Question

5 The diagram shows an experiment to measure the force exerted on a ball by a horizontal air flow.

The ball is suspended by a light string and weighs 0.15N.
The deflection of the string from vertical is 30°.
What is the force on the ball from the air flow?

A 0.075N B 0.087N C 0.26N D 0.30N

Answer 1

A. 0.075

Answer 2

F=0.15sin 30

Answer 3

The answer is B

Answer 4

how did u get b?

Answer 5

I don't know how to get it, but that is what the marking scheme says

Answer 6

i think u have to round off

Answer 7

Hmmm

Answer 8

do u know how i get this or not?

Answer 9

Yeah i do, but it's a wrong answer

Answer 10

no.. the answer is right
u just have to round off 0.075 to 0.08

Answer 11

Find T from vertical then find F from horizontal

Answer 12

Wow! Thanks alot!

Answer 13

bt that doen't give 0.08N

Answer 14

@jack.sparrow

Answer 15

It does,
\[Y:~~~ Tcos(30)=0.15 \\ ~~~~~~~~~~~~~~~~~~~~T=\frac{0.15}{\cos(30)}\]
\[X:~~~F=Tsin(30) \\ ~~~~~~~~~F=\frac{0.15}{\cos(30)}\sin(30)=0.15\tan(30)=0.087\text{N}\]