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3 years ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid.

Physics

1 answer:

kolbaska11 [484]3 years ago

5 0

Answer:

The terminal velocity of the diver is 115 m/s = 414 km/hr

Explanation:

At terminal velocity,

Fnet = mg - Fd = 0

Drag force, Fd = cρAv²/2

mg = cρAv²/2

Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by

v = √(2mg/ρcA)

where m = mass of body falling through fluid = 80 kg

g = acceleration due to gravity = 9.8 m/s²

ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³

c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7

A = the area of the diver facing the fluid = 0.14 m²

v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr

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if a student lifts their weight of 450 newtons up a set of stairs 5 meters high, how much work did they do?

irina1246 [14]

Answer:

2,250J

Explanation:

W = Fs = (450)(5) = 2,250

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2 years ago

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An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago

mario62 [17]

Answer:

An increase in air temperature because of its compression.

Explanation:

The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.

$P = kT$

<em>where P: is the gas pressure, T: is the gas temperature and k: is a constant.</em>

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.

I hope it helps you!

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3 years ago

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d

yulyashka [42]

Answer:

a) F = 3.2 10⁻¹⁰ N , b) v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

F = q E

The electric field is related to the potential reference

V = E d

E = V / d

Let's replace

F = e V / d

Let's calculate

F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

v² = v₀ + 2 a d

v = √ 2 ad

Acceleration can be found with Newton's second law

e V / d = m a

a = e / m V / d

a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

v = √ (98,448 10¹⁴)

v = 9.9 10⁷ m / s

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2 years ago

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belka [17]

Answer:

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Explanation:

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