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Wittaler [7]
3 years ago
11

Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (2 m²), what's t

he force of gravity acting on the column of water?
Physics
2 answers:
Anettt [7]3 years ago
5 0
He’s ight =(h) you know this because this is simple
Vesna [10]3 years ago
4 0

Answer:

196000 N

Explanation:

The following data were obtained from the question:

Height (h) = 10 m

Area (A) = 2 m²

Force (F) =.?

Next, we shall determine the pressure in the tank.

This can be obtained as follow:

P = dgh

Where

P is the pressure.

d is the density of the liquid.

g is acceleration due to gravity

h is the height.

Height (h) = 10 m

Density (d) of water = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Pressure (P) =...?

P = dgh

P = 1000 × 9.8 × 10

P = 98000 N/m²

Therefore, the pressure acting on the tank is 98000 N/m²

Finally, we shall determine the force of gravity acting on the column of water as follow:

Area (A) = 2 m²

Pressure (P) = 98000 N/m²

Force (F) =.?

Pressure (P) = Force (F) /Area (A)

P = F /A

98000 = F/ 2

Cross multiply

F = 98000 × 2

F = 196000 N

Therefore, the force of gravity acting on the column of water is 196000 N

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WILL MARK BRAINLIEST PLEASE HELP AND SHOW ALL WORK
jok3333 [9.3K]

Answer:

Same reading.

Explanation:

Assume that after the string breaks the ball falls through the liquid with constant speed. If the mass of the bucket and the liquid is 1.20 kg, and the mass of the ball is 0.150 kg,

A.) Before the string break, the total weight = weight of the can + weight of the water.

According to Archimedes' Principle which state that: “A body immersed in a liquid loses weight by an amount equal to the weight of the liquid displaced.” Archimedes principle also states that: “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it

B.) After the string break.

The scale will have the same reading as before the string break.

6 0
3 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
3 years ago
The force of friction always opposes the _______.
Ierofanga [76]

Answer:

motion

Explanation:

uh look it up?

8 0
3 years ago
Read 2 more answers
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Shkiper50 [21]
All compounds are molecules but not all molecules are compounds. A molecule is two atoms joined together. A compound is two different atoms joined together.
5 0
2 years ago
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