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3 years ago

how do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate

1 answer:

3 0

While skydiving, its not just freely falling under Earth's gravity. Additional force called drag acts against the gravity which slows down the rate of fall. Drag is caused by the air molecules which pushes against the body as it falls through them. This is actually a significant amount of force which slows down the rate of fall of the body. Drag depends on the contact surface area and weight. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position because of the less contact surface area of the body with the air molecules while in the former case. No two persons have identical body shape and weight. Hence, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

You might be interested in

Which forces tend to slow down an object

laila [671]

The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop.

Hope this helps!

8 0

3 years ago

Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

u = 0 m/s

g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

b) impact velocity

v = √(2gh)

v = √(2× 9.8 × 23.72)

v = √464.912

v = 21.56 m/s²

6 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector $X = 7.8*Cos(50) = 5.01 Km$ y $y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km)$ so that $Tan (\alpha ) = \frac{0.375}{5.01} = 4.28$ ; $\alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree)$ to get how far we use Pythagorean theorem so $R^{2} = x^{2}+y^{2}$ so that $R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)$

6 0

3 years ago

Read 2 more answers

What is Initial temperature and final temperature equations?? ...?

Neporo4naja [7]

One that can help you is:
ΔT=TFinal−TInitial
That is of course adding both tmepratures. There is one more that is a lil bit more complex
Tf=Ti−ΔHrxn∗nrxn/(Cp,water∗mwater)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps

8 0

3 years ago

-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the

Nina [5.8K]

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM given as

V(max) = ω A

ω=Angular speed

A=Amplitude

$\omega =\dfrac{3.9}{0.165}\ rad/s$

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

5 0

3 years ago