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Akimi4
1 month ago
14

The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver

was achieved in a jump from a height of 39,000 m. At this elevation, the density of the air is only 4.3% of the surface density of air at 20∘C. Estimate the terminal speed of a skydiver at this elevation. Suppose the mass of the skydiver is 90 kg and the cross section area of the skydiver is 0.72 m2. You could consider the diver as a cylinder traveling sideways for this problem.
Physics
1 answer:
serg [3.5K]1 month ago
5 0

Answer:

The terminal velocity v_t= 202.96 m/s, approximately 203 m/s.

Explanation:

The formula for terminal velocity is given by

v_t= \sqrt{\frac{2mg}{\rho AC_d} }

Here, C_d represents the drag coefficient for the cylindrical shape, which is 1.15.

The air's surface density at 20°C is

ρ_surface = 1.2041 kg/m³.

At an altitude of 39,000 m, the air density is calculated as

ρ = 4.3/100 × 39000 = 0.05177 kg/m³.

Substituting these values back into the formula yields

v_t= \sqrt{\frac{2\times90\times9.81}{0.05177\times0.72\times1.15} }

The found terminal velocity v_t is 202.96 m/s, or roughly 203 m/s.

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If you are anchored in a fixed spot, and a set of six waves pass underneath you during a 60 second time interval, what is the wa
Ostrovityanka [3204]

Answer:

Explanation:

Within a duration of 60 seconds, six waves are observed.

With a total of 6 waves,

this equates to 3 wavelengths.

As a result,

the period for each wavelength is calculated as 60 divided by 3.

Thus, period = 20 seconds.

According to the frequency-period relationship,

f = 1 / T

f = 1 / 20

f = 0.05 Hz

5 0
1 month ago
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
ValentinkaMS [3465]

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

4 0
1 month ago
A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
Softa [3030]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0
1 month ago
Select the areas that would receive snowfall because of the lake effect.
Keith_Richards [3271]

Result:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

8 0
1 month ago
A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
serg [3582]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We utilize the formula

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the angular momentum magnitude of the person 2 meters prior to jumping onto the merry-go-round?

The equation we apply is

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the angular momentum of the person just before she hops onto the merry-go-round?

We utilize the formula

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular velocity of the merry-go-round after the individual jumps on?

We can apply the Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

At this point, we can determine ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round moves at this new angular speed, what force must the person exert to hold on?

We must calculate the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) When the person is halfway around, they choose to simply release their hold on the merry-go-round to exit the ride.

What is the linear speed of the person the moment they exit the merry-go-round?

we can apply the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular velocity of the merry-go-round after the individual releases their hold?

ω₀ = 1.53 rad/s

It returns to its original angular speed

8 0
22 days ago
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