Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

s=1/2at^2 you have s and t find a. net force=ma. plug and solve.
What is s?
s-->distance travelled a--> acceleration due to gravity here t-->time taken

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Okay, thanks:)
cheers!
I 431.2N but that isn't an option..
43.12 N is the answer.
So this is what I did.. s=12m a=? t=1.75s 12=((1.75)^2a)/2 12=(3.0625a)/2 24=3.0625a a=7.84 netforce=55*7.84 netforce=431.2N
Net acceleration. a = 2*12/1.75² Net force ma = 5.5* 2*12/1.75² = 43.12 N
I get what your saying now. But the only problem is my choices are 120N, 54N, & 0N. But I can rule out 0N automatically
hmm. no idea. sorry.
Its okay, thanks for your help though :) I really do appreciate it
you're welcome. good luck. :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question