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algol13
3 years ago
9

write a short paragraph that explains why the earth would not be considered a closed system and be sure to include evidence or e

xamples for your answer.Also, suggest a system that would be considered the ultimate closed system.
Physics
2 answers:
Tomtit [17]3 years ago
8 0
The earth would not be considered a closed system. A closed system is a place or a habitat that is self sustaining and does not require any outside help or force. A closed system requires air food water or anything that any organism inside the system would need the earth is not a closed system because humans can easily leave or enter earth via a space ship as well as asteroids crash landing inside earth would be considered as entering. A biosphere is a ultimate closed system (A biosphere is a self sustaining sphere with plants to make air with any other organisms to create carbon dioxide *look up biospheres for more info*)
kati45 [8]3 years ago
6 0

Earth is considered a closed system because only energy, not matter, is exchanged. A closed system like Earth can be easily represented by a sealed jar. When the lid is sealed, matter like air or water can neither enter nor exit the jar. If you placed that same jar in a 100-degree oven, the air inside the jar would eventually increase in temperature to 100 degrees, representing an exchange of energy known as heat.  

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Please help!!
xz_007 [3.2K]

Displacement is the area under the velocity/time graph. So for example this object's displacement in the first 3 seconds is (1/2)(3sec)(12.5 m/s)= 18.75m. (and then it starts backing up, displacement decreasing, after 3sec when velocity is negative).

But This object is never speeding up. Its velocity is smoothly decreasing at (25/6) m/s^2 (the slope of the graph). So the answer to the question is actually zero.

3 0
4 years ago
Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicul
ANEK [815]

Answer:

4.14 cm.

Explanation:

Given,

For Coil 1

radius of coil, r₁ = 5.6 cm

Magnetic field, B₁ = 0.24 T

For Coil 2

radius of coil, r₂ = ?

Magnetic field, B₂ = 0.44 T

Using formula of maximum torque

\tau_{max}= NIAB

Since both the coil experience same maximum torques

now,

NIA_1B_1 = NIA_2B_2

A_1B_1 = A_2B_2

r_1^2 B_1 = r_2^2 B_2

5.6^2\times 0.24= r_2^2\times 0.44

r_2 = 4.14\ cm

Radius of the coil 2 is equal to 4.14 cm.

4 0
4 years ago
11. Will the cube in#10 float in water? Will it float in benzene?
densk [106]
Where is the cube I don't see any picture?
5 0
4 years ago
13) Fred is travelling 3.4 m/s and decelerates at a rate of 0.83 m/s until he comes to a stop
Finger [1]

Answer:

<h2>Tt will take 4.04 seconds</h2>

Explanation:

In this problem/exercise, we are going to apply the newtons first equation of motion to solve for the time taken

Given that

final velocity= 3.4m/s

initia velocity u= 0m/s

deceleration= 0.84 m/s^2

time t= ?

applying

v=u+at

Substituting our given data into the expression above we have

3.4=0+0.84t

3.4=0.84t

divide both sides by 0.84 we have

3.4/0.84= t

t= 4.04 seconds

3 0
3 years ago
Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when the tension in it is 50.0 n.
blsea [12.9K]

As we know that Nth harmonic of the string is given by

f = \frac{N}{2L}\sqrt{\frac{T}{m/L}}

now here we will have

m/L = mass density = 2 g/m

m/L = 0.002 kg/m

Length = L = 0.600 m

Tension = T = 50.0 N

now from above formula we have

f = \frac{N}{2(0.600)}\sqrt{\frac{50.0}{0.002}}

f = 131.8N

now for first harmonic N = 1

f_1 = 131.8 Hz

for second harmonic N = 2

f_2 = 263.5 Hz

for third harmonic N = 3

f_3 = 395.3 Hz

8 0
4 years ago
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