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marusya05 [52]
3 years ago
5

According to the particle theory of matter: (a) How is the motion of particles in a liquid different from the motion of particle

s in a solid? (b)How is the motion of particles in a gas different from the motion of particles in a liquid?
Physics
1 answer:
kolezko [41]3 years ago
3 0

liquid partcles move relatively freely, solid atoms move about fixed place, gas ats go anywhere they like

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If your brakes give out, why can't you just pull the keys out of the ignition?
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There should be a small amount of play in the wheel when the steering is locked. Gently pull the key from the ignition while you slowly jiggle the steering wheel back and forth. If this is the cause of the problem, the key should come out after a little effort.

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If you represent velocity with an arrow, the length of the arrow represents -
Mamont248 [21]

Answer:

direction

Explanation:

3 0
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The distance between Earth and Mars is 225 million km. When converted using the conversion factor 1 AU = 1.5 × 108 km, the dista
noname [10]
To convert km to AU, we divide 225,000,000 km by the factor of 1.5 x 10^8 = 150,000,000 km. This gives us 225,000,000 / 150,000,000 = 1.5 AU. Therefore, the distance between Earth and Mars in AU is 1.5 AU.
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4 0
4 years ago
How long will it take a person walking at 2.1 m/s to travel 13 m?
MrRissso [65]

Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

  • This is a speed formula so you would use the formula speed=distance/time
  • You need to rearrange it to time=distance/speed
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7 0
3 years ago
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where
const2013 [10]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:  

W = ^{x_{2}}_{x_{1}}\int F_{x} dx

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx  

W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}  

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}    

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}      

W = 29.85 J

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!                                

6 0
3 years ago
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