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Usimov [2.4K]
3 years ago
8

50pts: Fill this out.

Physics
2 answers:
Vikentia [17]3 years ago
6 0

the person who answered before me is correct

Bogdan [553]3 years ago
4 0

Push is in, pull is out

----------------------------------------------------------------------------------------------------

Contact Forces  

Frictional Force  

Tension Force  

Normal Force  

Air Resistance Force  

Applied Force

----------------------------------------------------------------------------------------------------

Action-at-a-Distance Forces  

Gravitational Force  

Electrical Force  

Magnetic forces

----------------------------------------------------------------------------------------------------

Very bottom, last part

First drawing: Arrow down above box

Second drawing: Arrow up from below box    

 


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Neon atoms at 245 K pass through a fan that gives each mole of neon gas an additional kinetic energy of 16.0 J. Part A What is t
hjlf

Answer:

246.28 K

Explanation:

The total energy of one mole of gas molecules can be calculated by the formula given below

E = \frac{3}{2}\times R\times T

Where R is gas constant and T is absolute temperature.

Put the value of R as 8.314 and temperature as 245 , we get

E = \frac{3}{2}\times 8.314\times 245

= 3055.4 J

Add 16 j to it

Total energy of gas molecules = 3055.4 + 16 = 3071.4 J.

If T be the temperature after addition of energy then

\frac{3}{2}\times 8.314\times T = 3071.4

T =\frac{2\times 3071.4}{3\times 8.314}

T = 246.28 K

7 0
3 years ago
What graph shape is this what does the snap tell you
vekshin1
The picture is hard to see but if you still need help message me
7 0
3 years ago
The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0
3 years ago
Read 2 more answers
Question 23 of 32
marshall27 [118]
Do you still need help?!?!
7 0
3 years ago
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
4 years ago
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