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____ [38]
3 years ago
12

A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the weight

) in water?
Physics
2 answers:
Blababa [14]3 years ago
3 0
d=\frac{m}{V}\\
\\
0.08 = \frac{m}{7,840}\rightarrow m=0.08 * 7,840=627.2 \ kg

\vec{F_g}=m.\vec{g}=627.2 *9.8=6,146.56 \ N
Anni [7]3 years ago
3 0
Density = 7840 kg per m³
Volume = 0.08 m³
Mass = 0.08 of 7840 kg = 627.2 kg

Now I'm going to give you the correct answer to your question, and then,
after that, I'm also going to give you the answer you expect.

The correct answer is:  (Ryan2 the Expert in Answer #1 is correct) . . . . .

The force of gravity on a mass doesn't depend on what it's sunk in, surrounded by,
hanging from, or resting on.  As long as it's on or near the Earth's surface, the force
of gravity acting on it is the same, no matter what else is around.

Force of gravity (weight) = (mass) x (gravity) = 627.2 kg x 9.8 m/s² = 6146.56 newtons
=============================================

The apparent weight is less, because when it's immersed in a fluid, there's
a buoyant force acting on it, which cancels part of the force of gravity.

The buoyant force = weight of the displaced water.

Assume the density of water is 1 kg per liter = 1,000 kg per m³.

Then the weight of 0.08 m³ of water is (80 kg x 9.8 m/s²) = 784 newtons.

This is the upward buoyant force on the steel when it's in water,
and it makes the steel seem 784 newtons lighter in water.  So the
apparent weight of the steel in water is

6146.56 minus 784 = 5362.56 newtons.

That's why you can pick up your big brother in the swimming pool.
The force of gravity on him doesn't change, but the buoyant force
in the water balances part of the force of gravity, and makes him
seem to weigh less.

If he could blow himself up like a balloon, and displace enough water
to weigh just as much as he really does, then the buoyant force would
cancel his weight completely, and he would totally float !

That's how steel ships float in water.  They're shaped in a shape that can
displace a lot of water.  The whole secret is in the shape.
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A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
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Answer:

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Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

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Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

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v^2-u^2 = 2as'

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u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

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Thank you for posting your Physics question here. I hope the answer helps.  Upon calculating the ramp with the horizontal the answer is 20.49 Deg. Below is the solution:

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