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lana66690 [7]
3 years ago
12

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

Physics
1 answer:
professor190 [17]3 years ago
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.

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fenix001 [56]
The effect on the climate is that the climate goes up
7 0
4 years ago
How can people reconcile technology with environmental preservation?​
MatroZZZ [7]
1. Developing renewable energy technology

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Hope this helps :)
5 0
3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
Suppose our apparatus will only allow us to distinguish the first order bright fringe from the central bright spot if they are s
ozzi

Answer:

λ = 8.716 mm

Explanation:

Given:

- d = 10 cm

- Q >= 5 degrees

Find:

- Find the shortest wavelength of light for which this apparatus is useful

Solution:

- The formula that relates the split difference and angle of separation between successive fringes is given by:

                                            d*sin(Q) = n*λ

Where,

λ: wavelength

d: split separation

Q: angle of separation between successive fringes

m: order number.

- Since this apparatus only shows the first order light so m =1

- the shortest possible wavelength corresponds to:

                                            d*sin(Q) = λ

                                            λ = 0.1*sin(5)

                                            λ = 8.716 mm

3 0
3 years ago
Ammonia, NH3, can be made by reacting nitrogen and hydrogen and the equation is N2 + 3H2 --&gt; 2NH3 How many moles of NH3 can b
horrorfan [7]
Make a proportion
3 H2 - 2 NH3
19H2 - x
x = (19x2)/3= 12,666666
8 0
3 years ago
Read 2 more answers
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