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Diano4ka-milaya [45]
4 years ago
13

The average human has a density of 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling. (a) Without making any swimming movem

ents, what percentage of the human body would be above the surface in the Dead Sea (a body of water with a density of about 1230 kg/m3) in each of these cases?
Physics
1 answer:
Gala2k [10]4 years ago
8 0

Answer:

20% of the human body float above the surface in the dead sea.

Explanation:

Given data,

The human density after inhaling, d₁ = 945 kg/m³

The human density after exhaling, d₂ = 1020 kg/m³

Therefore the average human density, d = 982.5 kg/m³

The density of the dead sea, D = 1230 kg/m³

The percentage of density of human body to the dead sea is,

                                   P % = (982.5 / 1230) x 100 %

                                           = 80 %

Therefore, the human body has 80% of dead sea density.

Hence, 20% of the human body float above the surface in the dead sea.

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When you let the air go out of a balloon the balloon moves or shoots like a rocket. Would this happen if there was no surroundin
Pie

Answer:

The balloon would still move like a rocket

Explanation:

The principle of work of this system is the Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.

This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).

3 0
3 years ago
Refer to the attached image!!!​
dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

  • T is time of motion
  • g is acceleration due to gravity
  • θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

6 0
2 years ago
Please help, and show steps. Thank you very much!
Vikentia [17]
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
6 0
4 years ago
A woman wearing athletic clothing outside. The woman is stretching her legs in a lunge position. Based on the nonverbal messages
Inessa05 [86]

Answer:

D

Explanation:

4 0
4 years ago
A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th
lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

7 0
3 years ago
Read 2 more answers
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