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3 years ago

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

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Contact Forces

Frictional Force

Tension Force

Normal Force

Air Resistance Force

Applied Force

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Action-at-a-Distance Forces

Gravitational Force

Electrical Force

Magnetic forces

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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

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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

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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.

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3 years ago

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Question 23 of 32

marshall27 [118]

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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

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

$W=-0.79 J$

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4 years ago