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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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Answer:

Bruh............

Explanation:

8 0
2 years ago
In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of
seraphim [82]

Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N   is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

8 0
2 years ago
Converting 15 miles to kilometer
Dmitry [639]
15 miles to kilometers would be: 24.14 kilometers
8 0
3 years ago
Read 2 more answers
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
3 years ago
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of
jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0
3 years ago
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