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Morgarella [4.7K]
2 years ago
5

What happens when an objects density is the same as the fluid

Physics
1 answer:
Dennis_Churaev [7]2 years ago
3 0

Answer:

Explanation: If an object is exactly the same density as the liquid, it will not move up or down. It will just stay right where it is (unless it is pushed around by water currents). If you put it on the surface, it will remain on the surface

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Simon is riding a bike at 12 km/h away from his friend Keesha. He throws a ball at 5 km/h back to Keesha, who is standing still
pychu [463]
So there are different ways this could be solved. I'll do try to explain it the way I was taught... 

Simon is riding his bike at 12 km/hr relative to the sidewalk, away from where Keesha is.

Simon throws the ball at Keesha, at 5 km/hr. 

Keesha sees the ball approaching her at (12-5) = 7 km/hr relative to the ground to her. 

Therefore the answer is: 7 km/hr


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3 years ago
To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th
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The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).<span>

The amount of weight displaced or the amount of buoyant force is: </span>

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

 

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

<span>average density = 1.07 kg/L = 1.07 g/mL</span>

5 0
3 years ago
Read 2 more answers
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Explanation:

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Answer:

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h = ut + 0.5at^2

<u>For first half distance</u>

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Object falls at half the distance, so h = h/2 where t = t1

Hence, we have

h/2 = at1^2/2 - equation 1

<u>For second half distance: </u>

Similarly,

h = a(t1 + t2)^2/2 - equation 2 where t = t1 + t2 and u= 0

Using equation 2 by equation 1

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Answer:

As I know it is in Jupiter the objects weight more than 2x than on Earth.

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