karymediaz17 karymediaz17

04-11-2020
Physics

contestada

A2.00 kg block is pulled across a
flat, frictionless floor with a 4.17 N
force directed 40.0° above
horizontal. What is the normal force
acting on the block?
(Hint: It is NOT = mg)
normal force (ND)

Respuesta :

zombiebrain0325 zombiebrain0325

05-11-2020

Answer:

Explanation:

R + 4.17sin(40) = mg

R = mg - 4.17sin(40)

= 2(9.8) - 4.17sin(40)

= 16.91957567

= 16.92 N

Answer Link

alexloveadoptme222 alexloveadoptme222

14-07-2021

Explanation:

R + 4.17sin(40) = mg

R = mg - 4.17sin(40)

= 2(9.8) - 4.17sin(40)

= 16.91957567

= 16.92 N

Answer Link

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