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AlexFokin [52]
3 years ago
11

You have a piece of cork with a volume of 2 cm^3 and a density of 210kg/m^3. You hold it under water and release it.

Physics
1 answer:
insens350 [35]3 years ago
7 0

I am sorry if it didn't helped

answers;

Calculate the buoyant force of a piece of cork of 8cm3 that floats in water. Density of cork is 207kg/m3. ?

I need the mass, in order to get the volume to apply t to the Buoyancy formula of: B=(W)object=(m)object(g)

Explanation:

From Archimedes Principle, any object partially or totally submerged in a fluid is buoyed upwards with a force equal to the weight of the displaced fluid.

∴

B

=

ρ

f

l

V

f

l

g

=

1000

k

g

/

m

3

×

8

×

10

−

6

m

3

×

9

,

8

m

/

s

2

=

0

,

0784

N

(assuming the density of water is at standard temperature and pressure, and that the cork is totally submerged as it floats in the water

it's not the answer of your question ⁉️ but it is similar ........

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A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute
levacccp [35]

Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

          m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

where,  m_{1} = mass of the projectile

            m_{2} = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

         75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v

                                  = \frac{45}{50.075}

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E = \frac{1}{2}mv^{2}

                  = \frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}

                  = 13500 J

Now, energy after the impact will be as follows.

             E' = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

           \Delta E = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n = \frac{\Delta E}{E}

             = \frac{13480}{13500} \times 100

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