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maxonik [38]
3 years ago
11

A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

s are accurate to three significant digits. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7.) If each skydiver has a frontal area of 0.14 m2, calculate their terminal velocities (in m/s). 52 kg skydiver m/s 95 kg skydiver m/s How long will it take (in s) for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)
Physics
1 answer:
Scilla [17]3 years ago
3 0

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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Sound and light are both found as _____, with a variety of ____. The sun, a source of light waves specifically, releases a type
Novosadov [1.4K]

Answer:

Waves; wavelength; electromagnetic energy; ultraviolet light.

Explanation:

Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.

Sound travels faster through solids than it does through either liquids or gases.

Light wave can be defined as an electromagnetic wave that do not require a medium of propagation for it to travel through a vacuum of space where no particles exist.

Hence, sound and light are both found as waves, with a variety of wavelength. The sun, a source of light waves specifically, releases a type of electromagnetic energy. It can be found as UVA or UVB types. These lights give off different levels of ultraviolet light, some of wich can be harmful.

Additionally, the ultraviolet spectrum is divided into three categories and these are; UVA, UVB and UVC.

6 0
2 years ago
One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri
maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0
3 years ago
Two soccer players try to kick one soccer ball with the same force at the same time in opposite directions. Describe the force a
slamgirl [31]

Answer:

It will do nothing. The forces are balanced and the ball's motion will not change

7 0
3 years ago
an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in
Lostsunrise [7]

Let volume of empty boat be = 100% = 1V

and mass of boat be M

In water 10%, 0.1V of the volume is submerged.

Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V

M leads to 0.1V boat submerging

boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

120kg = 0.6M

M = (120/0.6)kg

M = 200kg.

The mass of the boat is 200kg.

4 0
2 years ago
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction
Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed  = 25.5m/s

Acceleration  = 1.94m/s²

Time  = 2.3s

Unknown:

Final speed of the car  = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

    v = u  + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

 Now insert the parameters and solve;

      v  = 25.5 + (1.94 x 2.3)  = 29.96m/s

3 0
3 years ago
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