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Grace [21]
3 years ago
8

Gelatin has a density of 1.27 g/cm³. if you have a blob of gelatin dessert that fills a 2.0 liter bottle, what is its mass?

Physics
2 answers:
lyudmila [28]3 years ago
7 0
There are 1,000 cm³ in 1 liter, 2,000 cm³ in 2 liters.

Each cm³ of gelatin has  1.27 grams of mass in it.

There are 2,000 cm³ in the whole blob, so the total mass is

            (2,000 cm³) · (1.27 g/cm³)  =  2,540 grams

                                                         or    2.540 kilograms  .
Verdich [7]3 years ago
6 0
Density = mass / volume so mass = density x volume

2 litres = 2000cm^3


mass = 1.27 x 2000 = 2540g = 2.54 kg
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3 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
  • This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
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2 years ago
Will a seismic wave traveling through a solid go slower or faster than a seismic wave traveling through a liquid? Why?
Snezhnost [94]
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<span>It would travel faster because their speed depends on the density and composition of material that they pass through.</span>
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3 years ago
Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t
Tems11 [23]

Answer:

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After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

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