3 years ago

What forces were exerted during the air resistance experiment?

Physics

1 answer:

zepelin [54]3 years ago

6 0

The forces that were exerted during the air resistance was the gravity.

<u>Explanation:</u>

Air resistance is the frictional power air applies against a moving article. As an article moves, air opposition backs it off. The quicker the article's movement, the more prominent the air obstruction applied against it.

The force acting upon the objects dropped in the air was air resistance and gravity. The force acting upon the objects dropped in the vacuum was gravity only.

You might be interested in

A ball is hung from a rope, making a pendulum. When it is pulled 5° to the side, the restoring force is 1.0 N. What will be the

Trava [24]

Answer:

restoring force is 2 N

Explanation:

given data

angle pulled = 5°

force = 1 N

pulled = 10°

to find out

restoring force

solution

we know here force

force = m×g×sinθ ..........1

so here θ is very small so sinθ = θ

1 = mg(5)

mg = 1 /5 ..................2

and

now for 10 degree

we know here θ is small so sinθ = θ

so from equation 1

force = m×g×sinθ

put equation 2 here

force = 1/5 × 10

force = 2 N

so restoring force is 2 N

7 0

3 years ago

You hang a heavy ball with a mass of 94 kg from a platinum rod 3.2 m long by 2.2 mm by 3.1 mm. You measure the stretch of the ro

vlada-n [284]

Answer:

$1.67\times 10^{11} N/m^2$

Explanation:

Mass of the ball, $m=94 kg$

Force = weight of the ball

$F=mg = 94kg\times 9.81 m/s^2=922.14N$

For Young's modulus, substitute the values in the formula which is as follows:

$\frac{F}{A}=Y\frac{\Delta L}{L}\\\Rightarrow Y= \frac{FL}{A\Delta L}\\\Rightarrow Y =\frac{922.14 \times 3.2}{2.2\times 10^{-3}\times 3.1\times 10^{-3} \times 0.002588}=1.67\times 10^{11} N/m^2$