terminal speed

k = proportionality constant

Substituting the value of F from equation (1) in equation (2)

v = mg/k                                                         .......(3)

Given,

m = 30 g = 0.030 kg

k = 0.500 kg/s

g = 9.8 m/s²

To find,

v =?

Put the values in equation (3)

v = mg/k              

v = 0.03(9.8)/ 0.500

  = 0.294/0.500

  = 0.588 m/s

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Hence, the terminal speed of the marble is 0.588 m/s.

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Answer 2

0.588 m/s

0.588 m/s is the terminal speed of the marble.

Calculation:

We know that,

F = mg  ----> eq(1)

where,

F = force

m = mass

g = acceleration due to gravity

We also know,

v = F/k ----> eq(2)

where,

v = terminal speed

k = proportionality constant

On substituting the value of F from eq(1) in eq(2)

v = mg/k  

Given,

m = 30 g = 0.030 kg

k = 0.500 kg/s

g = 9.8 m/s²

we need to find,

v =?

substitute the values in equation (3)

v = mg/k            

v = 0.03(9.8)/ 0.500

 = 0.294/0.500

 = 0.588 m/s

Hence, the terminal speed of the marble is 0.588 m/s.

To learn more about calculation of force visit:

brainly.com/question/15562875

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