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Serhud [2]
3 years ago
11

An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object

Physics
1 answer:
Elden [556K]3 years ago
8 0

Answer:

1.8 = relative density (there are no units for relative density)

Explanation:

It displaces water equal to it's volume and gets buoyancy equal to that amount of water

2.7 - 1.2  = 1.5  N  of buoyancy

density of water = 1 gm /cc

 1.5 N = m (9.81)

    m of water displaced = .1529 kg

             152.9 cc of water will produce this buoyancy....this is the volume of the object

   

find mass of object     2.7 = m (9.81)     shows m = .2752 kg = 272.5 gm

density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc

Relative to water (which is 1 gm / cc) the relative density is 1.8

  ====> it is 1.8 times denser than water and will sink when in water....

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A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or sh
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Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast,  a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

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Answer:

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The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

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