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NemiM [27]
3 years ago
10

comparing the blank of an object with that of a liquid will tell you where or not the objects will float in that liquid

Physics
2 answers:
jenyasd209 [6]3 years ago
7 0

Answer: Density

Explanation: Recall Archimedes Principle. There are two forces acting an object submerged in a liquid: the force of gravity and the  (opposite directed) force of buoyancy. The buoyancy is proportional to the mass of the liquid displaced by the submerged part of the object.

Density is the ratio of mass to volume. Therefore if the density of the submerged object is higher than that of the displaced liquid, the net force will point in the direction of the gravity (object will sink). In the opposite case, the net force will point in the direction of the buoyant force (upward) and the object will float.  

tamaranim1 [39]3 years ago
5 0

Answer:

density

Explanation:

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The brainstem is to breathing and arousal as the limbic system is to memories and: (2 points)
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The brainstem is to breathing and arousal as the limbic system is to memories and emotion. See the explanation below, please.

Explanation:

The limbic system develops certain responses to emotional stimuli (joy, anger, sadness, fear, pleasure). It interacts with the autonomic endocrine and nervous system. It is also related to sexuality, memory, attention, memories. It includes some areas such as hypothalamus, tonsil, hippocampus, among others.

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Read the scenario below and answer the question that follows. Susan was in her algebra class preparing to take a test. Her instr
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You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

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