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Masteriza [31]
3 years ago
14

Compared to the buoyant force of the atmosphere on a 1-kilogram helium-filled balloon, the buoyant force of the atmosphere on a

nearby 1-kilogram solid iron block is _____.
Physics
2 answers:
bezimeni [28]3 years ago
7 0

Answer:

Explanation:

The buoyant force depends on the volume of the body

Buoyant force = Volume immersed x density of fluid x gravity

So, the density of fluid that means air is same for both the cases, but the volume of 1 kg helium balloon is more than the density of iron, so the buoyant force on helium is more than the buoyant force on iron.

Vladimir [108]3 years ago
6 0

It all depends on the SIZE of the balloon.

If the balloon is made of really tough rubber, and it holds the helium in the same volume as the solid iron block, then the buoyant force of the atmosphere is the same for both objects.

But if the balloon is just some flimsy stuff, and it lets the helium expand to a much bigger volume than the iron block, then the buoyant force on the balloon is greater than the buoyant force on the solid iron block.

In fact, it DOESN'T MATTER what's in the balloon and what's in the block.  It doesn't matter whether either one of them is solid, liquid, or gas, and it doesn't matter whether they have the same or different mass.  

Whichever one has greater VOLUME has a greater buoyant force of atmosphere acting on it.

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Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot
m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force F_1=430 N

other Pulls it with a rope of rope F_2=360 N

mass of crate m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

f_r=F_1+F_2

where f_r is the friction force

f_r=430+360

f_r=790 N

f_r=\mu N

where \muis the coefficient of static friction

N=mg

790=\mu 29\cdot 9.8

\mu =0.27

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Explanation:

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