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gavmur [86]
3 years ago
15

A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?

What could you do to reduce this pressure without removing the water?
Physics
1 answer:
allsm [11]3 years ago
7 0

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

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

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

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Explanation:

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How many atoms of oxygen in the chemical formula 2 Ca(CIO2)2?<br> O 2<br> O 4<br> O 6<br> O 8
alexdok [17]

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Explanation:

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3 years ago
The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 5.00 T/s. Part A) What is the electric field str
Veronika [31]

Answer:

(A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is 3.75\times10^{-2}\ V/m.

Explanation:

Given that,

Magnetic field = 2.0 T

Diameter = 5.0 cm

Rate of decreasing in magnetic field = 5.00 T/s

(A). We need to calculate the electric field strength inside the solenoid at a point on the axis

Using formula of electric field inside the solenoid

E=\dfrac{r}{2}|\dfrac{dB}{dt}|

Electric field on the axis of the  solenoid

Here, r = 0

E=\dfrac{0}{2}\times5.00

E = 0

The electric field strength inside the solenoid at a point on the axis is zero.

(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis

Using formula of electric field inside the solenoid

E=\dfrac{r}{2}|\dfrac{dB}{dt}|

E=\dfrac{1.50\times10^{-2}}{2}\times|5.00|

E=0.0375= 3.75\times10^{-2}\ V/m

Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.

(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is 3.75\times10^{-2}\ V/m.

4 0
4 years ago
A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
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Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

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ω(0) = ω(i) + α*t1

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t1 = 1.3/1.8

t1 = 0.722 s

5 0
3 years ago
The pressure at the bottom of a cylindrical container with a cross-sectional area of 45.5 cm2 and holding a fluid of density 420
iris [78.8K]

Answer:

3.33 m

Explanation:

Pressure is the distributed force applied to the surface of an object per unit area. The force is applied perpendicular to the surface of the object. The SI unit of pressure is N/m² or Pa.

Hydrostatic pressure is the pressure that a fluid exerts at a point due to the force of gravity.

The relationship between  pressure on the bottom of the container,  atmospheric pressure and the pressure due to the depth of the fluid is given by:

P_{bottom}-P_{atm}=P_{depth}\\\\where\ P_{bottom}=pressure\ at\ the \ fluid\ bottom,\ P_{atm}=atmospheric\ pressure\\P_{depth}=pressure\ due\ to\ fluid\ depth=\rho gh. \ Hence:\\\\P_{bottom}-P_{atm}=\rho gh\\\\Given \ that\ P_{bottom}=115\ kPa=115*10^3\ Pa, let\ us\ assume\ P_{atm}=101\ kPa=101*10^3\ Pa,\rho=420\ kg/m^3,g=acceleration\ due\ to \ gravity=10\ m/s^2.\\\\Therefore:\\\\115*10^3-101*10^3=420*10*h\\\\14*10^3=4200h\\\\h=3.33\ m\\\\

6 0
3 years ago
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