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nikklg [1K]
3 years ago
11

Two gliders collide on an air track moving in from opposite directions, and they bounce back. The masses of the gliders are 0.20

kg and 0.30 kg. The magnitudes of their initial velocities are 0.50 m/s and 0.40 m/s, respectively. After the collision, the magnitudes of their velocities are 0.40 m/s and 0.20 m/s, respectively. Assume the first glider is initially heading in the positive direction.The total momentum of the gliders before collision is closest to:
Physics
1 answer:
timurjin [86]3 years ago
3 0

Answer:

so initial momentum is 0.22kgm/s

Explanation:

m1=0.20kg

m2=0.30kg

initial velocity of m1=u1=0.50m/s

initial velocity of m2=u2=0.40m/s

total momentum of the system before collision

Pi=m1u1+m2u2

Pi=0.20kg×0.50m/s+0.30kg×0.40m/s

Pi=0.1kgm/s+0.12kgm/s

Pi=0.22kgm/s

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Pls help me i have to submit this due tomorrow :'D
aniked [119]

\textsf {a) As there is no change in the initial and final positions, the displacement is}\\\textsf {equal to 0.}

b) Finding total distance :

Distance travelled from 0 s to 5 s :

  • 1/2 x 5 x 30 = 75 m

Distance travelled from 5 s to 10 s :

  • 5 x 30 = 150 m

Distance travelled from 10 s to 15 s :

  • 150 + 1/2 x 5 x 5
  • 150 + 12.5 = 162.5 m

Distance travelled from 15 s to 20 s :

  • 36 x 5 + 1/2 x 5 x 4
  • 180 + 10 = 190 m

Distance travelled from 20 s to 25 s :

  • 45 x 5 + 1/2 x 5 x 5
  • 225 + 12.5 = 237.5 m

Distance travelled from 25 s to 30 s :

  • 40 x 5 + 1/2 x 10 x 5
  • 200 + 25 = 225 m

Distance travelled from 30 s to 35 s :

  • 20 x 5 + 1/2 x 20 x 5
  • 100 + 50 = 150 m

Distance travelled from 35 s to 40 s :

  • 1/2 x 20 x 5
  • 50 m

Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50

Total = 1240 m

c) velocity at t = 15 s

  • 36/15
  • 12/5
  • 2.4 m/s

d) average velocity

  • 0 m/s (as displacement is equal to 0)

e) average speed

  • 1240/40
  • 31 m/s

f) Part d uses displacement whereas part e uses distance

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ASHA 777 [7]

Answer:

twice as much work

Explanation:

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The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i
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Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed
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Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
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Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
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t1 = (56+24)/32 = 2.5 
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